# Further Worked Text Solutions by qingyunliuliu

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```									Chapter 1 – Organising and
displaying data
Solutions to Exercise 1A
1
a   A number that represents a quantity,       g   Is a quality ⇒ categorical
e.g. height, pulse rate
h   Is a discrete quantity ⇒ discrete
b   A number that represents a quality, e.g.       numerical
gender, sport preferences
i   Is a continuous quantity ⇒
2   Continuous and discrete variables              continuous numerical

3                                              j   Is a quality ⇒ categorical
a   Is a continuous quantity ⇒
continuous numerical                       k   Is a quality ⇒ categorical

b   Is a discrete quantity ⇒ discrete          l   Is a discrete quantity ⇒ discrete
numerical                                      numerical

c   Is a continuous quantity ⇒                 m   Is a quality ⇒ categorical
continuous numerical
n   Is a discrete quantity ⇒ discrete
d   Is a quality ⇒ categorical                     numerical

e   Is a discrete quantity ⇒ discrete          4   Height, weight, age and pulse rate are
numerical                                      quantities ⇒ numerical
Sex and plays sport are qualities ⇒
f   Is a discrete quantity ⇒ discrete              categorical
numerical
Solutions to Exercise 1B
1                                            3
a   The most frequently occurring value      a   The table is constructed by counting
the number of responses for each
b   i B occurs 5 times and is the most           category and then calculating
frequent value, so is the mode.           percentages of the whole.
ii 8 occurs 5 times and is the most
frequent value, so is the mode.

2
a   The table is constructed by counting
the number of responses for each
category and then calculating
percentages of the whole.
b   The bar chart is constructed by
counting the number of responses for
each category and then drawing the bar
for that response to the appropriate
height.

b   The bar chart is constructed by
counting the number of responses for
each category and then drawing the bar
for that response to the appropriate
height.

4
a   The total number of schools sums to
20 and the missing percentage must be
55, since the 3 percentages must sum
to 100.

b   Reading off the table, 5 schools

c   Summing the values, 20 schools

d   As calculated in a, 55%

e   Taking information from the table:
‘Report: 20 schools were classified
according to school type. The majority
of these schools, 55%, were found to
be Government schools. Of the
remaining schools, 25% were
Independent while 20% were
Catholic schools.’
5                                              b   The segmented bar chart is constructed
a   Is qualitative so therefore categorical.       by reading percentages of each
response from the completed table and
b   The segmented bar chart is constructed         then drawing the bar for that response
by reading percentages of each                 to the appropriate height.
response from the table and then
drawing the bar for that response to the
appropriate height.

7
a   The bar chart is constructed by reading
the count number of each response
from the table and then drawing the
c   Taking information from the table:             bar for that response to the appropriate
‘Five hundred Australians were                 height.
classified according to their place of
birth; the majority, 78.3%, were born
in Australia, while 21.8% were born
overseas.’

6
a   Sum the two count values to get the
total count, which represents 100%.
Divide each of the count values by the
total count and multiply by 100 to get
each of the two percentages.

b   The percentage segmented bar chart is
each response from the table and then
drawing the bar for that response to the
appropriate height.
c   Taking the count values and
percentages from the table, we can say:
‘The eye colours of 11 children were
brown eyes. Of the remaining children,
hazel eyes.’

8
a   The count for ‘rarely’ can be
calculated by subtracting the counts for
‘regularly’ and ‘sometimes’ from the
total. The total percentage must be
100% and the ‘sometimes’ percentage
will be 100% minus the percentages
for ‘regularly’ and ‘rarely’.

b   Taking the count values and
percentages from the completed table:
‘Report: When 22 students were asked
the question, “How often do you play
sport?”, the dominant response was
‘Sometimes’, given by 45.5% of the
students. Of the remaining students,
31.8% of the students responded that
they played sport ‘Rarely’ while
22.7% said that they played sport
‘Regularly’.’
Solutions to Exercise 1C
1   The mode of 1 means most cars            4
stopping at a traffic light had only 1   a   There are many possible answers, as
occupant, since 1 is the most                long as 5 intervals are used, each is
frequently occurring value.                  exactly the same size, and the values
20 106 and 63 579 are included.
2   Counting the number of times each            Intervals of 10 000 starting from
value appears, we can form the table         20 000 are one possible response, i.e.:
below. Dividing the count number for               20 000–29 999
each value by the total value gives us             30 000–39 999
the percentages.                                   40 000–49 999
50 000–59 999
60 000–69 999

b   Intervals of 5 starting from 5 are one
possible response, i.e.:
5–9
10–14
15–19
20–24
3                                                      25–29
a   The missing count can be found by                  30–34
subtracting the count numbers for all
the other values from the total. The     c   Intervals of 2 starting from 0 are one
total percentage is 100% and the other       possible response, i.e.:
three missing percentages can be                   0–1.99
calculated by dividing the count                   2–3.99
number for that number of children in              4–5.99
the family by the total for the count              6–7.99
numbers and multiplying by 100.                    8–9.99

b   Reading off the completed table, 10      5
families had only 1 child.               a   i Reading off the graph, 17%
ii Reading off the graph, 13%
c   Adding the counts for 2, 3 and 4             iii Reading off the graph, 46%
children from the completed table, 8         iv Summing the first three columns,
families had more than 1 child.                  33%

d   Reading off the completed table,         b   i Multiplying the percentage as read
14.3% of families had no children.              off the graph by 30 and dividing by
100% = 6
e   Adding the percentages for 0, 1 and 2        ii Multiplying the percentage as read
from the completed table, 90.5% of              off the graph by 30 and dividing by
families had less than 3 children.              100% = 4

c   15–19 is the highest bar and is
therefore the mode in this case.
6   Draw in the axes and scales, labelling       c   In the window menu, make Xmin = 60,
both. Draw in the histogram bars to the          Xmax = 95, Ymin = 0, Ymax = 6 and
appropriate height as given in the               Xscl = 5. The graph below will result:
table.

d   Using the Trace function, place the
cursor on the second class interval.
The answer of n = 3 will be displayed.

9
7   Draw in the axes and scales, labelling       a   Using the Stat List Editor of your
both. Draw in the histogram bars to the          graphics calculator, enter the 25
appropriate height as given in the               numbers of children into L1 and plot
table.                                           the final results.

b   i The calculator automatically chooses
the optimal interval width of 1.3.
ii The count is 7, reflecting the fact
that the data of 3 children in the
family appears seven times in our
histogram.
8
a   Using the Stat List Editor of your           c   In the window menu, make Xmin = 0.5,
graphics calculator, enter the 23 pulse          Xmax = 10.5, Ymin = 0, Ymax = 8 and
rates into L1 and plot the final results:        Xscl = 1. The graph below will result:

b   i The calculator automatically chooses       d   i Using the Trace function, 3
the optimal interval width of 5.3.            ii It includes the data of 5 children in
ii The count is 5, reflecting the fact              the family 3 times in that interval.
that 5 data are contained within this
interval, namely 70, 71, 72, 73 and 74.
Solutions to Exercise 1D
1                                             d   The mode is the highest bar. There are
a   The mode is the highest bar. There is         no potential outliers and the histogram
one potential outlier and the histogram       tails off to the right from the
appears approximately symmetric               approximate centre, so is positively
around the approximate centre.                skewed.

b   The mode is the highest bar. There are    2
no potential outliers and the histogram   a   All the distributions appear
tails off to the left from the                approximately symmetric around their
approximate centre, so is negatively          respective approximate midlines.
skewed.
b   There are no clear outliers in any of the
distributions.

c   In A the middle is 8–10, in B it is
24–26 and in C it is 40–42. Note that
the middle interval in A and B is also
the mode for those distributions.

d   The spread is lowest in B since the
range is only 8, compared to 14 for A
and 18 for C.
c   The mode is the highest bar. There is
e   The spread was greatest in C, with a
one potential outlier and the histogram
range of 18.
tails off to the left from the
approximate centre, so is negatively
3
skewed.
a   The mode is the highest bar. There is
one potential outlier and the histogram
tails off to the right from the
approximate centre, so is positively
skewed.
b   The mode is not clear since two bars      e   The mode is the highest bar. There are
appear to be the same size. There is          no potential outliers and the histogram
one potential outlier and the histogram       tails off to the left from the
appears approximately symmetric               approximate centre, so is negatively
around the approximate centre.                skewed

c   The mode is the highest bar. There are    f   There are two highest bars and thus
no potential outliers and the histogram       two modes. There are no potential
appears approximately symmetric               outliers and the histogram appears
around the approximate centre.                approximately symmetric around the
approximate centre.

d   The mode is the highest bar. There are
no potential outliers and the histogram
4   Reading the information from the
tails off to the right from the
histogram: ‘Report: For the 28
approximate centre, so is positively
students, the distribution of pulse rates
skewed.
is approximately symmetric with an
outlier. The centre of the distribution
lies in the interval 75–80 beats per
minute and the spread of the
distribution is 55 beats per minute. The
outlier lies in the interval 110–115
beats per minute.’
Solutions to Exercise 1E
1                                              3
a   Using stem values representing             a   Below are the two plots for the
multiples of 10, plot the 23 values.           different interval sizes.

b   The more appropriate plot is ii since it
shows a greater spread of the data,
whereas i is much more compact.

b   In a histogram, we see the frequency of    c   Reading off the second stem and leaf
various intervals but in a stem and leaf       plot:
plot we see this information as well as        ‘Report: For the 15 men, the
the individual data themselves, giving         distribution of their wrist
The centre of the distribution is at 17.6
2                                                  cm and the distribution has a spread of
a   A tails off to higher values from the          3.4 cm. There are no outliers.’
approximate centre, so is positively
skewed. There are no potential             4
outliers.                                  a   i The stem and leaf plot with stems 5,
B is approximately symmetric around              6, 7, 8, 9, 10 and 11:
its approximate centre, so is
symmetric. There are no potential
outliers.
C tails off to lower values from the
approximate centre, so is negatively
skewed. There are two potential
outliers in the values of 1 and 3.
ii The plot with the above stems
b   This is the same as the median value              halved:
and is found by counting the data and
finding the middle datum. It is 15 for
A, 32 for B and 58 for C.

c   This is calculated by subtracting the
lowest value from the highest value. It
is 52 for A, 52 for B and 68 for C.

d   The only outliers are 1 and 3 for C.
b   From the information in the second         7
stem and leaf plot & using Q. 3c:          a   Plotting the values as a dot plot:
‘Report: For the 22 students, the
distribution of weights is positively
skewed with an outlier. The centre of
the distribution is at 60 kg and the
distribution has a spread of 60 kg. The
outlier is a weight of 110 kg.’

5                                              b   75 is the highest line of dots, so is the
a   The back–to–back stem and leaf plot            mode.
for the two teams:
c   i The middle value of the plot falls at
the 74 line of dots, so 74 is the
centre.
ii The spread is the highest value
minus the lowest value, and so is 7.

8
a   Large data set size, hence histogram
b   From the information in the back–to–
back stem and leaf plot and from
b   Medium data set size, hence stem plot
appraising the shapes of the two sides
or histogram
of the plots:
‘Report: The distribution of the
c   Categorical data, hence bar chart
number of possessions is
approximately symmetric for both
d   Large data set size, hence histogram
teams. The two distributions have
similar centres, at 24 and 22
e   Medium data set size, hence stem plot
or histogram
the distribution is less for Carlton, 39
possessions, compared to 51
f   Categorical data, hence bar chart
possessions for Essendon.’
g   Large data set size, hence histogram
6
a   Plotting the values as a dot plot:
h   Small data set size, hence dot plot

b   4 is the highest line of dots, so is the
mode.

c   i The middle value of the plot falls at
the 4 line of dots, so 4 is the
centre.
ii The spread is the highest value
minus the lowest value, and so is 6.
Solutions to Multiple–choice questions
1   Number of cars owned is a numerical           9    If the outliers are ignored, the
variable but car size is categorical.              histogram is approximately symmetric
⇒D                                                 around its approximate centre interval
of 10–12. ⇒ D
2   It is a categorical data set so a bar chart
can be used. ⇒ D                              10   Ignoring the outliers, the 10–12
interval is at the centre of the
Note that none of the other plot types             histogram. ⇒ B
are compatible with categorical data
sets.                                         11   Spread is the top number minus the
bottom number, 28 – 6 = 22. ⇒ E
3   It is impossible to have anything other
than a positive whole number of cars          12   The modal interval is the one with the
that one owns; one cannot have half a              most values, which is 25–29. ⇒ B
car. Therefore, number of cars owned
is a discrete variable. ⇒ B                   13   The brown hair segment is 34
percentage points in size. Since there
4   The 4 percentages must sum to 100%.                are 200 students, this equates to 68
100 – 29.2 – 19.2 – 23.6 = 28.0 ⇒ B                students with brown hair. ⇒ D

5   The 4 counts must sum to a total of           14   Brown is the largest segment. ⇒ C
250.
250 – 73 – 70 – 59 = 48 ⇒ B                   15   The small number of data would make
a dot plot most suitable. ⇒ B
6   Sport has the highest count and
percentage. ⇒ A

7   Summing the heights of all the bars,
3 + 4 + 6 + 4 + 2 + 1 + 1 + 1 = 22
⇒E

8   Summing the bar heights less than 14,
3 + 4 + 6 + 4 = 17 ⇒ D
Chapter 2 – Summarising numerical
data: the median, range,
interquartile range and box plots
Solutions to Exercise 2A
1                                              4
a   The range is the spread of all values      a   After ordering the 9 values,
and is the difference between the              middle = median = 14.
largest and smallest value in any given        Two middle values of lower 50%
data set.                                      (4 lowest values) are 11 and 12,
average = Q1 = 11.5;
b   The median value is the middle value           two middle values of upper 50%
in the data set unless that data set has       (4 highest values) are 16 and 16,
an even number of values, in which               average = Q3 = 16;
case there is no middle value. To find         IQR = Q3 – Q1 = 4.5
the median, take the average of the two
b   After ordering the 11 values,
middle values.
middle = median = 21.
Two middle values of lower 50%
c   Quartiles are the three numbers that
(5 lowest values) are 14 and 14,
divide a distribution into four equal
average = Q1 = 14;
parts. Note that Q2 is the same as the
two middle values of upper 50%
median, while Q1 is the median of the
(5 highest values) are 28 and 28,
lower 50% of values and Q3 is the
average = Q3 = 28;
median of the upper 50% of values.
IQR = Q3 – Q1 = 14
d   The interquartile range is the spread of   c   After ordering the 14 values,
the middle 50% of values and is the            two middle values are 5 and 7,
difference between Q3 and Q1.                  average = median = 6.
Middle value of lower 50% (7 lowest
2                                                  values) = Q1 = 4;
a   After ordering the 6 values, the two           middle value of upper 50% (7 highest
middle values are 4 and 6,                     values) = Q3 = 9;
average = median = 5                           IQR = Q3 – Q1 = 5

b   After ordering the 5 values,               5
middle value = median = 12                 a   Of the 14 values, the two middle
values are 10 and 12;
c   After ordering the 8 values, the two           average = median = 11.
middle values are 100 and 103,             b   Middle value of the lower 50%
average = median = 101.5                       (7 lowest values) = Q1 = 10;
middle value of the upper 50%
d   After ordering the 7 values,                   (7 highest values) = Q3 = 15.
middle value = median = 0.59
c   IQR = Q3 – Q1 = 5
3   After ordering the 9 values,
middle value = \$850                        d   R = highest value – lowest value = 18

e   Writing the values in a line, check that
Q1 = 10, Q3 = 15 and M is in the
middle of 10 and 12.
6                                           7
a   Of the 20 values, the two middle        a   Of the 23 values,
values are 25 and 27;                       middle value = median = 20
average = median = 26.
b   Middle value of the lower 50% (11
b   Two middle values of lower 50%              lowest values) = Q1 = 8;
(10 lowest values) are 16 and 19,           middle value of the upper 50% (11
average = Q1 = 17.5;                        highest values) = Q3 = 26.
two middle values of upper 50%
(10 highest values) are 30 and 31,      c   IQR = Q3 – Q1 = 18
average = Q3 = 30.5.
d   R = highest value – lowest value = 54
c   IQR = Q3 – Q1 = 13

d   R = highest value – lowest value = 29
Solutions to Exercise 2B
1                                            b   Using the Trace function, we find that
a   Minimum = 7                                  minX = 3, Q1 = 10.5, Med = 21,
Maximum = 25                                 Q3 = 26.5 and maxX = 55.
Of the 14 values, two middle values          The outlier is 55.
are 10 and 12; average = median = 11.
Middle value of the lower 50%            4
(7 lowest values) = Q1 = 10;             a   Using the Stat List Editor function of
Middle value of the upper 50%                the graphics calculator, the box plot
(7 highest values) = Q3 = 15                 below can be graphed:

b   Using the Stat List Editor function of
the graphics calculator, the box plot
below can be graphed:

b   Using the Trace function, we find that
minX = 25, Q1 = 35, Med = 38.5,
Q3 = 43 and maxX = 64.
The outlier is 64.
2
a   Minimum = 136                            5
Maximum = 189                            a   i Median is middle vertical of
Of the 23 values, middle value                   coloured box = 10
= median = 158.                              ii Q1 is left edge of coloured box = 5,
Middle value of the lower 50%                    Q3 is right edge of coloured box = 21
(11 lowest values) = Q1 = 148;               iii IQR = Q3 – Q1 = 16
Middle value of the upper 50%                iv Minimum is left end of horizontal
(11 highest values) = Q3 = 169                   line = 0, maximum is right end of
horizontal line = 45
b   Using the Stat List Editor function of       v No outliers
the graphics calculator, the box plot
below can be graphed:                    b   i Median is middle vertical of
coloured box = 27
ii Q1 is left edge of coloured box = 12,
Q3 is right edge of coloured box = 42
iii IQR = Q3 – Q1 = 30
iv Minimum is left end of horizontal
line = 5, maximum is right end of
horizontal line = 50
3                                                v No outliers
a   Using the Stat List Editor function of
the graphics calculator, the box plot    c   i Median is middle vertical of
below can be graphed:                            coloured box = 38
ii Q1 is left edge of coloured box = 32,
Q3 is right edge of coloured box = 42
iii IQR = Q3 – Q1 = 10
iv Minimum is leftmost outlier = 5,
maximum is right end of horizontal
line = 50
v Outlier is 5
d   i Median is middle vertical of              6
coloured box = 16                       a   i Upper fence is right edge of
ii Q1 is left edge of coloured box = 14,           coloured box + 1.5 times the IQR
Q3 is right edge of coloured box = 21          = 40 + 1.5 × 10 = 55
iii IQR = Q3 – Q1 = 7                           ii Lower fence is left edge of coloured
iv Minimum is leftmost outlier = 5,                box – 1.5 times the IQR
maximum is rightmost outlier = 50              = 30 – 1.5 × 10 = 15
v Outliers are 5, 7, 36, 40, 50
b   i Upper fence is right edge of
e   i Median is middle vertical of                     coloured box + 1.5 times the IQR
coloured box = 41                              = 70 + 1.5 × 20 = 100
ii Q1 is left edge of coloured box = 29,        ii Lower fence is left edge of coloured
Q3 is right edge of coloured box = 44          box – 1.5 times the IQR
iii IQR = Q3 – Q1 = 15                             = 50 – 1.5 × 20 = 20
iv Minimum is left end of horizontal
line = 0, maximum is right end of
horizontal line = 47
v No outliers

f   i Median is middle vertical of
coloured box = 41
ii Q1 is left edge of coloured box = 38,
Q3 is right edge of coloured box = 45
iii IQR = Q3 – Q1 = 7
iv Minimum is leftmost outlier = 10,
maximum is right end of horizontal
line = 47
v Outliers are 10, 15, 20, 25
Solutions to Exercise 2C
Box plot 2 matches histogram D due to the        Box plot 4 has a narrower coloured box and
presence of one high–value outlier.              thus smaller IQR, representing a reduced
spread of values within the middle 50% of
Box plot 3 matches histogram C due to the        values.
presence of both a high–value and low–value      This correlates with a more centralised data
outlier.                                         set, which is histogram A; that is, the data
are clearly grouped around the approximate
Box plots 1 and 4 are very similar except that   centre so the spread of the middle 50% of
box plot 1 has a wider coloured box and thus     values is reduced.
larger IQR, representing a greater spread of
values within the middle 50% of values.
This correlates with a more spread-out data
set, which is histogram B.
Solutions to Exercise 2D
1                                               2
a   Since the median value of 38 is closer      a   Reading the median values of 76 and
to the right edge of the coloured box           73 off the graphs and noting that both
and there is one outlier, we can say:           medians are approximately in the
‘The distribution is negatively skewed          centres of their respective coloured
with an outlier. The distribution is            boxes, we can say:
centred at 38, the median value. The            ‘The distributions of pulse rates are
spread of the distribution, as measured         approximately symmetric for both men
by the IQR, is 10 and, as measured by           and women. There are no outliers. The
the range, 45. There is an outlier at 5.’       median pulse rate for females (M =76
beats/minute) is greater than for males
b   Since the median value of 16 is closer          (M =73 beats/minute). The IQR is also
to the left edge of the coloured box and        greater for females (IQR=14
there are 5 outliers, we can say:               beats/minute) than males (IQR=8
‘The distribution is positively skewed          beats/minute). The range of pulse rates
with outliers. The distribution is              is also greater for females (R =30
centred at 16, the median value. The            beats/minute) than males (R =19
spread of the distribution, as measured         beats/minute).’
by the IQR, is 6 and, as measured by
the range, 45. The outliers are at 5, 8,    b   Comparing the two graphs and noting
36, 40 and 50.’                                 that the female distribution is much
c   Since the median value of 41 is closer          ‘For this group of males and females,
to the right edge of the coloured box           the females on average had higher and
and there are no outliers, we can say:          more variable pulse rates.’
‘The distribution is negatively skewed
with no outliers. The distribution is       3
centred at 41, the median value. The        a   Reading the median values of 40 and
spread of the distribution, as measured         30 off the graphs and noting that the
by the IQR, is 15 and, as measured by           median for A is approximately in the
the range, 47.’                                 centre of its coloured box but the
median for B is closer to the left edge
d   Since the median value of 41 is                 of its coloured box, we can say:
approximately in the middle of the              ‘The distribution for Brand A is
coloured box and there are 4 outliers,          approximately symmetric with outliers.
we can say:                                     The distribution for Brand B is
‘The distribution is approximately              positively skewed. The median battery
symmetric but with outliers. The                lifetime for Brand A (M =40 hours) is
distribution is centred at 41, the              greater than for Brand B (M =30
median value. The spread of the                 hours). The IQR is for Brand A
distribution, as measured by the IQR,           (IQR=10 hours) is less than for Brand
is 7 and, as measured by the range, 37.         B (IQR=15 hours). The range of
The outliers are at 10, 15, 20 and 25.’         lifetimes for Brand A (R =30 hours) is
also less for Brand B (R =40 hours).
For Brand A there are two outliers at
58 and 60 hours.’

b   Comparing the two graphs and noting
that the brand B distribution is much
‘On average, Brand A batteries have
Solutions to Multiple–choice questions
1   Ordering the 9 values,                   9    Range = highest value – lowest value
5th value = middle value                            = 81 – 51 = 30 ⇒ D
= median = 12. ⇒ C
10   Median is closest to left edge of
2   Ordering the 9 values,                        coloured box = positively skewed.
the two middle values of the lower            ⇒D
50% (4 values) are 8 and 10;
average = Q1 = 9. ⇒ A                    11   Median in middle of coloured box
= symmetric. ⇒ A
3   Range = highest value – lowest value
= 19 – 4 = 15 ⇒ E                  12   Median is closest to right edge of
coloured box with an outlier
4   Ordering the 10 values, the two middle        = negatively skewed with an outlier.
values are 12 and 13;                         ⇒B
average = median = 12.5. ⇒ D
13   Median in middle of coloured box with
5   Ordering the 10 values,                       outliers = symmetric with outliers.
Q1 = 3rd value = 10;                          ⇒B
Q3 = 8th value = 16;
IQR = Q3 – Q1 = 6 ⇒ B                    14   Median is closest to left edge of
coloured box with outliers
6   Minimum value = 22,                           = positively skewed with outliers.
Q1 = 3rd value = 23,                          ⇒E
median = average of two middle
values = 24.5,                                (Note that E is not the correct answer,
Q3 = 8th value = 27,                          but the question asks for the best
maximum value = 29 ⇒ B                        answer.)

7   Median = middle value = 63 ⇒ C           15   IQR = Q3 – Q1
= 65 – 60 = 5;
8   IQR = Q3 – Q1                                 60 – 1.5 × 5 = 52.5 = lower fence,
= 75 – 65 = 10 ⇒ A                        65 + 1.5 × 5 = 72.5 = upper fence
⇒A
Chapter 3 – Summarising numerical
data: the mean and the standard
deviation
Solutions to Exercise 3A
1                                       3
a   n = number of values = 4;           a   The median cuts a distribution in half
∑   x = sum of values = 12;             by definition whereas the mean only
does so for a symmetric distribution.
x=
∑x    =3
n                             b   The median and mean are the same
only for a symmetric distribution.
b   n = number of values = 5;
∑
c   The median is the middle value so isn’t
x = sum of values = 104;
affected by the numerical values of

x=
∑x    = 20.8
outliers. The mean takes all values into
account and thus is affected.
n
d   The median would be more
c   n = number of values = 7;
appropriate as there are likely to be a
∑   x = sum of values = 21;             few very high salaries that would skew

x=
∑x    =3                       4
the mean value positively.

n                             a   After ordering and summing the 8
values, mean = 288.8 = 36.1;
2                                                             8
a   After ordering and summing the 11       median = average of two middle
33                      values (36.0 and 36.0) = 36.0
values, mean =     = 3;
11
median = 6th value = 3;             b   Since the mean and median are very
mode = most common value = 2            close to each other, this means the
patient’s temperature distribution is
b   After ordering and summing the 12       approximately symmetric.
60
values, mean =      = 5;
12                  5
median = average of two middle      a   After ordering and summing the 7
values (5 and 5) = 5;
mode = most common value = 5            values, mean = \$25.55 = \$3.65;
7
median = middle value = \$1.70

b   In this case, the median is a much
better marker of a typical amount
spent. This is because the mean has
been positively skewed by the large
positive outlier of \$16.55.
6                                           7
a   Mean shouldn’t be used, due to the      a   Since the distribution is approximately
distribution being very negatively          symmetric, either could be used.
skewed.
1979
b   Mean =        = 82.5;
b   No reason not to use the mean.                        24
median = average of two middle
c   Mean shouldn’t be used, due to the          values (82 and 83) = 82.5.
presence of outliers.
8
d   Mean shouldn’t be used, due to the      a   Since the distribution is strongly
distribution being very positively          negatively skewed, the median would
skewed.                                     be the best measure.

e   Mean shouldn’t be used, due to the                 1608
b   Mean =       = 69.9,
presence of outliers and the                         23
distribution being positively skewed.       median = middle value = 73

f   No reason not to use the mean.
Solutions to Exercise 3B
1                                                   2
a   Summing the 5 values,                           a   Range = highest value – lowest value
15                                                 = 21 – 3 = 18;
mean =    =3
5                                           estimated standard deviation
x        (x –   x)          (x –   x )2                  18
=     = 4.5 = 5
1               –2                   4                    4
2               –1                   1           (rounded to nearest whole number)
2               –1                   1           Summing the 11 values,
4                1                   1                   114
mean =       = 10.36
6                3                   9                    11
Sum               0                  16               x     (x – x )            (x – x )2

∑ (x − x )
2
3       –7.36               54.169
16                       4       –6.36               40.449
s=                         =      = 4 =2                6       –4.36               19.009
n −1                  4                       7       –3.36               11.289
8       –2.36                5.569
b   Summing the 5 values,                                   9       –1.36                1.849
50                                              10       –0.36                0.129
mean =    = 10
5                                              14         3.64              13.249
x       (x –   x)          (x –   x )2             15         4.64              21.529
6              –4                  16              17         6.64              44.089
8              –2                   4              21       10.64             113.209
10               0                   0             Sum            0            324.545
10               0                   0
∑ (x − x )
2
16               6                  36                                    324.55
s=                   =          = 32.46 = 5.7
Sum               0                  56                   n −1               10

∑ (x − x )
2
56               b   Range = highest value – lowest value
s=                         =      = 14 = 3.74
n −1                  4                          = 119 – 99 = 20;
estimated standard deviation
c   Summing the 5 values,                                        20
=     =5=5
25                                                   4
mean =       =5                                     (rounded up to nearest whole number)
5
Standard deviation is a measure of how              Summing the 7 values,
much the values deviate from the                            766
mean =       = 109.43
mean.                                                        7
Since all the values are the                           x     (x – x )           (x – x )2

∑ (x − x )
2                              99      –10.43            108.784
mean,                      = 0;
101       –8.43              71.064
standard deviation = 0                               106       –3.43              11.764
112         2.57              6.604
114         4.57             20.884
115         5.57             31.024
119         9.57             91.584
Sum       –0.01            341.714

∑ (x − x)
2
341.71
s=                   =          = 56.95 = 7.5
n −1               6
c   Range = highest value – lowest value           5
= 2.5 – 1.6 = 0.9;                      a   Summing the 10 values,
estimated standard deviation                              201
mean =     = 20.1
0.9                                              10
=     = 0.23 = 3
4                                            x       (x – x )                  (x –   x )2
(rounded up to nearest whole number)                  17           –3.1                         9.61
18           –2.1                         4.41
Summing the 7 values,                                 19           –1.1                         1.21
14.3                                           20           –0.1                         0.01
mean =      = 2.04
7                                             20           –0.1                         0.01
x       (x –    x)       (x –   x )2              20           –0.1                         0.01
1.6              –0.44           0.193             21            0.9                         0.81
1.9              –0.14           0.019             21            0.9                         0.81
2.0              –0.04           0.001             22            1.9                         3.61
2.0              –0.04           0.001             23            2.9                         8.41
2.1               0.06           0.003            Sum              0                         28.9

∑ (x − x )
2.2               0.16           0.025                                 2
28.9
2.5               0.46           0.211          s=                         =          = 3.21 = 1.8
Sum                0.02           0.457                   n −1                     9

∑ (x − x)
2
0.457                  b   Since the median and mean are very
s=                  =         = 0.076 = 0.28
n −1             6                        similar in value, the distribution must
be very symmetric in appearance.
3
a   The IQR, by definition, will always            6   TVs: Summing the 7 values,
incorporate 50% of the scores,                          mean = 3151 = 450
specifically the middle 50% of scores.                           7
x        (x – x )                 (x – x )2
b   Since range = highest score – lowest                 354         –96.1                  9235.21
score, the range only uses the smallest              378         –72.1                  5198.41
and largest scores.                                  381         –69.1                  4774.81
404         –46.1                  2125.21
c   The standard deviation is the average                471           20.9                   436.81
amount by which the scores differ                    539           88.9                 7903.21
from the mean.                                       624         173.9                 30241.21
Sum            0.3                59914.87

∑ (x − x)
4   It doesn’t make sense to calculate a                              2

s=                         59914.87
mean and standard deviation for sex                                       =            = 9985.8 = 100
n −1                  6
(b), year level (d) and weight (f), since
all three are categorical variables and
not numerical variables.                           Cars: Summing the 7 values,
Mean and standard deviation cannot be                    mean = 2632 = 376
7
calculated for categorical variables.
x        (x – x )             (x – x )2
217          –159                 25281
286           –90                  8100
357           –19                   361
370            –6                    36
417            41                  1681
435            59                  3481
550           174                 30276
Sum             0                 69216

∑ (x − x)
2
69216
s=                     =             = 11536 = 107
n −1                     6
Alcohol: Summing the 7 values,                        Females: Summing the 23 values,
104.2                                                 1734
mean =        = 14.9                                  mean =      = 75.4
7                                                    23
x        (x – x )            (x – x )2                   x      (x – x )      (x – x )2
9.5           –5.4               29.16                   65        –10.4         108.16
9.9            –5                   25                   73          –2.4          5.76
12.5           –2.4                5.76                   74          –1.4          1.96
14.6           –0.3                0.09                   81           5.6         31.36
16.0            1.1                1.21                   59        –16.4         268.96
17.6            2.7                7.29                   64        –11.4         129.96
24.1            9.2               84.64                   76           0.6          0.36
Sum            –0.1              153.15                   83           7.6         57.76
95          19.6        384.16
s
70          –5.4         29.16
∑ (x − x )
2
153.15                         73          –2.4          5.76
=                         =           = 25.53 = 5.1           79           3.6         12.96
n −1                     6
64        –11.4         129.96
77           1.6          2.56
7   Males: Summing the 23 values,                                 80           4.6         21.16
1642                                            82           6.6         43.56
mean =      = 71.4
23                                             77           1.6          2.56
x        (x –           x)   (x – x )2                  87          11.6        134.56
80                 8.6           73.96                   66          –9.4         88.36
73                 1.6            2.56                   89          13.6        184.96
73                 1.6            2.56                   68          –7.4         54.76
78                 6.6           43.56                   78           2.6          6.76
75                 3.6           12.96                   74          –1.4          1.96
65                –6.4           40.96                  Sum          –0.2      1707.48

∑ (x − x )
69                –2.4            5.76                               2
70                –1.4            1.96                                       1707.4
s=                 =          = 77.6 = 8.8
70                –1.4            1.96                        n −1             22
78                 6.6           43.56
58               –13.4          179.56
77                 5.6           31.36
64                –7.4           54.76
76                 4.6           21.16
67                –4.4           19.36
69                –2.4            5.76
72                 0.6            0.36
71                –0.4            0.16
68                –3.4           11.56
72                 0.6            0.36
67                –4.4           19.36
77                 5.6           31.36
73                 1.6            2.56
Sum                –0.2          607.48

∑ (x − x )
2

s=                             607.48
=           = 27.61 = 5.3
n −1                  22
Solutions to Exercise 3C
1                                             d   2.28 is 2 standard deviations above the
a   68% of values lie within 1 standard           mean.
deviation of the mean.                        ⇒ 2.5% of values are above 2.28.
134 + 20 = 154, 134 – 20 = 114,
⇒ 68% of values lie between 114 and       e   1.28 is 3 standard deviations below the
154.                                          mean.
⇒ 0.15% of values are below 1.28.
b   95% of values lie within 2 standard
deviations of the mean.                   f   1.88 is the mean.
134 + 40 = 174, 134 – 40 = 94,                ⇒ 50% of values are above 1.88.
⇒ 95% of values lie between 94 and
174.                                      3
a   i 8 and 20 are both 2 standard
c   99.7% of values lie within 3 standard             deviations from the mean.
deviations of the mean.                           ⇒ 95% of values are between 8
134 + 60 = 194, 134 – 60 = 74,                    and 20.
⇒ 99.7% of values lie between 74              ii 11 is 1 standard deviation below the
and 194.                                          mean.
⇒ 16% of values are below 11.
d   16% of values are greater than 1              iii 20 is 2 standard deviations above
standard deviation above the mean.                the mean.
134 + 20 = 154,                                   ⇒ 2.5% of values are above 20.
⇒ 16% of values are above 154.                iv 14 is the mean.
⇒ 50% of values are below 14.
e   2.5% of values are less than 2 standard       v 5 is 3 standard deviations below the
deviations below the mean.                        mean.
134 – 40 = 174,                                   ⇒ 0.15% of values are below 5.
⇒ 2.5% of values are below 174.               vi 11 and 17 are both 1 standard
deviation from the mean.
f   0.15% of values are less than 3                   ⇒ 68% of values are between 11
standard deviations below the mean.               and 17.
134 – 60 = 74,
⇒ 2.5% of values are below 74.            b   8 is 2 standard deviations below the
mean.
g   50% of values are greater than the            ⇒ 2.5% of values are below 8.
mean.                                                1000
⇒ 50% of values are greater than 134.         2.5 ×        = 25 walkers would be
100
expected to complete the circuit in less
2                                                 than 8 minutes if 1000 walkers
a   1.68 and 2.08 are both 1 standard             attempted it.
deviation from the mean.
⇒ 68% of values are between 1.68
and 2.08.

b   1.28 and 2.48 are both 3 standard
deviations from the mean.
⇒ 99.7% of values are between 1.28
and 2.48.

c   2.08 is 1 standard deviation above the
mean.
⇒ 16% of values are above 2.08.
4                                           5
a   i 155 and 185 are both 3 standard       a   i 66 is the mean.
deviations from the mean.                   ⇒ 50% of values are below 66.
⇒ 99.7% of values are between 155       ii 70 is 1 standard deviation above the
and 185.                                    mean.
ii 175 is 1 standard deviation above            ⇒ 16% of values are above 70.
the mean.                               iii 62 and 70 are both 1 standard
⇒ 16% of values are above 175.              deviation from the mean.
iii 170 is the mean.                            ⇒ 68% of values are between 62
⇒ 50% of values are above 170.              and 70.
iv 160 is 2 standard deviations below       iv 62 is 1 standard deviation below the
the mean.                                   mean.
⇒ 2.5% of values are below 160.             ⇒ 16% of values are below 62.
v 165 is 1 standard deviation below         v 58 and 74 are both 2 standard
the mean.                                   deviations from the mean.
⇒ 16% of values are below 165.              ⇒ 95% of values are between 58
vi 160 and 180 are both 2 standard              and 74.
deviations from the mean.               vi 70 is 1 standard deviation above the
⇒ 95% of values are between 160             mean.
and 180.                                    ⇒ 16% of values are above 70, and
thus 100 – 16 = 84% of values are
b   175 is 1 standard deviation above the           below 70.
mean.
⇒ 16% of values are above 175.          b   54 and 78 are both 3 standard
5000                                   deviations from the mean.
16 ×        = 800
100                                   ⇒ 99.7% of values are between 54
and 78.
2000
99.7 ×        = 1994
100
Solutions to Exercise 3D
1                           2
a   z = 120 – 100 = 1       a   English: z =
69 – 60
= 2.25
20                                    4
75 – 60
Biology: z =          =3
b   z = 140 – 100 = 2                            5
20                                   55 – 55
Chemistry: z =           =0
6
80 – 100                                  55 – 44
c   z=            = –1          Further Maths: z =           = 1.1
20                                        10
73 – 82
Psychology: z =          = –2.25
100 – 100                                  4
d   z=             =0
20
b   English: A z–score of 2.25 is at least 2
40 – 100                 standard deviations above the mean, so
e   z=          = –3
20                    the student was within the top 2.5% of
scores for English.
110 – 100
f   z=             = 0.5
20                  Biology: A z–score of 3 is 3 standard
deviations above the mean, so the
90 – 100               student was within the top 0.15% of
g   z=            = –0.5
20                  scores for Biology.

125 – 100              Chemistry: A z–score of 0 is the mean,
h   z=             = 1.25
20                  so the student was exactly average for
Chemistry.
85 – 100
i   z=            = –0.75
20                  Further Maths: A z–score of 1.1 is at
least 1 standard deviation above the
50 – 100               mean, so the student was within the top
j   z=            = –2.5
20                  16% of scores for Further Maths.

Psychology: A z–score of –2.25 is at
least 2 standard deviations below the
mean, so the student was within the
bottom 2.5% of scores for Psychology.
Solutions to Exercise 3E
1                                            2
a   The sample chosen by each person will    a   Create a suitable table to record all the
be different and, once chosen, should        details.
be keyed into the graphics calculator.
b   Answers will depend on the sample of
b   The mean IQ will be the sum of all the       12 people chosen.
IQs divided by 10.
c   Since the different sample will most
c   Since the different sample will most         likely be made up of different people,
likely be made up of different people,       it is unlikely that the statistics would
it is unlikely that the mean IQ would        be the same.
be the same.
d   i Summing the ages of the 33 males
in the population, we get 756.
756
Thus, mean =           = 22.9
33
ii Since this is a population parameter
(that is, it is the mean for the entire
male population), we use μ as its
symbol.
Solutions to Multiple–choice questions
1   Summing the 9 values, we get 108.                      50 – 55
7    z=
108                                                2.5
Mean =     = 12 ⇒ C
9                                              -5
=     = –2 ⇒ B
2.5
2   Range = highest value – lowest value
= 25 – 1 = 24.                          8    The rule only applies to normal
Standard deviation estimate = range / 4            distributions, which are bell–shaped.
= 24 = 6. ⇒ E                                    ⇒D
4
9    Using the 68–95–99.7% rule, 68% of
3   Mean = 12                                          the values lie within one SD of the
x         (x –   x)     (x –   x )2             mean. ⇒ A
1                –11           121
8                 –4            16         10   Using the 68–95–99.7% rule, 99.7% of
10                –2             4              the values lie within three SDs of the
10                –2             4              mean. ⇒ C
11                –1             1
12                 0             0         11   Using the 68–95–99.7% rule, 2.5% of
15                 3             9              the values lie more than two SDs
16                 4            16              below the mean. ⇒ E
25                13           169
Sum                 0           340         12   Using the 68–95–99.7% rule, 16% of

s=
(
∑ x−x     )2 =       340
= 42.5 = 6.52
the values lie more than one SD above
the mean. ⇒ D
n −1                8
⇒E                                            13   Since 17.8 and 19.0 are 2 standard
deviations from the mean, so 95% of
4   The mean is by definition the balance              the students will have ages between
point of the data.                                 these two values. ⇒ E
The word ‘average’ often refers to the
mean but there are in fact three types        14   18.4 is the mean, so 50% of students
of averages: mean, median and mode.                will have ages above this value.
⇒D                                                 50 ×
500
= 250 ⇒ C
100
5   Since phone numbers are categorical
variables and not numerical, the mean         15   18.1 is 1 standard deviation below the
and standard deviation of these cannot             mean, so 16% of students will have
be calculated. ⇒ B                                 ages below this value.
500
16 ×       = 80 ⇒ B
6   The median is not a very accurate                        100
measure of the centre of the
distribution when it is clearly skewed
or if it has outliers, since the mean is
affected by both skewing and outliers.
⇒D
Chapter 4 – Displaying and
describing relationships between
two variables
Solutions to Exercise 4A
1    Table 1:                                    3
‘Old–No’ = 85 – 15 = 70                     a   i ‘Male Total’ + ‘Female Total’
‘Total–Yes’ = 23 + 15 = 38                          = 45 + 55 = 100 people surveyed
‘Total–No’ = 22 + 70 = 92                       ii ‘Male Against’ = 20
‘Total–Total’ = 45 + 85 = 130                   iii ‘Female Total’ = 55
iv ‘Female For’ = 30
Table 2:                                             v ‘Male For’ + ‘Female For’
‘Young–Yes’ = 23 × 100% = 51.1%                    = 25 + 30 = 55
45
‘Young–No’ =   22 × 100% = 48.9%
45                                            25
b   ‘Male–For’ =     × 100% = 55.6%
45
‘Young–Total’ = 45 × 100% = 100.0%                                20
45                              ‘Male–Against’ =     × 100% = 44.4%
‘Old–Yes’ = 15 × 100% = 17.6%                                     45
85                                                  30
‘Female–For’ =     × 100% = 54.5%
55
2    When two variables are related, the                                 25
‘Female–Against’ =     × 100%
dependent variable is the one affected                              45
by the other (independent) variable.              = 45.5%
Changing the independent variable               ‘Male–Total’ = ‘Female–Total’
changes the dependent variable, but the           = 100.0%
dependent variable doesn’t affect the           We can thus draw the table below:
other variable, so changing it does not
change the independent variable.
For example, changing a person’s job
will change their salary, but giving a
person a pay rise and thus changing
their salary doesn’t change their job. In
this instance, salary is the dependent      c   Since the ‘for’ and ‘against’
variable and job is independent.                percentages were very similar for both
sexes, we can make the following
Keeping this in mind, we get the                statement:
following results:                              ‘The percentage of males and females
in support of Sunday races was similar,
a    Participates in regular exercise                55.6% to 54.5%. This indicates that a
b    Salary level                                    person’s support for Sunday races was
not related to their sex.’
c    Comfort level
d    Incidence of hay fever
e    Musical taste
f    AFL team supported
4                                             d   Since the ‘yes’ and ‘no’ percentages
a   It is more likely that enrolment status       were very similar for both full–time
determines drinking behaviour rather          and part–time students, we can make
than the other way around, making             the following statement:
drinking behaviour the dependent              ‘The percentage of full–time and part–
variable.                                     time students who drank alcohol is
similar, 80.5% to 81.8%. This
b   i ‘Total–Yes’ = 196                           indicates that drinking behaviour is not
ii ‘Part–time–Total’ = 88                     related to enrolment status.’
iii ‘Full–time–Yes’ = 124

124
c   ‘Full–time–Yes’ =       × 100%
154
= 80.5%
30
‘Full–time–No’ =        × 100%
154
= 19.5%
72
‘Part–time–Yes’ =      × 100%
88
= 81.8%
16
‘Part–time–No’ =      × 100% = 18.2%
88
‘Full–time–Total’ = ‘Part–time–Total’
= 100.0%
We can thus draw the table below:
Solutions to Exercise 4B
1                                                     We can thus draw the table below:
a   Using the information in the table, we
can draw the following segmented bar
chart, taking care to draw the bars up
to their correct heights as given in the
table.

c   Married = M, Widowed = W,
Divorced = D, Separated = S,
Never = N, Exciting = Ex,
Pretty routine = Pr, Dull = Du.

392
M–Ex =        × 100% = 47.6%
824
401
M–Pr =        × 100% = 48.7%
824
b   Since the heights of both the ‘For’ and                    31
M–Du =         × 100% = 3.8%
‘Against’ bars are very similar, we can                   824
draw the same conclusion as we did                         51
W–Ex =        × 100% = 33.8%
previously:                                               151
‘The approximately equal length of the                     82
W–Pr =        × 100% = 54.3%
segments representing those in favour                     151
of Sunday racing in the bar chart                          18
W–Du =         × 100% = 11.9%
indicates that the percentage of males                    151
who favour Sunday racing is similar to                     77
D–Ex =         × 100% = 46.7%
the percentage of females who favour                      165
Sunday racing, so there is no                             77
D–Pr =        × 100% = 46.7%
relationship between people’s attitude                   165
to Sunday racing and sex.’                                 11
D–Du =        × 100% = 6.7%
165
2                                                            18
S–Ex =      × 100% = 42.9%
a   i ‘Total–Total’ = 1461
42
20
ii ‘Divorced–Total’ = 165                         S–Pr =      × 100% = 47.6%
42
iii ‘Separated–Dull’ = 4
4
iv ‘Married–Pretty routine’ = 401                 S–Du =      × 100% = 9.5%
42
146
b   ‘Married–Total’ = 392 + 401 + 31                  N–Ex =        × 100% = 52.3%
279
= 824                                                  124
‘Widowed–Dull’ = 73 – 31 – 11 – 4 – 9             N–Pr =       × 100% = 44.4%
279
= 18                                                      9
‘Widowed–Exciting’ = 151 – 18 – 82                N–Du =        × 100% = 3.2%
279
= 51
‘Divorced–Pretty routine’ = 165 –11 – 77
= 77
‘Separated–Pretty routine’ = 42 – 4 – 18
= 20
‘Total–Exciting’ = 392 + 51 + 77 + 18 + 146
= 684
d   Using the information from the above   e   Comparing the heights of the ‘dull’,
percentage table, we can draw the          ‘pretty routine’ and ‘exciting’ bars, we
segmented bar chart below:                 can see a number of differences in
terms of a person’s attitude to life and
how this relates to their marital status.
We can thus say:
‘Yes. There are several ways that this
can be seen: for example, by
comparing the married and widowed
groups, we can see that a smaller
percentage of those widowed found
life exciting (33.8%) compared to
those who were married (47.6%). Or, a
bigger percentage of widowed people
found life pretty routine (54.3% to
48.7%) and dull (11.9% to 3.8%)
compared to those who were married.’

f   It is more likely that marital status
determines attitude to life than the
other way around, and so marital status
is most likely the independent variable.
Solutions to Exercise 4C
1                                           2
a   i Weight loss is numerical, level of    a   Battery life time is the numerical
exercise is categorical.                 variable in this graph while battery
ii Weight loss is most likely the           price is the categorical variable.
dependent variable.
b   This contention is supported by the
b   i Hours of study is categorical, test       box plots. As price increases, the
mark is numerical.                       median lifetime of the battery
ii Test mark is most likely the             increases. Also, a higher price gives a
dependent variable.                      battery with a less variable lifetime,
which is thus more consistently of
c   i State of residence is categorical,        longer lifetime.
number of sporting teams is
numerical.                           3
ii Number of sporting teams is most     a   i Pulse rate is the numerical variable
likely the dependent variable.              in this graph while sex is the
categorical variable.
d   i Temperature is numerical, season is       ii Pulse rate is the dependent variable
categorical.                                in this graph while sex is the
ii Temperature is most likely the              independent variable.
dependent variable.
b   This contention is supported by the
box plots. We can see that the median
pulse rate for females is 76 beats/min
while for males is 73 beats/min.
Furthermore, female pulse rates are
distributed across a much wider range
than male pulse rates. These both
indicate that pulse rate is related to sex.
Solutions to Exercise 4D
1   Using the Stat List Editor function of   3   Using the Stat List Editor function of
the graphics calculator, key the             the graphics calculator, key the
minimum temperature values into the          temperature values into the L1 column
L1 column and the maximum                    and the diameter values into the L2
temperature values into the L2 column.       column. The following plot will be
The following plot will be created:          created:

2   Using the Stat List Editor function of   4   Using the Stat List Editor function of
the graphics calculator, key the balls       the graphics calculator, key the time
faced values into the L1 column and          values into the L1 column and the
the runs scored values into the L2           number in theatre values into the L2
column. The following plot will be           column. The following plot will be
created:                                     created:
Solutions to Exercise 4E
1                                                  b   i Comparing with the sample graphs
a   No relationship would be expected.                     on page 99 of Essential Further
Mathematics, the correlation
b   A positive relationship would be                       coefficient is approximately +0.7,
expected, salary level should increase                 which is a moderate positive
with intelligence.                                     relationship.
ii Comparing with the sample graphs
c   A positive relationship would be                       on page 99 of Essential Further
expected, tax paid should increase with                Mathematics, the correlation
salary.                                                coefficient is approximately –0.4,
which is a weak negative
d   A positive relationship would be                       relationship.
expected, aggression should increase               iii Comparing with the sample graphs
with frustration.                                      on page 99 of Essential Further
Mathematics, the correlation
e   A negative relationship would be                       coefficient is approximately +0.9,
expected, population density should                    which is a strong positive
decrease with distance from the centre                 relationship.
of a city.                                         iv There is no relationship, so the
correlation coefficient will be
f   A negative relationship would be                       approximately 0.
expected, creativity should decrease
with time spent watching TV.                   3   As explained on page 100 of Essential
Further Mathematics, the three
a   i The dots drift upwards to the right,             1) The variables are both numeric
so it is a positive relationship.              2) The relationship between the two
ii The dots drift downwards to the                    variables is linear
right, so it is a negative relationship.       3) There are no outliers in the
iii The dots drift upwards to the right,              relationship.
so it is a positive relationship.
iv There is no relationship between
the variables.
Solutions to Exercise 4F
1                                                 3
a   Scatterplot A shows a positive, non-          a   Using the Stat List Editor and
linear relationship with no outliers.             LinReg(a+bX) function of the
Scatterplot B shows a negative, linear            graphics calculator, you will get:
relationship with one outlier.                    a = –0.502
Scatterplot C shows a negative, linear            b = 0.631
relationship with no outliers.                    r2 = 0.668
r = 0.818 (to 3 decimal places)
b   It wouldn’t be appropriate to use the
correlation coefficient for scatterplot A     b   Using the Stat List Editor and
(as the relationship is non-linear) or for        LinReg(a+bX) function of the
scatterplot B (as the relationship has an         graphics calculator, you will get:
outlier).                                         a = 8.2965
b = 1.0597
2   Using the information given in the                r2 = 0.7713
table, we can draw the following table:           r = 0.8782 (to 4 decimal places)
(x – x )   c   Using the Stat List Editor and
x      x–x     y       y–y    × (y – y )       LinReg(a+bX) function of the
2        –2    1         –4           8        graphics calculator, you will get:
3        –1    6          1          –1        a = 131.3552
6         2    5          0           0        b = –2.6022
3        –1    4         –1           1        r2 = 0.4525
6         2    9          4           8        r = –0.6727 (to 4 decimal places)
Sum        0               0          16
d   Using the Stat List Editor and
r=
(     )(
∑ x−x y− y        )                           LinReg(a+bX) function of the
(n − 1)s x s y                              graphics calculator, you will get:
a = 85.49
16                                      b = 0.73
=
4 × 1.871× 2.915                                r2 = 0.68
= 0.73                                            r = 0.82 (to 4 decimal places)
Solutions to Exercise 4G
1                                          3
a   Coefficient of determination = r2      a   The coefficient of determination = r2
= (0.675)2 = 0.456 = 45.6%                  = (–0.611)2 = 0.373 or 37.3%,
which can be interpreted as 37.3% of
b   Coefficient of determination = r2          the variability observed in hearing test
= (0.345)2 = 0.119 = 11.9%                scores being explainable by variation
in age.
c   Coefficient of determination = r2
= (–0.567)2 = 0.321 = 32.1%           b   The coefficient of determination = r2
= (0.716)2 = 0.513 or 51.3%,
d   Coefficient of determination = r2          which can be interpreted as 51.3% of
= (–0.673)2 = 0.453 = 45.3%               the variability observed in mortality
rates being explainable by variation in
e   Coefficient of determination = r2          smoking rates.
= (0.124)2 = 0.015 = 1.5%
c   The coefficient of determination = r2
2                                                = (–0.807)2 = 0.651 or 65.1%,
a   r2 = 0.8215,                               which can be interpreted as 65.1% of
r = 0.8215 = 0.906 (the relationship       the variability observed in life
is positive)                               expectancies being explainable by
variation in birth rates.
b   r2 = 0.1243,
d   The coefficient of determination = r2
r = 0.1243 = –0.353 (the                     = (0.818)2 = 0.669 or 66.9%,
relationship is negative)                  which can be interpreted as 66.9% of
the variability observed in daily
maximum temperature being
explainable by the variability in daily
minimum temperatures.

e   The coefficient of determination = r2
= (0.8782)2 = 0.771 or 77.1%,
which can be interpreted as 77.1% of
the variability in the runs scored by a
batsman being explainable by the
variability in the number of balls they
face.
Solutions to Exercise 4H
1   It cannot necessarily be said that taller   3   It cannot necessarily be said that eating
people are better at mathematics.               an ice cream at the beach increases the
The relationship is more likely to be           risk of drowning.
that as people get older, they have been        The relationship is more likely that
learning mathematics for longer and             more ice creams are eaten on warmer
thus perform better on tests of                 days and more people swim on warmer
mathematical ability.                           days, so there are more people in the
water who are at risk of drowning and
2   It cannot necessarily be said that              thus do drown.
religion drives people to drink alcohol.
The relationship is more likely to be       4   It cannot necessarily be said that taking
that as a town gets bigger, more                up a musical instrument will increase
churches are built and the increased            your performance in mathematics.
population means there are more                 The relationship is more likely to be
people who drink alcohol, meaning               that there are certain traits that those
that more alcohol is consumed by the            who do well in music and those who
people of the town.                             have an aptitude for maths tend to
share, such that those who are
competent at one are more likely to be
competent at the other.
Solutions to Exercise 4I
1                                           e   Two numerical variables = scatterplot
a   Two categorical variables = segmented
bar chart                               f   Two categorical variables = segmented
bar chart
b   Two numerical variables = scatterplot
g   Two categorical variables = segmented
c   Numerical dependent variable (hours         bar chart
spent at the beach) and categorical
independent variable (state of          h   Numerical dependent variable
residence) = parallel box plots             (cigarettes smoked per day) and
categorical independent variable (sex)
d   Two numerical variables = scatterplot       = parallel box plots
Solutions to Multiple–choice questions
1   Plays sport and sex are both               10   Using the Stat List Editor and
categorical variables. ⇒ A                      LinReg(a+bX) function of the
graphics calculator, you will get:
2   ‘Females–No’ = 175 – 79                         a = 39.62
= 96 ⇒ D                                      b = 5.36
r2 = 0.62
3   Percentage ‘Males–No’                           r = 0.7863 (to 4 decimal places)
34                                         ⇒C
=     × 100% = 33.3% ⇒ B
102
11   Coefficient of determination = r2
4   There appears to be a relationship               = (–0.7685)2 = 0.5906 ⇒ D
since the percentage of females
playing sport compared to males is         12   Coefficient of determination = r2
much lower. ⇒ D                                   = (0.765)2 = 0.585 = 58.5%.
Since heart weight is the dependent
5   Battery life is a numerical variable but        variable, this result tells us that 58.5%
brand is a categorical variable. ⇒ C            of the variation in heart weights can be
attributed to a mouse’s body weight.
6   The first statement supports the                ⇒A
contention because the difference in
median battery life shows one brand to     13   Since the dependent variable (weight)
be superior.                                    is numerical and the independent
The second and third statements                 variable (level of nutrition) is
support the contention because they             categorical, we would use parallel box
both show one brand to be more                  plots. ⇒ B
reliable than the other.
Therefore, all three statements support    14   Since the two variables are both
the contention. ⇒ E                             categorical, we would use a segmented
bar chart. ⇒ C
7   Weight at age 21 and weight at birth
are both numerical variables. ⇒ D          15   This relationship tells us that people on
higher salaries recycle more garbage
8   The dots drift upwards towards the              but we cannot assume that it is the
right of the graph and appear to be             higher salary that causes more
aggregated fairly closely around an             recycling of garbage. It may be that
invisible linear, making the                    those on higher salaries have more
relationship a strong positive linear           education and thus are more aware of
relationship. ⇒ E                               the environmental impact of recycling.
⇒E
9   If r = –0.9, this means the relationship
is a strongly negative linear one. This
means that as drug dosage increases,
response time will decrease. ⇒ C
Chapter 5 – Regression: fitting lines
to data
Solutions to Exercise 5A
1   A residual is the vertical difference    2   To find the least squares regression
between a value on a plot and the            line, we must minimise the sum of the
regression line drawn to fit the plot.       squares of the residual values. ⇒ C
Solutions to Exercise 5B
1                                              4
a   Traffic volume is the independent          a   If the slope is negative, the correlation
variable (IV) since it most likely             coefficient must also be negative.
determines the pollution level, which
is the dependent variable (DV).            b   If the correlation coefficient is zero,
this means the least squares regression
b   b = r × sx                                     line will be horizontal and thus will
sy                                     have a slope of zero.
= 0.94 × 97.87 = 49.2
1.87                            c   The correlation coefficient being zero
a= y –bx                                       means the line will be horizontal and
thus will have a constant y-value for its
= 230.7 – 49.2 × 11.38 = –329.2
entire length. This y-value will thus be
Therefore, Pollution level = a + b × x
the middle of all the y-values and will
= –329.2 + 49.2 × Traffic volume.
be the mean y-value, which is y .
2
a   Birth rate is the independent variable     5   Using the Stat List Editor function of
(IV) since it most likely determines the       the graphics calculator, entering in the
life expectancy, which is the dependent        values given and using the
variable (DV).                                 LinReg(a+bX) function, we get:
a = 16.45
b   b = r × sx                                     b = 0.57
sy                                     r2 = 0.48
= –0.8069 × 9.689 = –1.44                    r = 0.69
5.411
a= y –bx                                   6   Using the Stat List Editor function of
= 55.1 + 1.445 × 34.8 = 105.4                 the graphics calculator, entering in the
Therefore, Life expectancy = a + b × x         values given and using the
= 105.4 – 1.44 × Birth rate.                  LinReg(a+bX) function, we get:
a = 131.4
3                                                  b = –2.6
a   Age is the independent variable (IV)           r2 = 0.45
since it most likely determines the            r = 0.67
distance travelled, which is the               Note that hours worked is the
dependent variable (DV).                       independent variable.

7
b   b = r × sx                                 a   Using the Stat List Editor function of
sy
the graphics calculator, entering in the
= 0.947 × 42.61 = 11.09                      values given and using the
3.64
LinReg(a+bX) function, we get:
a= y –bx                                       a = –2.6
= 78.04 – 11.09 × 5.63 = 15.6                 b = 0.73
Therefore, Distance = a + b × x                r2 = 0.77
= 15.6 + 11.09 × Age.                         r = 0.88

b   Replacing x and y with balls faced and
runs, we get:
Runs = –2.6 + 0.73 × Balls faced.
8
a   If we are predicting the number of TVs
from the number of cars, then the
former is depending upon the latter,
meaning the number of TVs is the
dependent variable.

b   Using the Stat List Editor function of
the graphics calculator, entering in the
values given and using the
LinReg(a+bX) function, we get:
a = 61.2
b = 0.93
r2 = 0.68
r = 0.82

c   Replacing x and y with the number of
cars and number of TVs respectively,
we get:
Number of TVs = 61.2 + 0.93 ×
Number of cars
Solutions to Exercise 5C
1                                                 4
a   Since hand span is being determined           a   Using the Stat List Editor function of
by height, Height is the independent              the graphics calculator, entering in the
variable.                                         values given and plotting them in the
graph window of the calculator, we get
b   Slope is the gradient value of 0.33               the following scatterplot:
whereas intercept is the constant value
of 2.9
c   Hand span = 2.9 + 0.33 × 160
= 55.7 cm
d   Residual = actual – predicted
= 58.5 – 55.7 = 2.8 cm
b   Using the LinReg(a+bX) function, we
2                                                     get:
a   Since fuel consumption is being                   a = 4.71
determined by weight, Fuel                        b = 0.72
consumption is the dependent                      r2 = 0.6115
variable.                                         r = 0.7820
This translates to:
b   Slope is the gradient value of 0.01               Test A score = a + b × Test B score
whereas intercept is the constant value             = 4.71 + 0.72 × Test B score.
of –0.1
c   Running the LinReg(a+bX) function
c   Fuel consumption = –0.1 + 0.01 × 980              with the Y1 command, such that the
= 9.7 litres/100 km                             regression curve is inputted to Y1,
allows us to draw the following
d   Residual = actual – predicted                     regression line:
= 8.9 – 9.7 = –0.8 litres/100 km
3   For the first graph, the y-intercept is
80. The graph then intersects the point
(8,48). This represents a drop in
y-value of 32 marks over 8 days, or 4
marks per day, a gradient of –4.
Hence, Mark = 80 – 4 × Days absent.
d   Plotting a Stat Plot on the graphics
For the second graph, the two most                calculator with the Y list as RESID
clearly identifiable points on the graph          gives us the following residual plot:
are (10,6) and (30,14). This represents
a rise of 8 for a run of 20, or 0.4 death
rate per 1 unit of birth rate, which is a
Extrapolating to the left from the point
(10,6), we know the graph rises 0.4
death rate units per birth rate unit to
the right, so must drop 0.4 death rate
units per birth rate unit to the left. Over
10 units to the y–axis, the line must
drop 0.4 × 10 = 4 units to the left,
giving us a y–intercept of 6 – 4 = 2.
Hence, Death rate = 2 + 0.4 × Birth
rate.
5                                              c   r=    r 2 = 0.7569 = 0.87
a   Using the Stat List Editor function of
the graphics calculator, entering in the
d   The r2 value can be interpreted as the
values given and plotting them in the
proportion of the variation in a
graph window of the calculator, we get
dependent variable that is due to the
the following scatterplot:
independent variable.
Thus, since 0.7569 × 100% = 75.7%,
we can say that 75.7% of the variation
in energy content is due to the
variation in fat content.

e   Energy content = 27.8 + 14.7 × 8
= 145.4 calories
b   Using the LinReg(a+bX) function, we
get:
f   Residual = actual – predicted
a = 17.5
= 132 – 145.4 = –13.4 calories
b = –1.1
r2 = 0.7075
7   A: A linear model probably wasn’t
r = 0.8412
appropriate as the linear regression
This translates to:
residual graph shows a clear
Test score = a + b × Careless errors
logarithmic pattern in the residuals,
= 17.5 – 1.1 × Careless errors
which means that a polynomial model
with a higher power or a different
c   Running the LinReg(a+bX) function
model entirely, such as a logarithmic
with the Y1 command, such that the
model, would be more appropriate.
regression curve is inputted to Y1,
allows us to draw the following
C: A linear model probably wasn’t
regression line:
appropriate as the linear regression
residual graph shows a clear
parabolic–like pattern in the residuals,
which means that a polynomial model
with a higher power would be more
appropriate.

8
d   Plotting a Stat Plot on the graphics       a   The gradient of the graph is –0.278.
calculator with the Y list as RESID            Since the sign of the gradient is
gives us the following residual plot:
negative, the success rate decreases
with distance from the hole. Since the
value of the gradient is 0.278, the
success rate decreases by 0.278%
the hole.

b   Success rate = 98.5 – 0.278 × 90
6
= 98.5 – 25.02 = 73.5%
a   Slope is the gradient value of 14.7
whereas intercept is the constant value
c   0% = 98.5 – 0.278 × Distance
of 27.8.
98.5 = 0.278 × Distance
Distance = 354.3 cm = 3.54 m
b   Since the gradient of the equation = b
= 14.7, this translates directly to an
increase of 14.7 in the energy content
for every increase of 1 g of fat.
d   r = r 2 = 0.497 = 0.705.                   10
However, the relationship is a negative    a    r = r 2 = 0.370 = 0.6083. However,
one, since success rate decreases as            the relationship is a negative one, since
distance increases, so r = –0.705.              hearing test score decreases as age
increases, so r = –0.6083.
e   The r2 value can be interpreted as the
proportion of the variation in a           b    The r2 = coefficient of determination
dependent variable that is due to the           value can be interpreted as the
independent variable. Thus, since               proportion of the variation in a
0.497 × 100% = 49.7%, we can say                dependent variable that is due to the
that 49.7% of the variation in success          independent variable. Thus, since
rate is due to the variation in distance        0.370 × 100% = 37.0%, we can say
of the golfer from the hole.                    that 37.0% of the variation in a
person’s hearing score test is due to the
9
variation in their age.
a   It is appropriate in this case, as the
plots appear to distributed in a very
c    Substituting Age and Hearing test
linear fashion.
score for x and y respectively into y =
b   Coefficient of determination = r2               3.9 – 0.024 × x:
= (0.967)2 = 0.9351                               Hearing test score = 3.9 – 0.024 × Age

c   The r2 value can be interpreted as the     d    The gradient of the graph is –0.024.
proportion of the variation in a                Since the sign of the gradient is
dependent variable that is due to the           negative, the hearing test score
independent variable. Thus, since               decreases with age. Since the value of
0.9351 × 100% = 93.5%, we can say               the gradient is 0.024, the success rate
that 93.5% of the variation in a                decreases by 0.024 points with each
person’s Pay rate is due to the                 additional year of age.
variation in their years of Experience.
e    i Hearing test score = 3.9 – 0.024 × Age
d   Substituting Experience and Pay rate               = 3.9 – 0.024 × 20 = 3.9 – 0.48
for x and y respectively into y = 8.56 +           = 3.42
0.289 × x:                                      ii Residual = actual – predicted
Pay rate = 8.56 + 0.289 × Experience             = 2.0 – 3.42 = –1.42
e   The y-intercept is by definition the pay
rate when experience = 0, and thus is      f    i The plot point at 35 years is a
the pay for an employee who has just               vertical distance of 0.3 above the
begun work.                                        regression line, so the residual is 0.3.
ii The plot point at 55 years is a
f   The slope of 0.29 tells us that the                vertical distance of 0.4 below the
average increase in pay rate for a                 regression line, so the residual is –0.4.
person per year of experience is \$0.29.
g    Since there is no clear pattern in the
g   i Pay rate = 8.56 + 0.289 × Experience          residual plot and the residual values
= 8.56 + 0.289 × 8 = 8.56 + 2.312            appear to be randomly arranged, we
= \$10.87                                     can say that our initial assumption that
ii Residual = actual – predicted                the relationship is linear is correct.
= 11.20 – 10.87 = \$0.33
h   Since there is no clear pattern in the
residual plot and the residual values
appear to be randomly arranged, we
can say that our initial assumption that
the relationship is linear is correct.
11                                              g    i Adult weight = 38.4 + 5.9 × Birth weight
a    Since birth weight is more likely to                = 38.4 + 5.9 × 3.0 = 38.4 + 17.7
determine adult weight than the other               = 56.1
way around, birth weight is more                ii Adult weight = 38.4 + 5.9 × Birth weight
likely the independent variable with                = 38.4 + 5.9 × 2.5 = 38.4 + 14.8
adult weight the dependent variable.                = 53.2
iii Adult weight = 38.4 + 5.9 × Birth weight
b    Using the Stat List Editor function of              = 38.4 + 5.9 × 3.9 = 38.4 + 23.0
the graphics calculator, entering in the            = 61.4
values given and plotting them in the
graph window of the calculator, we get     h    This contention is supported by the
the following scatterplot:                      data, since the data suggests that
76.5% of the variation in a person’s
adult weight is due to the variation in
their birth weight.

i    Plotting a Stat Plot on the graphics
calculator with the Y list as RESID
gives us the following residual plot:
c
i The relationship can be seen to
positive, linear and strongly so.
There are no clear outliers.
ii Since the relationship is a strong
one but is clearly not perfect, it
would be reasonable to predict a
value of r between 0.7 and 0.9.         12   A: negative – We can see that the
relationship is a negative one since the
d    Using the LinReg(a+bX) function, we             graph drifts downwards towards the
get:                                            right of the x axis.
a = 38.4                                        B: Drug dosage – This is the
b = 5.9                                         independent variable.
r2 = 0.7653                                     C: –0.9475 – As read off the LinReg
r = 0.8748                                      calculator screen picture (value r).
This translates to:                             D and E: 55.2 and –9.3 – As read off
Adult weight = a + b × Birth weight             the LinReg calculator screen picture
= 38.4 + 5.9 × Birth weight                   (values a and b).
F: decreases – Smaller values are
e    The r2 = coefficient of determination           obtained for response time for higher
value can be interpreted as the                 drug dosages.
proportion of the variation in a                G: 9.3 – As read off the LinReg
dependent variable that is due to the           calculator screen picture (value b).
independent variable.                           H: 55.2 – As read off the LinReg
Thus, since 0.7653 × 100% = 76.5%,              calculator screen picture (value a).
we can say that 76.5% of the variation          I: 89.8 – Since r2 = 0.898 and 0.898 ×
in a person’s adult weight is due to the        100% = 89.8%.
variation in their birth weight.                J and K: Response time and Drug
dosage – Drug dosage determines
f    The slope of 5.9 tells us that the              response time since drug dosage is the
average increase in adult weight for a          independent variable.
person per kg of birth weight is 5.9.           I: clear pattern – The residual graph
can be seen to show a parabolic–like
pattern.
13   Recognising that there is a strong,
positive linear relationship between the
two variables and that Femur length is
the independent variable, we can read
the values for a, b, r and r2 off the
middle LinReg calculator screen
picture to say:
‘The scatterplot shows that there is a
strong positive linear relationship
length, r = 0.9988. There are no
outliers. The equation of the least
squares regression line is:
Radial length = 9.0 + 1.6 × Femur
length..
The slope of the regression line
predicts that, on average, radial length
increased by 1.6 cm for each
centimetre increase in femur length.
The coefficient of determination
indicates that 99.8% of the variation in
radial lengths can be explained by the
variation in femur lengths. The
residual plot shows no clear pattern,
supporting the assumption that the
length is linear.’
Solutions to Exercise 5D
1   A least squares line does not work as       b   For the scatterplot, the points and three
well when clear outliers are present            median line below may be drawn:
compared to a three median line.
Hence, a three median line should be
used when there are clear outliers.

2
a   ii – The plots make a linear
relationship but have an outlier, so
only the three median line would be
appropriate.
The equation of the above line can be
b   iii – The relationship is non-linear, so
seen to be approximately y = –2 + x.
neither approach would be appropriate.
c   For the scatterplot, the points and three
c   i – The plots make a linear relationship
median line below may be drawn:
with no clear outliers, so either the
least squares or three median line
would be appropriate.

d   iii – The relationship is non-linear, so
neither approach would be appropriate.

3
a   For the scatterplot, the points and three
median line below may be drawn:

The equation of the above line can be
seen to be approximately y = 2 + x.

d   For the scatterplot, the points and three
median line below may be drawn:

The equation of the above line can be
seen to be approximately y = 1 + 0.7x.

The equation of the above line can be
seen to be approximately y = 14 – 0.9x.
4                                               5   For the scatterplot, the points and three
a   For the scatterplot, the points and three       median line below may be drawn:
median line below may be drawn:

The line can be seen to pass through
The line can be seen to pass through            the points (27,70) and (41,45).
the points (4,0) and (22,30).                   This gives us a gradient of 45 – 70
41 – 27
This gives us a gradient of 30 – 0
22 – 4               = –  25 = –1.8.
30 = 1.7.                                          14
=
16                                         Our equation is now y = a + 1.7x.
Our equation is now y = a + 1.7x.
Substituting in the point (27,70), we get:
Substituting in the point (4,0), we get:          70 = a – 1.8 × 27
0 = a + 1.7 × 4                                 70 = a – 48.6
0 = a + 6.8                                      a = 70 + 48.6 = 119
a = –6.8
Therefore, we can say that Life
Therefore, we can say that Percentage           expectancy = 119 – 1.8 × Birth rate.
recycled = –6.8 + 1.7 × Median
income.

b   The slope means that, on average, for
every increase in salary of \$1000, the
percentage of waster recycled by that
person increases by 1.7%.
Solutions to Exercise 5E
1
a   Predicting within the data range is
known as interpolation.

b   Predicting outside the data range is
known as extrapolation.

2
a   4.5 is within the data range, so it is
interpolation.

b   15 is outside the data range, so it is
extrapolation.

c   0 is outside the data range, so it is
extrapolation.

d   10 is within the data range, so it is
interpolation.

e   13 is outside the data range, so it is
extrapolation.
Solutions to Multiple–choice questions
1   For a least squares model to work, it is     10   The line passes through the points
assumed that the variables being                  (20,2) and (25,6). This gives us a
modelled are linearly related. ⇒ C                gradient of
6–2      4
= = 0.8
25 – 20 5
2   The constant value = –1.2 = y-intercept;          Our equation is now y = a + 0.8x.
the gradient value = 0.52 = the slope.            Substituting in the point (20,2), we get:
⇒D                                                  2 = a + 0.8 × 20
0 = a + 16
3   r = r 2 = 0.25 = 0.5                                a = –16
However, since the gradient = –9 and              Of the available responses, this is
thus is negative, r must also be                  closest to y = –14 + 0.8x. ⇒ A
negative. Thus, r = –0.5. ⇒ A
11   The intercept is not 96 as the constant
4   y = 8 – 9x = 8 – 9 × 5                            term in the equation is –96, meaning
= 8 – 45 = –37 ⇒ B                              the intercept is in fact –96. All the
other statements are true. ⇒ D
5   Using the LinReg(a+bX) function, we
get:                                         12   The gradient of 0.95 predicts that the
a = 24.4                                          dependent variable, weight, increases
b = –0.69                                         by 0.95 kg per 1 cm increase in the
r2 = 0.686                                        independent variable, Height. ⇒ E
r = –0.828
This can be expressed as y = 24.4 – 0.69x.   13   If r = 0.79, then r2 = 0.6241. The r2
⇒B                                                value can be interpreted as the
proportion of the variation in a
6   Using the LinReg(a+bX) function, we               dependent variable that is due to the
get:                                              independent variable.
a = 7.5                                           Since 0.6241 × 100% = 62%, we can
b = 0.5                                           say that 62% of the variation in weight
r2 = 0.5                                          is due to the variation in height. ⇒ A
r = 0.71
This can be expressed as y = 7.5 + 0.5x.     14   Weight = –96 + 0.95 × Height
⇒D                                                  = –96 + 0.95 × 179
= –96 + 170.05 = 74
7   Substituting the given values into the            Residual value = actual value –
equation, we get:                                 predicted value = 82 – 74 = 8 kg. ⇒ C
b = 0.733 × 3.391
1.871                               15   For the scatterplot, the points and three
= 1.33 ⇒ C                                      median line below may be drawn:

8   Residual value = actual value –
predicted value
–5.4 = actual value – 78.6
actual value = 78.6 – 5.4
= 73.2 ⇒ A

9   The y-intercept, and thus the constant
term in the equation, is clearly 8.7 and
not 0.9. The gradient is clearly
negative. It can be seen that the y-value
of the line decreases from 8.7 to 8.0 as
the x-value increases from 0 to 1. This           The line can be seen to pass through
is much closer to a gradient of –0.9              the points (21,5) and (26,7).
than –0.1. ⇒ A                                    This gives us a gradient of
7–5
26 – 21
2
= = 0.4 ⇒ B
5
Chapter 6 – Data transformation
Solutions to Exercise 6A
1
a
value      1     2         3        4         5        6        7
(value)2     1     4         9       16        25       36       49
log(value)    0   0.301     0.477    0.602     0.698    0.778    0.845
1/value     1    0.5      0.333    0.25       0.2     0.167    0.143

b   i A squared transformation stretches larger values.
ii A log transformation compresses larger values.
iii A reciprocal transformation compresses larger values.
2
a
value         1         2          4          8          16           32          64
(value)2        1         4         16         64          256         1024        4096
log(value)       0       0.301      0.602      0.903       1.204        1.505       1.806
1/value        1        0.5       0.25       0.125       0.063        0.031       0.016

b   i A squared transformation stretches larger values.
ii A log transformation compresses larger values.
iii A reciprocal transformation compresses larger values.
3
a
value         1   10           100        1000            10000             100000
(value)2        1   100         10000      1000000        100000000         10000000000
log(value)       0    1             2          3                4                 5
1/value        1   0.1          0.01       0.001           0.0001            0.00001

b   i A squared transformation stretches larger values.
ii A log transformation compresses larger values.
iii A reciprocal transformation compresses larger values.
4
a
value           20         10        5          2.5         1.25        0.625
(value)2         400        100       25         6.25         1.563       0.391
log(value)       1.301        1       0.699       0.398        0.097       –0.204
1/value        0.05        0.1       0.2         0.4          0.8         1.6

b   i A squared transformation stretches larger values.
ii A log transformation compresses larger values.
iii A reciprocal transformation compresses larger values.
5
a
value           2          20       200         2000          20000          200000
(value)2          4         400      40000      4000000       400000000     40000000000
log(value)       0.301      1.301     2.301       3.301          4.301           5.301
1/value         0.5       0.05      0.005       0.0005        0.00005        0.000005

b   i A squared transformation stretches larger values.
ii A log transformation compresses larger values.
iii A reciprocal transformation compresses larger values.
Solutions to Exercise 6B
1                                                 b   Creating an x2 list in the List Editor,
a   Using the List Editor and Statplot                the Statplot function can be used to
functions of the graphics calculator,             plot the following graph of y vs x2:
keying in the x and y values in the
table and plotting the graph of y vs x
returns the following graph:

The relationship can be seen to be in a
perfectly straight line and is thus linear.

The relationship can be seen to not be        3
in a perfectly straight line and is thus      a   Using the List Editor and Statplot
non-linear.                                       functions of the graphics calculator,
keying in the x and y values in the
b   Creating an x2 list in the List Editor,           table and plotting the graph of y vs x
the Statplot function can be used to              returns the following graph:
plot the following graph of y vs x2:

The relationship can be seen to not be
The relationship can be seen to be in a           in a perfectly straight line and is thus
perfectly straight line and is thus linear.       non-linear.

2                                                 b   Creating an x2 list in the List Editor,
a   Using the List Editor and Statplot                the Statplot function can be used to
functions of the graphics calculator,             plot the following graph of y vs x2:
keying in the x and y values in the
table and plotting the graph of y vs x
returns the following graph:

The relationship can be seen to be in a
perfectly straight line and is thus linear.

The relationship can be seen to not be
in a perfectly straight line and is thus
non-linear.
4                                                 b   Creating a log(x) list in the List
a   Using the List Editor and Statplot                Editor, the Statplot function can be
functions of the graphics calculator,             used to plot the following graph of y vs
keying in the x and y values in the               log(x):
table and plotting the graph of y vs x
returns the following graph:

The relationship can be seen to be in a
perfectly straight line and is thus linear.
The relationship can be seen to not be
in a perfectly straight line and is thus
6
non-linear.
a   Using the List Editor and Statplot
functions of the graphics calculator,
b   Creating a log(x) list in the List
keying in the x and y values in the
Editor, the Statplot function can be
table and plotting the graph of y vs x
used to plot the following graph of y vs
returns the following graph:
log(x):

The relationship can be seen to be in a           The relationship can be seen to not be
perfectly straight line and is thus linear.       in a perfectly straight line and is thus
non-linear.
5
a   Using the List Editor and Statplot            b   Creating a log(x) list in the List
functions of the graphics calculator,             Editor, the Statplot function can be
keying in the x and y values in the               used to plot the following graph of y vs
table and plotting the graph of y vs x            log(x):
returns the following graph:

The relationship can be seen to be in a
perfectly straight line and is thus linear.
The relationship can be seen to not be
in a perfectly straight line and is thus
non-linear.
7                                                 b   Creating a 1/x list in the List Editor,
a   Using the List Editor and Statplot                the Statplot function can be used to
functions of the graphics calculator,             plot the following graph of y vs 1/x:
keying in the x and y values in the
table and plotting the graph of y vs x
returns the following graph:

The relationship can be seen to be in a
perfectly straight line and is thus linear.
The relationship can be seen to not be        9
in a perfectly straight line and is thus      a   Using the List Editor and Statplot
non-linear.                                       functions of the graphics calculator,
keying in the x and y values in the
b   Creating a 1/x list in the List Editor,           table and plotting the graph of y vs x
the Statplot function can be used to              returns the following graph:
plot the following graph of y vs 1/x:

The relationship can be seen to not be
The relationship can be seen to be in a           in a perfectly straight line and is thus
perfectly straight line and is thus linear.       non-linear.
b   Creating a 1/x list in the List Editor,
8
the Statplot function can be used to
a   Using the List Editor and Statplot
plot the following graph of y vs 1/x:
functions of the graphics calculator,
keying in the x and y values in the
table and plotting the graph of y vs x
returns the following graph:

The relationship can be seen to be in a
perfectly straight line and is thus linear.
c   As explained on page 157 of Essential
Further Mathematics, log(x) and 1/x
The relationship can be seen to not be            transformations can be used for the
in a perfectly straight line and is thus          same shape of graph, so a log(x)
non-linear.                                       transformation should also work for
this data.
d   As explained on page 157 of Essential
Further Mathematics, x2 and 1/x
transformations cannot be used for the
same shapes of graphs, so an x2
transformation should not work for this
data.
10                                                 b    Creating a log(x) list in the List
a    Using the List Editor and Statplot                 Editor, the Statplot function can be
functions of the graphics calculator,              used to plot the following graph of y vs
keying in the x and y values in the                log(x):
table and plotting the graph of y vs x
returns the following graph:

The relationship can be seen to not be
in a perfectly straight line and is thus
The relationship can be seen to not be             non-linear.
in a perfectly straight line and is thus
non-linear.                                   12
a    Using the List Editor and Statplot
b    Creating an x2 list in the List Editor,            functions of the graphics calculator,
the Statplot function can be used to               keying in the x and y values in the
plot the following graph of y vs x2:               table and plotting the graph of y vs x
returns the following graph:

The relationship can be seen to be in a
perfectly straight line and is thus linear.        The relationship can be seen to not be
in a perfectly straight line and is thus
11                                                      non-linear.
a    Using the List Editor and Statplot
functions of the graphics calculator,         b    Creating a 1/x list in the List Editor,
keying in the x and y values in the                the Statplot function can be used to
table and plotting the graph of y vs x             plot the following graph of y vs 1/x:
returns the following graph:

The relationship can be seen to not be             The relationship can be seen to not be
in a perfectly straight line and is thus           in a perfectly straight line and is thus
non-linear.                                        non-linear.
Solutions to Exercise 6C
1                                                 b   Creating a y2 list in the List Editor,
a   Using the List Editor and Statplot                the Statplot function can be used to
functions of the graphics calculator,             plot the following graph of y2 vs x:
keying in the x and y values in the
table and plotting the graph of y vs x
returns the following graph:

The relationship can be seen to be in a
perfectly straight line and is thus linear.

The relationship can be seen to not be        3
in a perfectly straight line and is thus      a   Using the List Editor and Statplot
non-linear.                                       functions of the graphics calculator,
keying in the x and y values in the
b   Creating a y2 list in the List Editor,            table and plotting the graph of y vs x
the Statplot function can be used to              returns the following graph:
plot the following graph of y2 vs x:

The relationship can be seen to not be
The relationship can be seen to be in a           in a perfectly straight line and is thus
perfectly straight line and is thus linear.       non-linear.

2                                                 b   Creating a y2 list in the List Editor,
a   Using the List Editor and Statplot                the Statplot function can be used to
functions of the graphics calculator,             plot the following graph of y2 vs x:
keying in the x and y values in the
table and plotting the graph of y vs x
returns the following graph:

The relationship can be seen to be in a
perfectly straight line and is thus linear.

The relationship can be seen to not be
in a perfectly straight line and is thus
non-linear.
4                                                 b   Creating a log(y) list in the List
a   Using the List Editor and Statplot                Editor, the Statplot function can be
functions of the graphics calculator,             used to plot the following graph of
keying in the x and y values in the               log(y) vs x:
table and plotting the graph of y vs x
returns the following graph.

The relationship can be seen to be in a
perfectly straight line and is thus linear.
The relationship can be seen to not be
in a perfectly straight line and is thus      6
non-linear.                                   a   Using the List Editor and Statplot
functions of the graphics calculator,
b   Creating a log(y) list in the List                keying in the x and y values in the
Editor, the Statplot function can be              table and plotting the graph of y vs x
used to plot the following graph of               returns the following graph:
log(y) vs x:

The relationship can be seen to not be
The relationship can be seen to be in a           in a perfectly straight line and is thus
perfectly straight line and is thus linear.       non-linear.

5                                                 b   Creating a log(y) list in the List
a   Using the List Editor and Statplot                Editor, the Statplot function can be
functions of the graphics calculator,             used to plot the following graph of
keying in the x and y values in the               log(y) vs x:
table and plotting the graph of y vs x
returns the following graph:

The relationship can be seen to be in a
perfectly straight line and is thus linear.
The relationship can be seen to not be
in a perfectly straight line and is thus
non-linear.
7                                                 b   Creating a 1/y list in the List Editor,
a   Using the List Editor and Statplot                the Statplot function can be used to
functions of the graphics calculator,             plot the following graph of 1/y vs x:
keying in the x and y values in the
table and plotting the graph of y vs x
returns the following graph:

The relationship can be seen to be in a
perfectly straight line and is thus linear.
The relationship can be seen to not be        9
in a perfectly straight line and is thus      a   Using the List Editor and Statplot
non-linear.                                       functions of the graphics calculator,
keying in the x and y values in the
b   Creating a 1/y list in the List Editor,           table and plotting the graph of y vs x
the Statplot function can be used to              returns the following graph:
plot the following graph of 1/y vs x:

The relationship can be seen to be in a           The relationship can be seen to not be
perfectly straight line and is thus linear.       in a perfectly straight line and is thus
non-linear.
8                                                 b   Creating a 1/y list in the List Editor,
a   Using the List Editor and Statplot                the Statplot function can be used to
functions of the graphics calculator,             plot the following graph of 1/y vs x:
keying in the x and y values in the
table and plotting the graph of y vs x
returns the following graph:

The relationship can be seen to be in a
perfectly straight line and is thus linear.
The relationship can be seen to not be        c   As explained on page 163 of Essential
in a perfectly straight line and is thus          Further Mathematics, log(y) and 1/y
non-linear.                                       transformations can be used for the
same shape of graph, so a log(y)
transformation should also work for
this data.
d   As explained on page 163 of Essential
Further Mathematics, y2 and 1/y
transformations cannot be used for the
same shapes of graphs, so a y2
transformation should not work for this
data.
10                                                 b    Creating a log(y) list in the List
a    Using the List Editor and Statplot                 Editor, the Statplot function can be
functions of the graphics calculator,              used to plot the following graph of
keying in the x and y values in the                log(y) vs x:
table and plotting the graph of y vs x
returns the following graph:

The relationship can be seen to not be
in a perfectly straight line and is thus
non-linear.
The relationship can be seen to not be
in a perfectly straight line and is thus
12
non-linear.
a    Using the List Editor and Statplot
functions of the graphics calculator,
b    Creating a y2 list in the List Editor,
keying in the x and y values in the
the Statplot function can be used to
table and plotting the graph of y vs x
plot the following graph of y2 vs x:
returns the following graph:

The relationship can be seen to be in a            The relationship can be seen to not be
perfectly straight line and is thus linear.        in a perfectly straight line and is thus
non-linear.
11
a    Using the List Editor and Statplot            b    Creating a y2 list in the List Editor,
functions of the graphics calculator,              the Statplot function can be used to
keying in the x and y values in the                plot the following graph of y2 vs x:
table and plotting the graph of y vs x
returns the following graph:

The relationship can be seen to not be
The relationship can be seen to not be             in a perfectly straight line and is thus
in a perfectly straight line and is thus           non-linear.
non-linear.
Solutions to Exercise 6D
1                                              b   Creating a log(y) list in the List
a   As explained on page 166 of Essential          Editor, the Statplot function can be
Further Mathematics, log(x), 1/x,              used to plot the following graph of
log(y) and 1/y transformations can be          log(y) vs x:
used for plots where the points form a
line that mirrors the bottom left

b   Since the trend in the plots is both
decreasing and then increasing, this is
incompatible with any of the                   Using the LinReg(a+bX) function, we
transformations described so far, and          get a = 1.86, b = 0.048, r2 = 0.947,
so no transformation could be applied.         r = 0.973.
Plotting a residual plot shows reduced
c   As explained on page 166 of Essential          curvature as shown by the more
Further Mathematics, log(y), 1/y and           random residual plot below:
x2 transformations can be used for
plots where the points form a line that
mirrors the bottom right quadrant of a
circle.

d   As explained on page 166 of Essential
Further Mathematics, y2 and x2                 Creating a 1/y list in the List Editor,
transformations can be used for plots          the Statplot function can be used to
where the points form a line that              plot the following graph of 1/y vs x:
mirrors the upper right quadrant of a
circle.

2
a   Using the List Editor and Statplot
functions of the graphics calculator,
keying in the x and y values in the
table and plotting the graph of y vs x         Using the LinReg(a+bX) function, we
returns the following graph:                   get a = 0.008, b = –0.0003, r2 = 0.640,
r = –0.80.
Plotting a residual plot shows
increased curvature as shown by the
less random residual plot below:

The relationship can be seen to not be
in a perfectly straight line and is thus
non-linear.
Using the LinReg(a+bX) function, we
get: a = –619.8, b = 80.29, r2 = 0.933,
r = 0.966.
We know that log(y), 1/y and x2
transformations can be used for plots
where the points form a line that
mirrors the bottom right quadrant of a
circle.
Creating an x2 list in the List Editor,    b   Creating a log(y) list in the List
the Statplot function can be used to           Editor, the Statplot function can be
plot the following graph of y vs x2:           used to plot the following graph of
log(y) vs x:

Using the LinReg(a+bX) function, we
get a = 5.366, b = 2.029, r2 = 0.975,          Using the LinReg(a+bX) function, we
r = 0.987.                                     get a = 1.13, b = –0.059, r2 = 0.991,
Plotting a residual plot shows reduced         r = 0.996.
curvature as shown by the more                 Plotting a residual plot shows reduced
random residual plot below:                    curvature as shown by the more
random residual plot below:

We can thus see that the best
regression model would be the x2               Creating a 1/y list in the List Editor,
transformation.                                the Statplot function can be used to
Using the x2 model above, we can say           plot the following graph of 1/y vs x:
the best regression model is given by:
Yield = 5.366 + 2.029 × (Length)2.

3
a   Using the List Editor and Statplot
functions of the graphics calculator,
keying in the x and y values in the            Using the LinReg(a+bX) function, we
table and plotting the graph of y vs x         get a = 0.037, b = 0.022, r2 = 0.998,
returns the following graph:                   r = 0.998.
Plotting a residual plot shows
increased curvature as shown by the
less random residual plot below:

The relationship can be seen to not be
in a perfectly straight line and is thus
non-linear.
Creating a log(x) list in the List
Using the LinReg(a+bX) function, we            Editor, the Statplot function can be
get a = 11.27, b = –0.867, r2 = 0.977,         used to plot the following graph of y vs
r = –0.988.                                    log(x):

We know that log(y), 1/y, log(x) and
1/x transformations can be used for
plots where the points form a line that
mirrors the bottom left quadrant of a
circle.
Using the LinReg(a+bX) function, we           The relationship can be seen to not be
get a = 14.69, b = –11.21, r2 = 0.995,        in a perfectly straight line and is thus
r = 0.996.                                    non–linear.
Plotting a residual plot shows reduced        Using the LinReg(a+bX) function, we
curvature as shown by the more                get a = 567772.96, b = 589.21,
random residual plot below:                   r2 = 0.796, r = 0.892.
We know that log(y), 1/y and x2
transformations can be used for plots
where the points form a line that
mirrors the bottom right quadrant of a
circle.
b   Creating a log(y) list in the List
Creating a 1/x list in the List Editor,       Editor, the Statplot function can be
the Statplot function can be used to          used to plot the following graph of
plot the following graph of y vs 1/x:         log(y) vs x:

Using the LinReg(a+bX) function, we           Using the LinReg(a+bX) function, we
get a = 1.539, b = 26.15, r2 = 0.996,         get a = 4.755, b = 0.004, r2 = 0.808,
r = 0.998.                                    r = 0.899.
Plotting a residual plot shows                Plotting a residual plot shows reduced
increased curvature as shown by the           curvature as shown by the more
less random residual plot below:              random residual plot below:

We can thus see that the best                 Creating a 1/y list in the List Editor,
regression model would be the 1/y             the Statplot function can be used to
transformation.                               plot the following graph of 1/y vs x:
Using the 1/y model above, we can say
the best regression model is given by:
1/Cigarettes = 0.037 + 0.022 × Cost

4
a   Using the List Editor and Statplot
functions of the graphics calculator,
keying in the x and y values in the           Using the LinReg(a+bX) function, we
table and plotting the graph of y vs x        get a = 1.75×10–5, b = –1.55×10–7,
returns the following graph:                  r2 = 0.819, r = –0905.
Plotting a residual plot shows reduced         Using the LinReg(a+bX) function, we
curvature as shown by the more                 get a = 0.795, b = –0.006, r2 = 0.793,
random residual plot below:                    r = 0.891.
We know that y2 and x2 transformations
can be used for plots where the points
form a line that mirrors the top right

b   Creating a y2 list in the List Editor,
Creating an x2 list in the List Editor,        the Statplot function can be used to
the Statplot function can be used to           plot the following graph of y2 vs x:
plot the following graph of y vs x2:

Using the LinReg(a+bX) function, we
get a = 0.630, b = –0.008, r2 = 0.798,
r = 0.893.
Using the LinReg(a+bX) function, we            Plotting a residual plot shows reduced
get a = 58178.2, b = 43.163,                   curvature as shown by the more
r2 = 0.884, r = 0.940.                         random residual plot below:
Plotting a residual plot shows reduced
curvature as shown by the more
random residual plot below:

Creating an x2 list in the List Editor,
the Statplot function can be used to
plot the following graph of y vs x2:
We can thus see that the best
regression model would be the x2
transformation.
Using the x2 model above we can say
the best regression model is given by:
Population = 58178 + 43.163 × (Year)2          Using the LinReg(a+bX) function, we
get a = 0.776, b = –3.004×10–4,
5                                                  r2 = 0.837, r = –0.915.
a   Using the List Editor and Statplot             Plotting a residual plot shows reduced
functions of the graphics calculator,          curvature as shown by the more
keying in the x and y values in the            random residual plot below:
table and plotting the graph of y vs. x
returns the following graph:

We can thus see that the best
The relationship can be seen to not be         regression model would be the x2
in a perfectly straight line and is thus       transformation.
non-linear.                                    Using the x2 model above we can say
the best regression model is given by:
Rate = 0.776 – 0.0003 × (Month)2
6                                                Creating a log(x) list in the List
a   Using the List Editor and Statplot           Editor, the Statplot function can be
functions of the graphics calculator,        used to plot the following graph of y vs
keying in the x and y values in the          log(x):
table and plotting the graph of y vs.x
returns the following graph:

Using the LinReg(a+bX) function, we
get a = –44.202, b = 33.321,
The relationship can be seen to not be       r2 = 0.876, r = 0.936.
in a perfectly straight line and is thus     Plotting a residual plot shows reduced
non-linear.                                  curvature as shown by the more
random residual plot below:
Using the LinReg(a+bX) function, we
get a = 49.79, b = 0.0026, r2 = 0.661,
r = 0.813.
We know that y2, log(x) and 1/x
transformations can be used for plots
where the points form a line that
mirrors the top left quadrant of a circle.
Creating a 1/x list in the List Editor,
b   Creating a y2 list in the List Editor,       the Statplot function can be used to
the Statplot function can be used to         plot the following graph of y vs 1/x:
plot the following graph of y2 vs x:

Using the LinReg(a+bX) function, we
Using the LinReg(a+bX) function, we
get a = 84.91, b = –9917.8, r2 = 0.644,
get a = 2832.7, b = 0.357, r2 = 0.721,
r = 0.802. 3
r = 0.849.
Plotting a residual plot shows
Plotting a residual plot shows reduced
increased curvature as shown by the
curvature as shown by the more
less random residual plot below:
random residual plot below:

We can thus see that the best
regression model would be the log(x)
transformation.
Using the log(x) model above we can say
the best regression model is given by:
Literacy Rate = –44.202 + 33.321 ×
log(GDP).
Solutions to Multiple–choice questions
1   Since 12 = 1, a = 1;                        6    We know that for a graph that
since log(2) = 0.301, b = 0.301. ⇒ C             resembles the top right quadrant of a
circle, we can use a y2 or x2
2   A log transformation has a                       transformation to linearise the points.
compressing effect upon the high                 The only one of these amongst the
values in a data distribution. ⇒ D               answers is y2. ⇒ B

3   For a residual plot, the x-axis will        7    In order to linearise a set of points
represent the same thing as in the               using the transformations learnt so far,
original graph, ‘Number of weeks on a            the trend must be consistently
diet’ in this case.                              increasing or decreasing.
The y-axis will represent residual.              Since this trend increases but then
The regression line will be represented          decreases, none of the transformations
as a perfectly horizontal line of y-value        learnt so far can be used. ⇒ E
0.00.
If a point on the original graph is         8    The best transformation to use is the
above the regression line, it will be            one that returns the most linear graph
above the regression line on the                 with the most random residual plot.
residual graph, and if below on the              A graph is most linear when its r2
original graph, will be below on the             value = 100% and least linear when the
residual graph.                                  r2 value = 0%.
Taking ‘Number of weeks on a diet’               Thus, the best transformation will be
= 2, we can see there to be one point            the one with the highest r2 value,
above the regression line, and two               which is the y2 vs x graph. ⇒ B
below.
Thus, in the residual graph at ‘Number      9    For a x2 transformation, we can write
of weeks on a diet’ = 2, we should also          the general equation y = a + b × x2.
see one point above the line and two             y is the dependent variable = Weight,
below. ⇒ E                                       a is the y-intercept = 10,
b is the slope of the regression line = 5,
4   We know that for a graph that                    x is the independent variable = Width.
resembles the top left quadrant of a             Substituting all these values in gives us
circle, we can use a y2, log(x) or 1/x           Weight = 10 + 5 × (Width)2. ⇒ C
transformation to linearise the points.
The only one of these amongst the           10   Mark = 20 + 40 × log(Hours)
answers is y2. ⇒ B                                    = 20 + 40 × log(20)
= 20 + 40 × 1.301
5   We know that for a graph that                         = 20 + 52.04
resembles the bottom left quadrant of a               = 72.04 = 72 ⇒ D
circle, we can use a log(y), 1/y, log(x)
or 1/x transformation to linearise the
points. ⇒ A
Chapter 7 – Time Series
Solutions to Exercise 7A
1   Graph A: There does not appear to be       3   There does not appear to be a way to
a way to predict values at any point on        predict values at any point on the
the graph, meaning it exhibits random          graph, meaning it exhibits random
variation. The trend line of the graph         variation. Different sections of the
would show a generally decreasing              graph show different trends, with the
trend.                                         trend being increasing from 1920 to
1930, flat until 1940, decreasing until
Graph B: There does not appear to be           1945, increasing until 1960 and then
a way to predict values at any point on        decreasing until 1985. There is no
the graph, meaning it exhibits random          cyclic or seasonal variation evident.
variation. There appears to be
seasonal variation shown by the                Considering the graph, along with a
graph, since the troughs seem to               knowledge of history, we can say:
always occur halfway or three-quarters         ‘The number of whales caught
of the way through a year.                     increased rapidly between 1920 and
1930 but levelled off during the 1930s,
Graph C: There does not appear to be           which was the time of the Depression.
a way to predict values at any point on        In the period 1940–1945, which was
the graph, meaning it exhibits random          the time of the Second World War,
variation. There appears to be cyclical        there was a rapid decrease in the
variation shown by the graph, since            number of whales caught and numbers
the troughs are regularly spaced but are       fell to below the 1920 catch. In the
much greater than 1 year apart.                period 1945–1965 the numbers
increased again but then fell again until
2   Graph A: There does not appear to be           1985 when numbers were back to
a way to predict values at any point on        around the 1920 level. (This was a
the graph, meaning it exhibits random          time when people were becoming
variation. The trend line of the graph         more environmentally aware and
would show a generally decreasing              realised that whales were becoming an
trend. There appears to be cyclical            endangered species.)’
variation shown by the graph, since
the troughs are regularly spaced but are   4   There does not appear to be a way to
much greater than 1 year apart.                predict values at any point on the
graph, meaning it exhibits random
Graph B: There does not appear to be           variation. There is no clear increasing
a way to predict values at any point on        or decreasing trend, but instead the
the graph, meaning it exhibits random          graph oscillates. The fact that peaks
variation. There appears to be                 appear to occur every March with
seasonal variation shown by the                troughs every June, means the graph
graph, since the troughs are regularly         exhibits seasonal variation.
spaced and just over 1 year apart.

Graph C: There does not appear to be
a way to predict values at any point on
the graph, meaning it exhibits random
variation. The trend line of the graph
would show a generally increasing
trend.
5                                              8   Plotting the points with Year as the x-
a   Noting the flat and decreasing trends in       axis and Teachers as the y-axis, and
the two graphs, we can say:                    drawing in a trend line, we get the plot
‘The percentage of males who smoke             below:
has consistently decreased since 1945,
while the percentage of females who
smoke increased from 1945 to 1975
but then decreased at a similar rate to
males over the period 1975–1992.’

b   Since the two graphs have become
closer and closer together between
1945 and 1992, we can say that the
difference in smoking rates has
decreased over that period.
9
a   Plotting the points with Year as the
6   Plotting the points with Year as the x-
x-axis and Teachers as the y-axis, and
axis and Rate as the y-axis, and
drawing in a trend line, we get the plot
drawing in a trend line, we get the plot
below:
below:

7   Plotting the points with Year as the x-    b   Noting the flat trend of the red line but
axis and Population as the y-axis, and         the increasing trend of the blue line,
drawing in a trend line, we get the plot       we can say:
below:                                         ‘The number of male school teachers
has remained relatively constant over
the years 1991–2001, whereas the
number of female school teachers has
increased over this time.’
Solutions to Exercise 7B
1
a   This is the 3-mean of the y values for t = 3, 4, 5, which are 5, 3, 1 respectively.
Since 5 + 3 + 1 = 3, the 3-mean for t = 4 is 3.
3
b   This is the 3-mean of the y values for t = 5, 6, 7, which are 1, 0, 2 respectively.
1 + 0 + 2
Since              = 1, the 3-mean for t = 6 is 1.
3
c   This is the 3-mean of the y values for t = 1, 2, 3, which are 5, 2, 5 respectively.
Since 5 + 2 + 5 = 4, the 3-mean for t = 2 is 4.
3
d   This is the 5-mean of the y values for t = 3, 4, 5, 6, 7 which are 5, 3, 1, 0, 2 respectively.
5 + 3 + 1 + 0 + 2
Since                         = 2.2, the 5-mean for t = 3 is 2.2.
5
e   This is the 5-mean of the y values for t = 5, 6, 7, 8, 9 which are 1, 0, 2, 3, 0 respectively.
1 + 0 + 2 + 3 + 0
Since                          = 1.2, the 5-mean for t = 7 is 1.2.
5
f   This is the 5-mean of the y values for t = 2, 3, 4, 5, 6 which are 2, 5, 3, 1, 0 respectively.
Since 2 + 5 + 3 + 1 + 0 = 2.2, the 5-mean for t = 4 is 2.2.
5
2 + 5                                    5 + 3
g   The 2-mean of t = 2, 3 is       = 3.5. The 2-mean of t = 3, 4 is       = 4.
2                                       2
3.5 + 4
The centred 2-mean of the two 2-means is          = 3.75.
2
2 + 5                                    5 + 3
h   The 2-mean of t = 2, 3 is       = 3.5. The 2-mean of t = 3, 4 is       = 4.
2                                       2
3.5 + 4
The centred 2-mean of the two 2-means is          = 3.75.
2
5 + 2 + 5 + 3
i   The 4-mean of t = 1, 2, 3, 4 is               = 3.75.
4
2 + 5 + 3 + 1
The 4-mean of t = 2, 3, 4, 5 is               = 2.75.
4
3.75 + 2.75
The centred 2-mean of the two 4-means is              = 3.25.
2
5 + 1 + 0 + 2
j   The 4-mean of t = 4, 5, 6, 7 is               = 2.
4
The 4-mean of t = 5, 6, 7, 8 is 1 + 0 + 2 + 3 = 1.5.
4
2 + 1.5
The centred 2-mean of the two 4-means is         = 1.75.
2
2   For the 3-mean of a particular t-value, add the y-value of that t-value, the y-value of the t-
value one to the left and the y-value of the t-value one to the right. Divide the sum by 3 to
get the 3-mean for that t-value.

For the 5-mean of a particular t-value, add the y-value of that t-value, the two y-values of the
two t-values immediately to the left and the two y-values of the two t-values immediately to
the right. Divide the sum by 5 to get the 5-mean for that t-value.

t          1         2          3            4            5            6            7       8       9
y          10        12         8            4           12            8            10      18      2
3-mean         –        10         8            8            8            10           12      10      –
5-mean         –        –         9.2          8.8          8.4          10.4          10       –      –
3
a   Plotting the values on a graph with Day as the x-axis and Temperature as the y-axis gives us
the following graph:

b   For the 3-mean of a particular Day-value, add the Temperature-value of that Day-value, the
Temperature-value of the Day-value one to the left and the Temperature-value of the Day-
value one to the right. Divide the sum by 3 to get the 3-mean for that Day-value.

For the 5-mean of a particular Day-value, add the Temperature-value of that Day-value, the
two Temperature-values of the two Day-values immediately to the left and the two
Temperature-values of the two Day-values immediately to the right. Divide the sum by 5 to
get the 5-mean for that Day-value.

Day        1         2        3         4            5            6             7       8       9     10
Temp        24        27       28        40           22           23            22      21      25    26
3-mean        -       26.3     31.7      30.0         28.3         22.3          22.0    22.7    24.0    -
5-mean        -         -      28.2      28.0         27.0         25.6          22.6    23.4      -     -

c     Noting that the smoothed graphs show a much smaller temperature variation from day to
day, we can say:
‘The smoothed plot show that the ‘average’ maximum temperature changes relatively
slowly over the 10-day period (the 5-day average varies by only 5 degrees) when
compared to the daily maximum, which can vary quite widely (for example, nearly 20
degrees between the 4th and 5th day) over the same period of time.’
4
a   Plotting the values on a graph with Day as the x-axis and Exchange rate as the y-axis gives
us the following graph:

b   For the 3-mean of a particular Day-value, add the Exchange rate-value of that Day-value,
the Exchange rate-value of the Day-value one to the left and the Exchange rate-value of the
Day-value one to the right. Divide the sum by 3 to get the 3-mean for that Day-value.

For the 5-mean of a particular Day-value, add the Exchange rate-value of that Day-value,
the two Exchange rate-values of the two Day-values immediately to the left and the two
Exchange rate-values of the two Day-values immediately to the right. Divide the sum by 5
to get the 5-mean for that Day-value.
Day         1        2        3        4       5        6        7        8        9      10
Ex rate     0.743    0.754    0.737    0.751   0.724    0.724    0.712    0.735    0.716   0.711
3-mean        -      0.745    0.747    0.737   0.733    0.720    0.724    0.721    0.721     -
5-mean        -        -      0.742    0.738   0.730    0.729    0.722    0.720      -       -

c   Noting that the downwards trend is more obvious in the smoothed graphs than the original
plot, we can say:
‘The exchange rate is dropping steadily over the 10-day period. This is most obvious from
the smoothed plots, particularly the 5-moving mean plot.’
5   Calculating the 2-moving means by taking the mean of each pair of two adjacent months and
calculating the centred means by taking the 2-moving means of the original 2-moving means
gives us the following table:
Month                 Number of births       2-moving mean       Centred means
January                    10                                          -

10 + 12 = 11
2
February                     12                                     11 + 9 = 10
2
12 + 6 = 9
2
March                       6                                    9 + 5.5 = 7.25
2
6 + 5 = 5.5
2
April                       5                                    5.5 + 13.5 = 9.5
2
5 + 22 = 13.5
2
May                        22                                  13.5 + 20 = 16.75
2
22 + 18 = 20
2
June                       18                                  20 + 15.5 = 17.75
2
18 + 13 = 15.5
2
July                       13                                  15.5 + 10 = 12.75
2
13 + 7 = 10
2
August                      7                                      10 + 8 = 9
2
7+9=8
2
September                     9                                    8 + 9.5 = 8.75
2
9 + 10 = 9.5
2
October                      10                                   9.5 + 9 = 9.25
2
10 + 8 = 9
2
November                      8                                   9 + 11.5 = 10.25
2
8 + 15 = 11.5
2
December                      15                                          -
6   Calculating the 2-moving means by taking the mean of each pair of two adjacent months and
calculating the centred means by taking the 2-moving means of the original 2-moving means
gives us the following table:
Month                Number of births      2-moving mean         Centred means
April                     21

21 + 40 = 30.5
2
May                        40                                  30.5 + 46 = 38.25
2
40 + 52 = 46
2
June                       52                                   46 + 47 = 46.5
2
52 + 42 = 47
2
July                      42                                   47 + 50 = 48.5
2
42 + 58 = 50
2
August                      58                                  50 + 68.5 = 59.25
2
58 + 79 = 68.5
2
September                    79                                  68.5 + 80 = 74.25
2
79 + 81 = 80
2
October                     81                                  80 + 67.5 = 73.75
2
81 + 54 = 67.5
2
November                     54                                  67.5 + 52 = 59.75
2
54 + 50 = 52
2
December                     50
Solutions to Exercise 7C
1
a   The y-values for t = 3, 4, 5 are y = 5, 3, 1 respectively.
The median of these and thus the 3-median smoothed y value for t = 4 is 3.

b   The y-values for t = 5, 6, 7 are y = 1, 0, 2 respectively.
The median of these and thus the 3-median smoothed y value for t = 6 is 1.

c   The y-values for t = 1, 2, 3 are y = 5, 2, 5 respectively.
The median of these and thus the 3-median smoothed y value for t = 2 is 5.

d   The y-values for t = 1, 2, 3, 4, 5 are y = 5, 2, 5, 3, 1 respectively.
The median of these and thus the 5-median smoothed y value for t = 3 is 3.

e   The y-values for t = 5, 6, 7, 8, 9 are y = 1, 0, 2, 3, 0 respectively.
The median of these and thus the 5-median smoothed y value for t = 7 is 1.

f   The y-values for t = 2, 3, 4, 5, 6 are y = 2, 5, 3, 1, 0 respectively.
The median of these and thus the 5-median smoothed y value for t = 4 is 2.

g   The median of the y-values for t = 2, 3, which are y = 2, 5 respectively, is 3.5.
The median of the y-values for t = 3, 4, which are y = 5, 3 respectively, is 4.
The median of 3.5 and 4 is 3.75.

h   The median of the y-values for t = 7, 8, which are y = 2, 3 respectively, is 2.5.
The median of the y-values for t = 8, 9, which are y = 3, 0 respectively, is 1.5.
The median of 2.5 and 1.5 is 2.

i   The median of the y-values for t = 1, 2, 3, 4, which are y = 5, 2, 5, 3 respectively, is 4.
The median of the y-values for t = 2, 3, 4, 5, which are y = 2, 5, 3, 1 respectively, is 2.5.
The median of 4 and 2.5 is 3.25.

j   The median of the y-values for t = 4, 5, 6, 7, which are y = 3, 1, 0, 2 respectively, is 1.5.
The median of the y-values for t = 5, 6, 7, 8, which are y = 1, 0, 2, 3 respectively, is 1.5.
The median of 1.5 and 1.5 is 1.5.

2   For the 3-median of a particular t-value, find the median of the y-values of that t-value and
the t-values immediately either side of that t-value.

For the 5-mean of a particular t-value, find the median of the y-values of that t-value and the
two t-values to the left and the two t-values to the right of that t-value.
t              1         2        3          4          5         6         7          8    9
y             10        12        8          4         12         8        10         18    2
3-median          -        10        8          8          8        10        10         10    -
5-median          -         -        10         8         8         10        10          -    -
3
a   For the 3-median of a particular t-value, find the median of the y-values of that t-value and
the t-values immediately either side of that t-value.

For the 5-median of a particular t-value, find the median of the y-values of that t-value and
the two t-values to the left and the two t-values to the right of that t-value.
Day          1        2        3        4         5         6        7        8        9      10
Temp          24       27       28       40       22        23       22       21       25      26
3-median         -       27       28       28       23        22       22       22       25       -
5-median         -        -       27       27       23        22       22       23        -       -

b   Plotting the points of the 3-moving median and joining them together gives us the following
graph:

From the plot we can see that temperature varies over a small and consistent range over the
1-day time period.

4
a   For the 3-median of a particular t-value, find the median of the y-values of that t-value and
the t-values immediately either side of that t-value.

For the 5-mean of a particular t-value, find the median of the y-values of that t-value and the
two t-values to the left and the two t-values to the right of that t-value.

Day           1        2        3         4        5        6        7         8        9      10
Ex rate      0.743    0.754    0.737     0.751    0.724    0.724    0.712     0.735    0.716   0.711
3-median        -      0.743    0.751     0.737    0.724    0.724    0.724     0.716    0.716     -
5-median        -        -      0.743     0.737    0.724    0.724    0.724     0.716      -       -

b   Plotting the points of the 3-moving median and joining them together gives us the following
graph:

From the plot we can see that although exchange rate rises slightly in the first day, it goes on
to drop steadily for the rest of the time period.
5
a   Calculating the 3-median and 5-median smoothing plots and superimposing them over the
original plot gives us the graph below:

Note that the median smoothed graphs have no extreme values. All smoothed y-values are
the same as a y-value on the original plot, though at a particular x-value the 3-median or 5-
median plot doesn’t necessarily have the same y-value as the original plot does at that x-
value.

b   Calculating the 3-median and 5-median smoothing plots and superimposing them over the
original plot gives us the graph below:

Note that the median smoothed graphs have no extreme values. All smoothed y-values are
the same as a y-value on the original plot, though at a particular x-value the 3-median or 5-
median plot doesn’t necessarily have the same y-value as the original plot does at that x-
value.
6
a   Calculating the 3-median and 5-median smoothing plots and superimposing them over the
original plot gives us the graph below:

Note that the median smoothed graphs have no extreme values. All smoothed y-values are
the same as a y-value on the original plot, though at a particular x-value the 3-median or 5-
median plot doesn’t necessarily have the same y-value as the original plot does at that x-
value.

b   Noting the high variability of the GDP growth plot compared to the less variable 3-moving
and 5-moving median plots, we can say:
‘The plot of the raw data, which is essentially a plot of GDP growth smoothed over one year,
shows a great deal of variability over the whole time period. No clear trend is apparent.
When smoothed over a 3-year period, GDP growth is still variable but to a lesser extent. No
clear trend is apparent, but GDP appears to be going through a period of below average
growth during the time period Year 7 to Year 9. When smoothed over a 5-year period, GDP
growth is much less variable but clearly shows the period of below average growth during
the period Year 7 to Year 9.’
Solutions to Exercise 7D
1
a   We know that the average seasonal index must be 1 by definition. Since there are 12 seasons
(months) in this case, the total sum of seasonal indices must be 12.
Since 12 – 1.2 – 1.3 – 1.0 – 1.0 – 0.9 – 0.8 – 0.7 – 0.9 – 1.0 – 1.1 = 1.0, the seasonal index
for December must be 1.0.
b   i The February seasonal index of 1.3 means that the February monthly sales are 30%
greater than in an average month.
ii The August seasonal index of 0.7 means that the August monthly sales are 30% lower
than in an average month.
2
a   We know that the average seasonal index must be 1 by definition.
Since there are 4 seasons (quarters) in this case, the total sum of seasonal indices must be 4.
Since 4 – 1.04 – 1.18 – 1.29 = 0.49, the seasonal index for Q4 must be 0.49.
b   We know that the average seasonal index must be 1 by definition.
Since there are 3 seasons (terms) in this case, the total sum of seasonal indices must be 3.
Since 3 – 0.67 – 1.33 = 1.0, the seasonal index for Term 2 must be 1.0.
3
Quarter 1          Quarter 2         Quarter 3             Quarter 4
Year 1              1256               1060              1868                  1642
Deseasonalised     1256 = 1570        1060 = 1514       1868 = 1437           1642 = 1368
0.8                0.7               1.3                   1.2
Seasonal index          0.8                0.7               1.3        4 – 0.8 – 0.7 – 1.3 = 1.2

4
Quarter 1               Quarter 2      Quarter 3     Quarter 4
Year 1                      68                     102            115            84
Deseasonalised               68 = 80                102 = 93      115 = 100      84 = 93
0.85                    1.1           1.15           0.9
Seasonal index     1 – 1.1 – 1.15 – 0.9 = 0.85        1.10           1.15          0.90

5
a   Seasonal average = 48 + 41 + 60 + 65 = 53.5
4
Q1              Q2              Q3               Q4
48 = 0.90       41 = 0.77       60 = 1.12        65 = 1.21
53.5            53.5            53.5             53.5

b   Seasonal average = 60 + 56 + 75 + 78 = 67.25
4
Q1               Q2               Q3                   Q4
60 = 0.89        56 = 0.83        75 = 1.12            78 = 1.16
67.25            67.25            67.25                67.25

6
a   Seasonal average = 12 + 13 + 14 + 17 + 18 + 15 + 9 + 10 + 8 + 11 + 15 + 20 = 13.5.
12
The indices in the table below are calculated by dividing the Sales figure for each month by
13.5 and rounding to 2 decimal places.
Month    Jan     Feb    Mar    Apr      May      Jun     Jul    Aug    Sep    Oct    Nov    Dec
Sales     12      13     14     17       18       15      9      10     8      11     15     20
Index    0.89    0.96   1.04   1.26     1.33     1.11    0.67   0.74   0.59   0.81   1.11   1.48
b   Seasonal average = 22 + 19 + 25 + 23 + 20 + 18 + 20 + 15 + 14 + 11 + 23 + 30 = 20.
12
The indices in the table below are calculated by dividing the Sales figure for each month by
20 and rounding to two decimal places.
Month    Jan     Feb    Mar    Apr      May      Jun      Jul     Aug    Sep    Oct     Nov     Dec
Sales     22      19     25     23       20       18       20      15     14     11      23      30
Index    1.10    0.95   1.25   1.15     1.00     0.90     1.00    0.75   0.70   0.55    1.15    1.50

7
a   We know that the average seasonal index must be 1 by definition.
Since there are 4 seasons (quarters) in this case, the total sum of seasonal indices must be 4.
Since 4 – 1.30 – 0.58 – 1.10 = 1.02, the seasonal index for Q4 must be 1.02.

b   The seasonal index of 1.30 means that in Quarter 1 there are 30% more waiters employed by
the restaurant than are normally employed in an average quarter.

c
Quarter 1      Quarter 2            Quarter 3      Quarter 4
Waiters           198            145                   86            168
Deseasonalised    198 = 152      145 = 142             86 = 148      168 = 153
1.30           1.02                 0.58           1.10
Seasonal index       1.30           1.02                 0.58           1.10

8
Summer        Autumn               Winter               Spring
Students            56           125                 126                   96
Deseasonalised    56 = 112      125 = 125            126 = 97             96 = 80
0.5           1.0                  1.3                  1.2
Seasonal index       0.5           1.0                 1.3        4 – 0.5 – 1.0 – 1.3 = 1.2

9
a   i We know that the average seasonal index must be 1 by definition.
Since there are 12 seasons (months) in this case, the sum of seasonal indices must be 12.
Since 12 – 1.0 – 1.1 – 1.0 – 1.0 – 1.0 – 0.9 – 1.2 – 1.2 – 1.1 – 1.0 – 0.7 = 0.8, the seasonal
index for Feb is 0.8.

ii The deseasonalised data is calculated as follows:
Jan – 166 = 166                       Feb – 215 = 269               Mar – 203 = 185
1.0                                     0.8                       1.1
Apr – 209 = 209           May – 178 = 178                  Jun – 165 = 165
1.0                      1.0                              1.0
Jul – 156 = 173           Aug – 256 = 213                  Sep – 243 = 203
0.9                       1.2                              1.2
Oct – 207 = 188           Nov – 165 = 165                  Dec – 106 = 151
1.1                      1.0                               0.7

Month          Jan   Feb    Mar      Apr      May      Jun     Jul    Aug    Sep    Oct     Nov     Dec
Sales         166   215    203      209      178      165     156    256    243    207     165     106
Deseasonalised    166   269    185      209      178      165     173    213    203    188     165     151
Index         1.0   0.8    1.1      1.0      1.0      1.0     0.9    1.2    1.2    1.1     1.0     0.7

b   The seasonal index of 0.90 means that in July there are 10% fewer sales made by VMAX
than in an average month.
Solutions to Exercise 7E
1                                               3
a   From the plot we can see a steady           a   Plotting the points on a graph gives us
increasing trend in the data plots,             the following plot graph:
meaning the number of Australian
university students has increased
b   Using the LinReg(a+bX) function of
the graphics calculator, we get:
a = 520.4, b = 10.091, r2 = 0.925,
r = 0.962.
The regression line for the plot is thus:
Students(000s) = 520.4 + 10.091 × Year.
b   From the plot, a generally decreasing
The positive gradient of 10.091 means
trend in the number of questions asked
that every year the average increase in
is evident.
Australian university students
compared to the previous year is            c   Using the LinReg(a+bX) function of
10 091 students per year.                       the graphics calculator, we get:
a = 17.6, b = –0.4, r2 = 0659,
c   If 1992 is Year 1, then 2008 must be
r = 0.812.
Year 17.
The regression line for the plot is thus:
Students(000s) = 520.4 + 10.091 × Year
Questions = 17.6 – 0.4 × Year.
= 520.4 + 10.091 × 17
= 520.4 + 171.547                            The negative gradient of –0.4 means
= 691.947                                    that every year the average decrease in
Therefore we can say that there will be         the average number of questions asked
approximately 692 000 students                  by members of parliament during
enrolled at an Australian university in         question time compared to the
2008, rounded to the nearest thousand           previous year is 0.4 questions per year.
students.
d   Drawing in the trend line onto the plot
2                                                   graph gives us the following graph:
a   Using the LinReg(a+bX) function of
the graphics calculator, we get:
a = 50.9, b = 1.59, r2 = 0.815,
r = 0.903.
The regression line for the plot is thus:
Purchases = 50.9 + 1.59 × Quarter
b   The 4th quarter of Year 4 is quarter 16.
Substituting Quarter = 16 into the
equation, we get 50.9 + 1.59 × 16 =
76.34 purchases. This data is, however,
deseasonalised.                             e   If 1976 is Year 1, then 2010 will be
year 35.
Since to deseasonalise data we must             Substituting Year = 35 into the
divide by the seasonal index, to                equation we get:
reseasonalise data we must multiply by          Questions = 17.6 – 0.4 × 35
the season index. The seasonal index               = 3.6 questions.
for quarter 4 is 1.18, so since 76.34 ×
1.18 = 90 (rounded to the nearest           f   Forecasting, by definition, involves
whole number), we can forecast sales            extrapolating to predict information for
in the 4th quarter of Year 4 to be 90.          which there are no data values.
4                                               5
a   Plotting the points on a graph gives us     a   Using the LinReg(a+bX) function of
the following plot graph:                       the graphics calculator, we get:
a = 27.20, b = 0.199, r2 = 0.992,
r = 0.996.
The regression line for the plot is thus:
Age = 27.20 + 0.199 × Year

The positive gradient of 0.199 means
that every year the average increase in
the average age of mothers having
b   From the plot, a generally decreasing           their first child in Australia compared
trend in the percentage of sales made           to the previous year is 0.199 per year.
in department stores is evident.
b   If 1989 is Year 1, then 2010 will be
c   Using the LinReg(a+bX) function of              year 22. Substituting Year = 22 into
the graphics calculator, we get:                the equation we get:
a = 12.5, b = –0.26, r2 = 0.943,                Age = 27.20 + 0.199 × 22
r = 0.971.                                         = 31.6
The regression line for the plot is thus:
Percentage of sales = 12.5 – 0.26 × Year

The negative gradient of –0.26 means
that every year the average decrease in
the percentage of total retail sales
made in department stores compared to
the previous year is 0.26% per year.

d   Drawing in the trend line onto the plot
graph gives us the following graph:

e   Substituting Year = 15 into the
equation we get:
Percentage of sales = 12.5 – 0.26 × 15
= 8.6%
Solutions to Multiple–choice questions
1   Since the fluctuations in the graph
occur within a time period of a year or     7    If 1995 is Year 1, then 2010 will be
less, i.e. the troughs are every half a          Year 16.
year, the variation in the graph is              Substituting Year = 16 into the
seasonal. ⇒ C                                    equation Age = 27.06 + 0.236 × Year,
we get Age = 27.06 + 0.236 × 16
2   The 3-moving mean at time period = 4                = 30.8. ⇒ E
is the mean of the values for time
period = 3, 4, 5, which are 4.4, 2.7, 5.1   8    The positive gradient of 0.236 means
respectively.                                    that the average age of marriage for
4.4 + 2.7 + 5.1 = 4.1 ⇒ B                        males is increasing.
3                                       Since there are 12 months in a year and
12 × 0.236 = 2.8, we can see that the
3   The 5-moving median at time period = 3 is        average increase in age of marriage for
the median of the values for time                males is about 3 months per year. ⇒ A
period = 1, 2, 3, 4, 5, which are 99, 74,
103, 92, 88.                                9    Using the LinReg(a+bX) function of
The median of these values is 92. ⇒ D            the graphics calculator with Day as the
x-list and deseasonalised Price as the
4   We know that the average seasonal                y-list, we get a = 84.34, b = 1.246,
index must be 1 by definition. Since             r2 = 0.749, r = 0.865.
there are 12 seasons (months) in this            The deseasonalised regression line for
case, the total sum of seasonal indices          the plot is thus Price = 84.34 + 1.246 ×
must be 12.                                      Day. ⇒ A
Since 12 – 1.0 – 1.1 – 0.9 – 1.0 – 1.0 –
1.2 – 1.1 – 1.1 – 1.1 – 1.0 – 0.7 = 0.8,    10   If Sunday of Week 1 is Day 1, then
the seasonal index for Feb must be 0.8           Friday of Week 2 will be Day 13.
⇒E                                               Substituting Day = 13 into the
equation Price = 84.34 + 1.246 × Day,
5   To deseasonalise a data value, we must           we get Price = 84.34 + 1.246 × 13 =
divide the data value by its seasonal            100.5. This data is, however,
index.                                           deseasonalised.
423                                              Since to deseasonalise data we must
= 240 ⇒ B
1.8                                             divide by the seasonal index, to
reseasonalise data we must multiply by
6   Seasonal average                                 the season index.
1048 + 677 + 593 + 998                        The seasonal index for Friday is 1.2, so
=                              = 829.
4                                since 100.5 × 1.2 = 120.6, we can
Seasonal index for Autumn is the value           forecast Price on Friday of Week 2 to
for Autumn divided by the seasonal               be 120.6 cents/litre. ⇒ C
677
average =       = 0.82. ⇒ D
829
Chapter 8 – Revision of the Core
Solutions to Exercise 8.1
1   There are 7 numbers, the median of            6    Of the 17 values, the median value will
which is the middle and thus the 4th               be the 9th value, which in this case is
highest number. In this case, –4.4% is             24. ⇒ D
the 4th highest number and thus the
median. ⇒ D                                   7    Range = highest value – lowest value
= 35 – 11 = 24. ⇒ D
2   Mean =
-4.6 – 4.7 + 2.9 + 0.3 – 5.5 – 4.4 – 1.1      8    Of the 17 values, Q1 will be the mean
7                               of the 4th and 5th values, which is
= –2.44.                                     20 + 21
= 20.5;
Standard deviation is found by the                      2
below table:                                       Q3 will be the mean of the 13th and 14th
x      (x – x )       (x – x )2                                  28 + 28
value, which is             = 28.
–4.6       –2.16          4.6656                                       2
–4.7       –2.26          5.1076                 IQR = Q3 – Q1 = 28 – 20.5 = 7.5 ⇒ C
2.9         5.34        28.5156
0.3         2.74         7.5076            9    Since the histogram tails off to the left
–5.5       –3.06          9.3636                 from the peak more than it tails off to
–4.4       –1.96          3.8416                 the right, it is negatively skewed. ⇒ B
–1.1         1.34         1.7956
Sum        –0.02         60.7972            10   The coloured box would be bunched
up to the higher end of the scale, since
∑ (x − x )
2
60.8                       the middle 50% of values occurs in a
s=                  =        = 10.13 = 3.18
n −1            6                         narrow band at the higher end of the
⇒A                                                 scale. A negatively skewed distribution
has the median vertical line closer to
3   Range = highest value – lowest value               the right border of the coloured box
= 2.9 – –5.5 = 2.9 + 5.5                          than the left. ⇒ D
= 8.4% ⇒ E
11   The histogram tells us that 12 people
4   Since both variables only have two                 scored less than 30.
categories, male and female for ‘Sex’,             We know that 63 students sat the test.
and yes or no for ‘Mobile phone                           12
Since     × 100% = 19.0%, we can say
owner’, they are both categorical                         63
variables. ⇒ A                                     that 19% of students failed the test.
⇒B
5   Each of the horizontal black lines
represents a quarter of the observations      12   The histogram tells us that 4 students
and each of the two parts of the                   scored between 30 and 35, 7 students
coloured box either side of the vertical           between 35 and 40, and 9 students
median line represents a quarter of the            between 40 and 45.
observations.                                      4 + 7 + 9 = 20 ⇒ E
Since the lower left–most horizontal
black line starts to the right of the         13   The mode is the most commonly
value 30, we can safely say that less              occurring value and in a histogram is
than a quarter of all observations are             the highest bar, which in this case is
less than 30. ⇒ B                                  50–55. ⇒ E
14   Since there are 63 values, the median       21   We can see that since 9 people waited
value is the 32nd value.                         between 0–2 minutes and 10 people
From the histogram, we can see that              waited between 2–4 minutes, 19
there are 31 values in the data ranges           people waited less than 4 minutes.
45–60. Thus, the 32nd value must be in           ⇒C
the 40–45 data range. ⇒ D
22   We can see that 11 people waited more
15   If everyone’s weight has increased by            than 10 minutes.
2 kg, the mean weight will increase by                 11
Since    × 100% = 22, 22% of
2 kg from 72 kg to 74 kg, since the                    50
mean weight is the average of the                customers waited 10 or more minutes.
weights of all the 12 people.                    ⇒C
The standard deviation is a measure of
the spread of values and increasing all     23   The median would be the best
the weights by 2 kg will not change the          indicator of a typical waiting time.
spread at all, so the standard deviation         This is because the distribution is
will remain 5 kg. ⇒ C                            positively skewed, meaning that the
mean will be affected and be too far to
16   IQR = Q3 – Q1 = 110 – 60 = 50 ⇒ B                the left to give a good indication of
typical values. The interquartile range,
17   The box plot for a wet Saturday can be           range and standard deviation are all
seen to be more to the left than the plot        measures of spread so don’t give any
for a dry Saturday, so is generally              information on a typical value. ⇒ A
lower than a dry Saturday.
The wet Saturday plot is also much          24   Since 25 and 43 are 1 standard
wider and spread out than that for a dry         deviation either side of the mean, we
Saturday, so is more variable. ⇒ E               know that 68% of students will have
scores within this range. ⇒ D
18   Of the 16 students in class A, the
median mark will be the mean of the         25   95 is 2 standard deviations above the
8th and 9th values, which is                     mean and we know that for the 50% of
71 + 73                                          values above the mean, 47.5% of values
= 72. ⇒ A
2                                           will be between the mean (85 g) and 2
standard deviations above it (95 g).
19   The interquartile range is the length of         Therefore, 2.5% of eggs will be above
the coloured box, which is longest for           95g and thus Extra Large. ⇒ C
Class B. ⇒ B
26   100 is 2 standard deviations below the
20   IQR = Q3 – Q1 = 60 – 50 = 10.                    mean and we know that for the 50% of
Lower limit for outliers                         values below the mean, 47.5% of values
= Q1 – 1.5 × IQR                               will be between the mean (104 g) and 2
= 50 – 1.5 × 10 = 35.                          standard deviations below it (100 g).
Upper limit for outliers                         Therefore, 2.5% of single scoops will
= Q3 + 1.5 × IQR                               below 100 g. ⇒ B
= 60 + 1.5 × 10 = 75.
Thus, outliers are defined as values        27   The standard score is found by
lower than 35 or higher than 75. ⇒ A             subtracting the mean from the value
and dividing by the standard deviation.
45 – 65
Since           = –2.0, the student’s
10
standard score is –2.0. ⇒ B
Solutions to Exercise 8.2
1   The r value of 0.64 does not                5    Total = 44 + 9 + 8 = 61 ⇒ C
correspond to a percentage of students
who fulfilled a certain criteria, it is     6    11–12 ‘sometimes used’ = 73 – 47 – 8
merely the strength of the association             = 18
between hours spent studying and                 3–6 ‘total’ = 44 + 16 + 10 = 70
expenditure on junk food. We cannot              7–10 ‘sometimes used’ = 58 – 18 – 16
infer anything about junk food aiding              = 24
studying.                                        7–10 ‘total’ = 217 – 73 – 70 = 74
Since the r value is positive, there is a        Percentage of 7–10 who sometimes
positive correlation between hours                                   24
used the internet =    × 100% = 32%.
spent studying and expenditure on junk                               74
food. ⇒ D                                        ⇒E

2   All we are able to infer from the           7    If the outlier at (7,25) were changed to
relationship is that there is a positive         (7,5), the graph scatterplot would be much
correlation between reading scores and           closer to a negative straight line, which
height, meaning taller students tend to          would mean the r value would remain
read better and vice versa. ⇒ D                  negative and would be closer to –1.
⇒E
3   The relationship is negative and is very
close to being a perfectly straight line.   8    A back–to–back stem plot is useful for
Since a straight negative line has an            displaying relationships between a
r value of –1, this line can be seen to          dependent numerical variable and a
have an r value of –0.9. ⇒ E                     binomial categorical variable. Height
in centimetres versus sex fulfils these
4   The r value is closest to zero when              requirements. ⇒ D
there is no relationship between the
variables being compared. This will         9    Since both data are categorical, a table
show up as a more random scatterplot.            would be most useful. ⇒ A
The most random scatterplot is C, with
the other scatterplots being decidedly      10   Since a numerical dependent variable
linear. ⇒ C                                      is being graphed against a categorical
independent variable, parallel box
plots would be the best graphing
method. ⇒ D
Solutions to Exercise 8.3
1    A least squares regression line by          11   In this case, the median of the lowest 3
definition is the line that minimises the        values is the 2nd value from the left,
sum of the squares of the deviations of          while the median of the highest 3
the y–values from the regression line.           values is the 8th value from the left.
⇒A                                               The median of the middle 3 values is
the 5th value from the left.
rs y       0.675 × 4.983                    Thus, the 3-median line should be
2    b=          =                 = 1.3 ⇒ C          drawn between the 2nd and 8th value,
sx             2.567
and shifted downwards slightly,
towards the 5th value. ⇒ B
3    Using the Stat List Editor and
LinReg(a+bX) function of the                12   The 3-median regression line will be
graphics calculator, you will get:               derived from the line through the
a = 227.3, b = –4.3, r2 = 0.889,                 median of the lowest 5 values and the
r = –0.943.                                      median of the highest 5 values.
b = slope = –4.3 ⇒ A                             The (x,y) median of the lowest 5 values
can be seen to be (2,5), while the (x,y)
4    Coefficient of determination = r2                median of the highest 5 values can be
= 0.89 ⇒ D                                       seen to be (8,2).
The slope of the 3-median regression
5    Errors = 8.8 – 0.120 × 35                        line will be the same as the slope of the
= 8.8 – 4.2 = 4.6 ⇒ B                     line joining these two points.
The line joining (2,5) and (8,2) drops 3
6    r=    r 2 = 0.8198                               units vertically over a horizontal run of
= –0.905 = –0.91 ⇒ A                   6 units.
-3
Thus, gradient =      = –0.5. ⇒ C
7    The y-intercept can be seen to be                                   6
approximately 210. The slope of the
graph is negative and the line drops        13   The y-intercept can be seen to be
around 11 cm of rainfall for every 1             approximately 110. The slope of the
degree increase in temperature.                  graph is negative and the line drops 90
Hence, average rainfall = 210 – 11 ×             units vertically over a horizontal run of
temperature range. ⇒ A                           120 units.
-90
= –0.75 = –0.8.
120
8    r2 = (–0.9260)2
Hence, equation = y = 110 – 0.8x.
= 0.8575 ⇒ C
⇒E
9    Average rainfall = 210 – 11 × 10
14   Fuel consumption = 15.2 – 0.005 × 1000
= 210 – 110 = 100 cm ⇒ B
= 15.2 – 5 = 10.2 ⇒ A
10   The 3-median regression line will be
15   Since the y-value of the line at x = 12
derived from the line through the
is about 160, we would predict \$160
median of the lowest 5 values and the
would be the student’s weekly income.
median of the highest 5 values.
The (x,y) median of the lowest 5 values
⇒D
can be seen to be (4,3), while the (x,y)
16   Since for 12 hours a week the student
median of the highest 5 values can be
receives \$160, we can say that for 1
seen to be (13,9). ⇒ E
hour, the student receives \$160 = \$13.
12
⇒D
17   The dot represented that student is        19   Since the plots resemble the upper
about \$50 above the red regression              right quadrant of a circle, a y2 or x2
line, meaning that student’s income             transformation would be appropriate.
has a residual of \$50 relative to the           ⇒C
regression line. ⇒ D
20   Since the plots resemble the lower left
18   The residual plot for the scatterplot           quadrant of a circle, a log(y), 1/y,
will show the value at x = 0 to have a          log(x) or 1/x transformation would be
residual below the regression line; the         appropriate. ⇒ A
residuals for x = 1, 2, 3, 3.5 will be
above the regression line and the
residuals for x = 4.5, 5 will be below
the regression line.
All the residuals are below 0.5, except
for x = 3, which is slightly above. This
gives us residual plot D.
Solutions to Exercise 8.4
2100
1    Deseasonalised figure =        = 2625 ⇒ D
0.8

2    There is clearly a seasonal pattern in the time series, with a trough every Quarter 3.
There is, however, no longer-term linear trend. ⇒ A

3    Since the seasons are quarters, the seasonal indices will sum to 4.
Hence, w = 4 – 0.2 – 1.3 – 0.3 = 2.2. ⇒ E

4    There is a clear upwards trend but no variation that can be said to be cyclical or seasonal.
⇒A

5    The 3-moving mean at t = 5 will be the mean of 4, 6 and 8.
4 + 6 + 8
= 6 ⇒C
3

6    If the seasonal index is 0.85 = 85%, it is 0.15 = 15% below the quarterly average for the
year. ⇒ B

150
7    Deseasonalised figure =       = 125 ⇒ B
1.2

8    Smoothing a time series aims to remove all random variations such that seasonal variation,
cyclical variation and trends can be more readily seen and analysed. ⇒ D

9    There is a clear seasonal variation with a very slight upwards trend. ⇒ E

18187
10   Deseasonalised figure =         = 23022 ⇒ E
0.79

11   The two share prices have been steadily increasing but the yellow shares are increasing
consistently faster than the blue shares, meaning the difference in share price has shown a
consistently decreasing trend. ⇒ A

4 – 20
12   The line goes from (0,20) to (10,4), which is a y-intercept of 20 and a gradient of
10
-16
=      = –1.6.
10
Hence, y = 20 – 1.6t. ⇒ A

13   The 5-moving median at t = 6 will be the median of 103, 92, 110, 109 and 118, which is 109.
⇒C

14   There is a clear seasonal variation will an upwards trend. ⇒ C

15
Year      1960   1965   1970    1975   1980     1985   1990   1995
Price     3500   7500   5000    6000   4000     6000   4000   4200
3–median     –     5000   6000    5000   6000     4000   4200    –
This table shows us the correct graph is graph C.
Chapter 9 – Arithmetic and
geometric sequences
Solutions to Exercise 9A
1                                           2
a   Rule based.                             a   Next term = previous term × 10.
Next term = previous term + 1.              Next term = 1000.
Next term = 5.
b   Next term = previous term – 2.
b   Rule based.                                 Next term = 8.
Next term = previous term – 1.
Next term = 96.                         c   Next term = previous term × 3.

c   Random.                                     Next term = 27.

d   Random.                                 d   Next term = previous term + 1 .
7
e   Rule based.                                 Next term = 4
7
Next term = previous term.
Next term = 1.
e   Next term = (previous term)2.
Next term = 54.
f   Rule based.
Next term = 0 if previous term = 1 OR
f   Next term = previous term – 4.
next term = 1 if previous term = 0.
Next term = –10.
Next term = 1.
g   Next term = (previous term)2.
g   Rule based.
1
Next term = previous term + 2.              Next term =
256
Next term = 10.
h   Next term = 10 if previous term = –10
h   Rule based.
OR next term = –10 if previous term =
Next term = previous term × 2.
10.
Next term = 32.
Next term = 10.
i   Rule based.
i   Next term = previous term – 4.
Next term = previous term × 2.
Next term = –6.
Next term = 160.
Solutions to Exercise 9B
1
a   Yes, difference of +1.     h   No

b   Yes, difference of –1.     i   Yes, difference of +9.9.

c   No                         j   Yes, difference of +40.

d   No                         k   Yes, difference of +5.1.

e   No                         l   No

f   Yes, difference of +4.     3
a   The terms differ from their neighbours
g   No                             by a constant amount of 4.

h   No                         b   Common difference is 4.

i   No                         c   36 + 4 = 40.

j   Yes, difference of –4.     d   Since 20 is the 1st term, you must add
the common difference 7 times to get
k   Yes, difference of –50.        the 8th term.
20 + 7 × 4 = 20 + 28 = 48.
l   Yes, difference of +5.
e   Since 20 is the 1st term, you must add
m   No                             the common difference 12 times to get
the 13th term.
n   Yes, difference of –4.         20 + 12 × 4 = 20 + 48 = 68.

o   No                         4
a   The terms differ from their neighbours
2                                  by a constant amount of 2.
a   Yes, difference of +1.
b   Common difference is –2.
b   Yes, difference of –1.
c   –3 – 2 = –5.
c   No
d   Since 5 is the 1st term, you must add
d   No                             the common difference 6 times to get
the 7th term.
e   Yes, difference of +0.7.       5 + 6 × –2 = 5 – 12 = –7.

f   No                         e   10th term = 5 + 9 × –2 = 5 – 18 = –13.
50th term = 5 + 49 × –2 = 5 – 98 = –93.
g   No
Solutions to Exercise 9C
1                                               4
a   a = starting value = 3                      a   d = 47 – 37 = 5.
d = difference = 3                                   9 – 7
t2 = term 2 = 6                                 a = 37 – (7 – 1) × 5
= 37 – 30 = 7.
b   a = starting value = 23                         First 3 terms are thus 7, 12 and 17.
d = difference = 7
t3 = term 3 = 37                            b   d = 43 – 31 = 3.
15 – 11
c   a = starting value = 100                        a = 31 – (11 – 1) × 3
d = difference = –25                              = 31 – 30 = 1.
t4 = term 4 = 25                                First 3 terms are thus 1, 4 and 7.
d   a = starting value = 13
d = difference = –2                         c   d = -20 – 0 = –4.
11 – 6
t1 = term 1 = 13                                a = 0 – (6 – 1) × –4
= 0 + 20 = 20.
e   a = starting value = –20                        First 3 terms are thus 20, 16 and 12.
d = difference = –5
t2 = term 2 = –25
d   d = 159 – 134 = 5.
13 – 8
f   a = starting value = –12
a = 134 – (8 – 1) × 5
d = difference = 2
= 134 – 35 = 99.
t2 = term 2 = –10
First 3 terms are thus 99, 104 and 109.
2
a   t3 = 200 + 2 × 10 = 200 + 20 = 220          e   d = (10 – 60) / (12 – 7) = –10.
a = 60 – (7 – 1) × –10 = 60 + 60 = 120.
b   t20 = 2 + 19 × 5 = 2 + 95 = 97                  First 3 terms are thus 120, 110 and 100.

c   t5 = 250 + 4 × 50 = 250 + 200 = 450         f   d = 75 – 20 = 5.
21 – 10
a = 20 – (10 – 1) × 5
d   t12 = 26 + 11 × –2 = 26 – 22 = 4
= 20 – 45 = –25.
First 3 terms are thus –25, –20 and –15.
e   t7 = 25 + 6 × –5 = 25 – 30 = –5
5   Note that tn = a + (n – 1)d
f   t53 = 0 + 52 × 2 = 0 + 104 = 104
a   tn = 3 + (n –1) ×2 = 3 + 2n – 2
3                                                      = 2n + 1
a   t11 = 5 + 10 × 5 = 5 + 50 = 55
b   tn = 12 + (n – 1) ×4 = 12 + 4n – 4
b   t8 = 12 + 7 × –4 = 12 – 28 = –16                   = 4n + 8

c   t27 = 0.1 + 26 × 0.01 = 0.1 + 0.26 = 0.36   c   tn = 5 + (n – 1) ×3 = 5 + 3n – 3
= 3n + 2
d   t13 = –55 + 12 × 13 = –55 + 156 = 101
d   tn = 10 + (n – 1) × –2 = 10 – 2n + 2
e   t10 = –1.0 + 9 × –0.5 = –1.0 – 4.5 = –5.5          = 12 – 2n
f   t95 = 130 + 94 × –7 = 130 – 658 = –528      e   tn = 25 + (n – 1) × –5 = 25 – 5n + 5
= 30 – 5n
g   t7 = 1 + 6 × – 1 = 1 – 6 = –1
2          4        2   4              f   tn = –4 + (n – 1) × 2 = –4 + 2n – 2
= 2n – 6
6                                               7
a   tn = 2 + (n – 1) × 2 = 2 + 2n – 2           a   a = 5, d = 2,
= 2n                                         t12 = 5 + (12 – 1) × 2
= 5 + 22 = 27,
b   tn = 30 + (n – 1) × 4 = 30 + 4n – 4             12 terms must be written down.
= 4n + 26
b   a = 132, d = 50,
c   tn = 12 + (n – 1) × 3 = 12 + 3n – 3             t19 = 132 + (19 –1) × 50
= 3n + 9                                        = 132 + 900 = 1032,
19 terms must be written down.
d   tn = 10 + (n – 1) × –2 = 10 – 2n + 2
= 12 – 2n                                c   a = 100, d = –4,
t9 = 100 + (9 – 1) × –4
e   tn = 100 + (n – 1) × –25 = 100 – 25n + 25          = 100 – 32 = 68,
= 125 – 25n                                  9 terms must be written down.

f   tn = –5 + (n – 1) × 5 = –5 + 5n – 5         d   a = 10, d = 6,
= 5n – 10                                    t8 = 10 + (8 – 1) × 6
= 10 + 42 = 52,
8 terms must be written down.

e   a = 0.33, d = 0.33,
t7 = 0.33 + (7 – 1) × 0.33
= 0.33 + 1.98 = 2.31,
7 terms must be written down.

f   a = –17, d = 2,
t10 = –17 + (10 – 1) × 2
= –17 + 18 = 1,
10 terms must be written down.

g   a = 127, d = –5,
t27 = 127 + (27 – 1) × –5
= 127 – 130 = –3,
27 terms must be written down.
Solutions to Exercise 9D
1   Note that Sn = n (2 × a + (n – 1)d)         3
n
Note that Sn = (2 × a + (n – 1)d)
2                                            2
6                                      a   a = 8, d = 8,
a   S6 = (2 × 5 + (6 – 1) × 3)
2                                               n
= 3(10 + 15) = 75                             Sn = (16 + (n –1) × 8)
2
5                                      b   a = 20, d = 5,
b   S5 = (2 × 12 + (5 – 1) × –2)
2                                               n
Sn = (40 + (n – 1) × 5)
= 2.5(24 – 8) = 40                                 2

7                                      c   a = 0, d = 15,
c   S7 = (2 × –5 + (7 – 1) × 5)                          n
2                                          Sn = (0 + (n – 1) × 15)
= 3.5(–10 + 30) = 70                               2
n
= 15 (n – 1)
4                                                 2
d   S4 = (2 × 0.1 + (4 – 1) × 0.2)
2
= 2(0.2 + 0.6) = 1.6                      d   a = 2, d = 4,
n
Sn = (4 + (n – 1) × 4)
6                                               2
e   S6 = (2 × –9 + (6 – 1) × 3)
2
e   a = 100, d = 100,
= 3(–18 + 15) = –9                                 n
Sn = (200 + (n – 1) × 100)
n                                   2
2   Note that Sn =     (2 × a + (n – 1)d)
2                          f   a = 200, d = –50,
20                                             n
a   S20 =    (2 × 4 + (20 – 1) × 4)                 Sn = (400 + (n – 1) × –50)
2                                             2
= 10(8 + 76) = 840
4
8                                      a   a = 4, d = 3,
b   S8 = (2 × 10 + (8 – 1) × –3)
2                                          S5 = 5 (2 × 4 + (5 –1) × 3)
= 4(20 – 21) = –4                                  2
= 2.5(8 + 12) = 50,
9                                          5 terms must be summed.
c   S9 = (2 × 120 + (9 – 1) × –10)
2
= 4.5(240 – 80) = 720                     b   a = 10, d = 5,
S5 = 5 (2 × 10 + (5 – 1) × 5)
1000                                          2
d   S1000 =     (2 × 1 + (1000 – 1) × 1)              = 2.5(20 + 20) = 100,
2
= 500(2 + 999) = 500500                       5 terms must be summed.

100                                   c   a = 0.2, d = 0.2,
e   S100 =    (2 × 1.000 + (100 – 1) × 0.005)       S7 = 7 (2 × 0.2 + (7 – 1) × 0.2)
2                                             2
= 50(2 + 0.495) = 124.75                        = 3.5(0.4 + 1.2) = 5.6,
15                                        7 terms must be summed.
f   S15 =    (2 × –8 + (15 – 1) × 2)
2                                    d   a = 100, d = 100,
= 7.5(–16 + 28) = 90
S14 = 14 (2 × 100 + (14 – 1) × 100)
2
= 7(200 + 1300) = 10 500,
14 terms must be summed.
e   a = 1, d = 0.05,
S9 = 9 (2 × 1 + (9 – 1) × 0.05)
2
= 4.5(2 + 0.4) = 10.8,
9 terms must be summed.
5                                               e   a = 4, d = 4,
a   Counting the number of counters in                   n
Sn = (8 + (n – 1) × 4)
each square:                                         2
Square number        1    2    3     4            n
= (8 + 4n – 4)
Number of counters   4    8    12    16           2
in square                                         n
= (4n + 4)
2
b   Since we must add 4 counters to a                   4n
=     (n + 1) = 2n(n + 1)
square to make the next one, Nn = 4n                 2

c   i N6 = 4 × 6 = 24                           f   i S6 = 2 × 6 × (6 + 1)
ii N10 = 4 × 10 = 40                                  = 12 × 7 = 84
ii S10 = 2 × 10 × (10 + 1)
d   Summing the total number of counters                   = 20 × 11 = 220
required for a certain number of
squares:                                    g   S6 = 84;
S7 = 2 × 7 × (7 + 1)
Square number   1    2    3     4                = 14 × 8 = 112.
Total number    4    12   24    40
Therefore, 6 squares are the maximum
of counters
that can be made with 100 counters.
Solutions to Exercise 9E
1                          3
a   No, arithmetic.        a   The terms are separated from their
neighbours by a common ratio of 10.
b   Yes, ratio = 2.
b   The common ratio = 10.
c   Yes, ratio = 10.
c   2000 × 10 = 20 000.
d   No, random.
d   To get the 5th term, you must multiply
e   No, random.                the 1st term, 2, by the common ratio of
10, 4 times, to give you 20 000.
1
f   Yes, ratio =     .
2       e   To get the 15th term, you must multiply
the 1st term, 2, by the common ratio of
g   No, arithmetic.            10, 14 times, to give you 2 × 1014.

h   Yes, ratio = 2.        4
a   The terms are separated from their
i   Yes, ratio = 2.            neighbours by a common ratio of 1 .
4
j   Yes, ratio = –2.
b   The common ratio = 1 .
4
k   Yes, ratio = – 1 .
2
c   16 × 1 = 4
l   Yes, ratio = 2.                 4

2                          d   To get the 7th term, you must multiply
a   Yes, ratio = 1.1.          the 1st term, 1024, by the common
ratio of 1 , 6 times, to give you 1 .
4                       4
b   No, arithmetic.

c   No, random.            e   The 10th term is 1024 × ( 1 )9
4
= 1024 = 1
d   No, arithmetic.                262144 256

e   Yes, ratio = 1.5.      5
a   The terms are separated from their
f   Yes, ratio = 2.            neighbours by a common ratio of –5.

g   No, arithmetic.        b   The common ratio = –5.

h   Yes, ratio = –2.       c   125 × –5 = –625

i   Yes, ratio = 10.       d   To get the 6th term, you must multiply
the 1st term, –1, by the common ratio
of –5, 5 times, to give you 3125.
Solutions to Exercise 9F
1                                     5
a   a = 3, r = 2, t3 = 12             a    2187 = 27.
81
b   a = 15, r = 3, t2 = 45                8 – 5 = 3.
27 = 33 = r3, r = 3.
c   a = 200, r = 1 , t4 = 25              a = 81 = 81 = 1.
4
2                          r           81
Thus, first three terms = 1, 3 and 9.
d   a = 130, r = 1 , t1 = 130
10                        1250 = 0.125.
b
10000
e   a = 20, r = –1.2, t3 = 28.8           5 – 2 = 3.
0.125 = 0.53 = r3, r = 0.5.
f   a = –120, r = – 1 , t4 = 1.875        a = 10000 = 10000 = 20000.
4                              r        0.5
Thus, first three terms = 20000, 10000
2                                         and 5000.
a   t3 = 20 × 102 = 20 × 100 = 2000
c   -320 = –8.
b   t7 = 3 × 26 = 3 × 64 = 192             40
6 – 3 = 3.
c                   4
t5 = 512 × 0.5 = 32                   –8 = –23 = r3, r = –2.
a = 40 = 40 = 10.
2
d                       7
t8 = 1024 × –0.5 = –8                      r           4
Thus, first three terms = 10, –20 and 40.
e   t4 = 1000 × –1.013 = –1030.301
d   250 = 1.5625.
f                       2
t3 = –100 × –2.2 = –484               160
4 – 2 = 2.
3                                         1.5625 = 1.252 = r2, r = 1.25.
a   t3 = 7.5 × 42 = 120                   a = 160 = 160 = 128.
r       1.25
6                         Thus, first three terms are 128, 160 and
b   t7 = 7.5 × 4 = 30720
200.
c   t15 = 7.5 × 414 = 2 013 265 920
6   Note that tn = a × rn–1
4
a   tn = 3 × 2n–1
a   t7 = 1 × 56 = 15625
b   tn = 12 × 1.3n–1
b   t8 = 10 000 × 0.27 = 0.128
c   tn = 5 × 0.2n–1
c   t10 = –1 × –29 = 512
d   tn = –10 × 1.75n–1
d   t9 = –20 × 38 = –131 220
e   tn = 25 × –0.5n–1
e   t8 = 2 × 0.57 = 0.015 625
f   tn = –4 × –2n–1
7   Note that tn = a × rn–1       8   Note that tn = a × rn–1

4                                 4
a   a = 2, r =     = 2,           a   a = 2, r =     = 2,
2                                 2
tn = 2 × 2n–1                     t8 = 2 × 28–1
= 2 × 128 = 256.
45                    Hence, 8 terms are required to get a
b   a = 30, r =      = 1.5,
30                    term greater than 250.
tn = 30 × 1.5n–1
1.1
b   a = 1, r =       = 1.1,
6                                 1
c   a = 24, r =     = 0.25,                        9–1
24                    t9 = 1 × 1.1
n–1
tn = 24 × 0.25                       = 1 × 2.144 = 2.144.
Hence, 9 terms are required to get a
-0.8                  term greater than 2.
d   a = –1, r =      = 0.8,
-1
tn = –1 × 0.8n–1                                80
c   a =100, r =        = 0.8,
100
12–1
-500                t12 = 100 × 0.8
e   a = 1000, r =       = –0.5,
1000                   = 100 × 0.0859 = 8.59.
n–1
tn = 1000 × –0.5                  Hence, 12 terms are required to get a
term less than 10.
1.21
f   a = –1.1, r =       = –1.1,
-1.1                               -16
tn = –1.1 × –1.1n–1           d   a = –8, r =        = 2,
-8
t10 = –8 × 210–1
= –8 × 512 = –4096.
Hence, 10 terms are required to get a
term equal to –4096.

0.81
e   a = 0.9, r =        = 0.9,
0.9
22–1
t22 = 0.9 × 0.9
= 0.9 × 0.109 = 0.098.
Hence, 22 terms are required to get a
term less than 0.1.
Solutions to Exercise 9G
1                                      2
a   Note that r = 1 + R                a   Bacteria at start of 2nd minute = 10 × 2
100                   = 20.
i r=1+       10 = 1.10
100
b   Multiplying the previous number of
ii r = 1 + 5 = 1.05                    bacteria by 2 we get the following
100
table:
iii r = 1 + 13 = 1.13
100                          Time (minutes)        1      2     3
iv r = 1 + 20 = 1.20                     Number of bacteria    10     20    40
100
v r = 1 + 1 = 1.01                 c   a = 10, r = 20 = 2.
100                                       10
vi r = 1 + 100 = 2.00
100
d   Since tn = a × rn–1, we can see that
vii r = 1 + 150 = 2.50                 bn = 10 × 2n–1.
100
viiir = 1 – 10 = 0.90
100                    e   Using bn = 10 × 2n–1
ix r = 1 – 13 = 0.87                   i b5 = 10 × 25–1
100                           = 10 × 16 = 160
x r = 1 – 20 = 0.80                    ii b21 = 10 × 221–1
100
= 10 × 1 048 576 = 10 485 760
xi r = 1 – 5 = 0.95
100
3
xii r = 1 – 1 = 0.99               a   Multiplying the previous thickness by
100
xiiir = 1 + 15 = 1.15                  2 we get the following table:
100                          Number of folds      1      2      3     4
xiv r = 1 + 25 = 1.25                    Thickness (mm)      0.2    0.4    0.8   1.6
100
xv r = 1 – 25 = 0.75                                 0.4
100                    b   a = 0.2, r =      = 2.
0.2
b   Note that R = (r – 1) × 100%
n–1
tn = a × r , therefore t15 = 0.2 × 215–1
i R = (1.10 – 1) × 100% = 10%             = 0.2 × 16384
ii R = (1.20 – 1) × 100% = 20%            = 3276.8 mm = 3.2768 m
iii R = (1.05 – 1) × 100% = 5%
iv R = (1.50 – 1) × 100% = 50%
v R = (1.13 – 1) × 100% = 13%
vi R = (0.95 – 1) × 100% = –5%
vii R = (0.90 – 1) × 100% = –10%
viiiR = (0.87 – 1) × 100% = –13%
ix R = (0.50 – 1) × 100% = –50%
x R = (0.99 – 1) × 100% = –1%
Solutions to Exercise 9H
n
1   Note that Sn = a 1 – r ,               d   a = 2, r = 1 = 0.5,
1–r                                 2
20
S20 = 2 1 – 0.5
n
or Sn = a r – 1                                     1 – 0.5
r–1
5                                     = 2 × 0.9999 = 4.00
a   S5 = 5 3 – 1                                          0.5
3–1
242
=5×
2
= 605                   e   a = 1.0, r = 1.05 = 1.05,
1.0
10
4
S10 = 1 1.05 – 1
b   S4 = 10 1 – 0.1                                     1.05 – 1
1 – 0.1
= 1 × 0.6289 = 12.58
= 10 × 0.9999 = 11.11                              0.05
0.9
3
f   a = 1.1, r = 1.21 = 1.1,
c   S3 = –5 1.2 – 1                                           1.1
1.2 – 1                                        5
= –5 × 0.728 = –18.2                    S5 = 1.1 1.1 – 1
0.2                                    1.1 – 1
3                   = 1.1 × 0.611 = 6.72
0.1
d   S3 = 10 000 1.06 – 1
1.06 – 1
n
= 10 000 × 0.191 = 31 836           3   Note that Sn = a 1 – r ,
0.06                                  1–r
n
5
or Sn = a r – 1
e   S5 = 512 1 – 0.5                                     r–1
1 – 0.5
= 512 × 0.96 875 = 992              a   a = 10, r = 1.5,
0.5                                  n
Sn = 10 1.5 – 1
2                                                       1.5 – 1
a   a = 2, r = 4 = 2,                                       n
2                                   = 10 1.5 – 1 = 20(1.5n – 1)
20                                            0.5
S20 = 2 2 – 1
2–1                           b   a = 50, r = 0.2,
= 2 × 1 048 575 = 2 097 150                            n
1                        Sn = 50 1 – 0.2
1 – 0.2
n
b   a = 1000, r = 100 = 0.1,                      = 50 1 – 0.2 = 62.5(1 – 0.2n)
1000                                 0.8
15
S15 = 1000 1 – 0.1
1 – 0.1                    c   a = 4, r = 20 = 5,
= 10 000 × 1 = 11 111.11                           4
0.9                           n
Sn = 4 5 – 1
5–1
c   a = 1, r = 1.2 = 1.2,                             5n – 1 = (5n – 1)
1                              =4
9                                          4
S9 = 1 1.2 – 1
1.2 – 1
d   a = 8, r = 4 = 0.5,
= 1 × 4.1598 = 20.80                                8
0.2                                            n
Sn = 8 1 – 0.5
1 – 0.5
n
= 8 1 – 0.5 = 16(1 – 0.5n)
0.5
e   a = 0.9, r = 0.3 = 0.333,              c   a = 0.25, r = 0.5 = 2,
0.9                                              0.25
n                                    6
Sn = 0.9 1 – (1/ 3)                        S6 = 0.25 2 – 1
1 – 1/ 3                                  2–1
n
= 0.25 × 63 = 15.75.
= 0.9 1 – (1/ 3)                                       1
2/ 3                           Hence, 6 terms are required to get a
n
= 1.35 (1 – (1/ 3) )                     sum greater than 10.
4
a   a = 2, r = 4 = 2,                      d   a = 2000, r = 2100 = 1.05,
2                                                    2000
5                                                          26
S5 = 2 2 – 1                               S26 = 2000 1.05 – 1
2–1                                            1.05 – 1
= 2 × 31 = 62.                            = 2000 × 2.556 = 102 226.91.
1                                                0.05
Hence, 5 terms are required to get a       Hence, 26 terms are required to get a
sum greater than 50.                       sum greater than 100 000.

b   a = 10, r = 100 = 10,                  e   a = 0.1, r = 0.5 = 5,
10                                           0.1
6                                        4
S6 = 10 10 – 1                             S4 = 0.1 5 – 1
10 – 1                                      5–1
= 10 × 999 999 = 1 111 110                 = 0.1 × 624 = 15.6.
9                                        4
Hence, 6 terms are required to get a       Hence, 4 terms are required to get a
sum greater than 1 000 000.                sum greater than 12.

f   a = 0.1, r = 1 = 10,
0.1
5
S5 = 0.1 10 – 1
10 – 1
= 0.1 × 99999 = 1111.1.
9
Hence, 5 terms are required to get a
sum greater than 1000.
Solutions to Exercise 9I
1   Note that for sum to infinity, S =    a     c   a = 0.09, r = 0.01,
1–r
S = 0.09
a   a = 4, r = 1 = 0.25,                                1 – 0.01
4
4
= 0.09 = 9 = 1
S=                                                  0.99 99 11
1 – 0.25
= 4 = 5.3333                              d   a = 0.03, r = 0.1,
0.75
S = 0.03
1 – 0.1
b   a = 1000, r = 100 = 0.1,                          = 0.03 = 3
1000                                     0.9    90
1000
S=
1 – 0.1                                      But we must add 0.1 to give 1 + 3
10     90
= 1000 = 1111.11                                 = 9 + 3 = 12 = 2
0.9                                             90 90 90 15

c   a = 1, r = 0.9 = 0.9,                       e   a = 0.5, r = 0.1,
1
1                                        S = 0.5
S=                                                  1 – 0.1
1 – 0.9
= 0.5 = 5
= 1 = 10                                          0.9     9
0.1
f   a = 0.03, r = 0.1,
d   a = 6, r = 2 = 1 ,
6     3                              S = 0.03
1 – 0.1
S=    6
1 – 1/ 3                                       = 0.03 = 3
0.9    90
= 6 =9                                         But we must add 0.8 to give 8 + 3
2/ 3                                                                    10     90
= 72 + 3 = 75 = 5
e   a = 0.01, r = 0.001 = 0.1,                          90 90 90      6
0.01
S=  0.01                                    3
1 – 0.1                                  a   a = 20, r = 0.5,
= 0.01 = 0.0111                                            n
0.9                                         Sn = a 1 – r ,
1–r
2   A repeated decimal can be written as                            6

the sum of a geometric sequence where           S6 = 20 1 – 0.5
1 – 0.5
a is the highest place value digit in the
= 20 × 0.984375 = 39.375 cm.
decimal and each successive term is                             0.5
found by multiplying the previous term
by 0.1, i.e. r = 0.1.                       b   S=     20 = 20 = 40.
1 – 0.5 0.5
a   a = 0.1, r = 0.1,
4
S = 0.1
1 – 0.1                                 a   a = 30, r = 24 = 0.8,
30
= 0.1 = 1                                                n
0.9 9
Sn = a 1 – r ,
1–r
b   a = 0.06, r = 0.1,                                              6

S = 0.06                                        S6 = 30 1 – 0.8
1 – 0.1                                              1 – 0.8
= 0.06 = 6                                       = 30 × 0.737856 = 110.6784 cm.
0.9    90                                                  0.2
But we must add 0.1 to give 1 + 6           b   S=     30 = 30 = 150 cm.
10     90
1 – 0.8 0.2
= 9 + 6 = 15 = 1
90    90        90   6
5   a = 1, r = 0.75,                         7   Since the circumference of a circle = 2
S=      1    = 1 = 4.                        × π × (radius), a = 20 π .
1 – 0.75   0.25                         The radius of the next circle is half the
But we must add the starting height of       previous so the circumference will be
3 m to give us 7 m.                          half the previous too, giving r = 0.5.
Thus, S = 20π = 20π = 40 π .
6   a = 1, r = 0.9 = 0.9,                                   1 – 0.5    0.5
1
S=     1
1 – 0.9
= 1 = 10 kg weight loss.
0.1
Since he was originally 82 kg, the
lowest weight he can get down to is
72 kg.
Solutions to Exercise 9J
1   Set your graphics calculator to Seq         c   Enter in u(n) = 100 + (n – 1) × –5 for
mode for this question.                         the arithmetic sequence and v(n) = 100
a   Enter in u(n) = 32 + (n – 1) × 0.5 for          × –5n–1 for the geometric sequence to
the arithmetic sequence and v(n) = 32           get the graph below:
× 0.5n–1 for the geometric sequence to
get the graph below:

From the graph, it can be seen that the
arithmetic sequence is linear and is
From the graph, it can be seen that the         decreasing. The geometric sequence is
arithmetic sequence is linear and is            exponential and is decreasing.
increasing. The geometric sequence is
exponential and is decreasing.              d   Enter in u(n) = 100 + (n – 1) × 5 for
the arithmetic sequence and v(n) = 1 ×
b   Enter in u(n) = 1 + (n – 1) × 2 for the         5n–1 for the geometric sequence to get
arithmetic sequence and v(n) = 1 × 2n–1         the graph below:
for the geometric sequence to get the
graph below:

From the graph it can be seen that both
the arithmetic and geometric sequences
From the graph it can be seen that both         are increasing. However, while the
the arithmetic and geometric sequences          arithmetic sequence increases linearly,
are increasing. However, while the              the geometric sequence increases in an
arithmetic sequence increases linearly,         exponential manner such that, though
the geometric sequence increases in an          it is increasing at a slower rate
exponential manner such that though it          initially, its rate of increase rapidly
is increasing at a slower rate initially,       grows so that it soon overtakes the
its rate of increase rapidly grows so           arithmetic sequence.
that it soon overtakes the arithmetic
sequence.
Solutions to Multiple–choice questions
1   Option D is a geometric sequence with   10   tn = 4 × 1.5n–1 ⇒ B
a ratio of 3. ⇒ D
11   a = 2, r = 2.5 = 1.25,
2   Option E is an arithmetic sequence                       2
with a difference of –5. ⇒ E
5
S5 = 2 1.25 – 1
1.25 – 1
3   Arithmetic sequence, a = 2, d = 5.             = 2 × 2.052 = 16.4 ⇒ D
8th term = 2 + 5 × 7                                  0.25
= 2 + 35 = 37 ⇒ C
12   a = 25, r = 5 = 0.2,
25
4   a = 2, d = 5,                                     a
tn = 2 + (n – 1)5                            S=
1–r
= 2 + 5n – 5
= 25
= 5n – 3 ⇒ C                                 1 – 0.2
= 25 = 31.25 ⇒ C
5   t5 = 100 × (0.4)4                               0.8
= 100 × 0.0256
= 2.56 ⇒ B                           13   a = 8, r = 64 = 8
8
4–1=3
6   S8 = 8 (2 × 25 + (8 – 1)(–4))                8 = 23 = r3; r = 2.
2
= 4(50 – 28)                               t6 = 8 × 25
= 4 × 22 = 88 ⇒ B                             = 8 × 32 = 256 ⇒ D

14   a = 250, d = 26,
7   r = 1.5 = 5 ⇒ D                              t11 = 250 + (11 – 1)26
0.3
= 250 + 260 = 510
8   r=1+ R                                       Hence, the 11th term is the first to
100                                  exceed 500. ⇒ D
= 1 – 7.5
100
= 1 – 0.075 = 0.925 ⇒ C               15   r=1+ R
100
= 1 + 12 = 1.12 ⇒ D
9   t9 = 100 × 0.88                                      100
= 100 × 0.1678
= 16.8 ⇒ C
Chapter 10 – Difference equations
Solutions to Exercise 10A
1                                                 2
a   Since t1 = 2 and each term is 3 greater       a   Since tn+1 = 5tn + d, we can rearrange
than the last, the first 5 terms are 2, 5,        this to give us (substituting in
8, 11 and 14.                                     appropriate values from the sequence):
d = tn+1 – 5tn = 6 – 5(2) = 6 – 10 = –4.
b   Since t1 = 50 and each term is 5 less
than the last, the first 5 terms are 50,      b   Since Pn = rPn–1 + 2, we can rearrange
45, 40, 35 and 30.                                this to give us (substituting in
appropriate values from the sequence):
c   Since u1 = 27 and each term is 3 less             r = Pn – 2 = (7 – 2) = 5 = 1
than the last, the first 5 terms are 27,                Pn – 1       5       5
24, 21, 18 and 15.                            c   Since tn+1 = 0.25tn + 8, and t2 = 64, let
n = 1 to give 64 = 0.25t1 + 8.
d   Since t1 = 1 and each term is 3 times             This can be rearranged to give:
the last, the first 5 terms are 1, 3, 9, 27
and 81.                                           t1 = a = 64 – 8
0.25
=  56 = 56 × 4 = 224.
e   Since t1 = 1 and each term is –2 times                0.25
the last, the first 5 terms are 1, –2, 4,
–8 and 16.                                    d   Since vn = rvn–1 + d, we can use n = 2
to make the equation 650 = 500r + d,
f   Since t1 = 32 and each term is half the           and we can use n = 3 to make the
last, the first 5 terms are 32, 16, 8, 4          equation 875 = 650r + d.
and 2.                                            Subtracting the first equation from the
second we get 875 – 650 = 650r – 500r
g   Since R0 = 2 and each term is 4 times             + d – d, which gives us 225 = 150r or
the last plus 3, the first 5 terms are 2,         r = 1.5.
11, 47, 191 and 767.                              Substituting back into the first
equation, we get 650 = 500 × 1.5 + d,
h   Since t1 = 64 and each term is one                which gives us d = 650 – 500 × 1.5
quarter the last minus 1, the first 5               = 650 – 750 = –100
terms are 64, 15, 2.75, –0.3125 and           e   Since Tn+1 = rTn + d, we can use n = 1
–1.078 125.                                       to make the equation 100 = 1000r + d,
and we can use n = 2 to make the
i   Since v0 = 5 and each term is 4 times             equation –350 = 100r + d.
the last minus 5, the first 5 terms are 5,        Subtracting the second equation from
15, 55, 215 and 855.                              the first we get 100 – –350 = 1000r –
100r + d – d, which gives us 450 =
j   Since t1 = 2 and each term is 5 times             900r or r = 0.5.
the last minus 8, the first 5 terms are 2,        Substituting back into the first
2, 2, 2 and 2.                                    equation, we get 100 = 1000 × 0.5 + d,
which gives us d = 100 – 1000 × 0.5
= 100 – 500 = –400.
f   Since tn+1 = 4tn + 2, and t2 = 22, let us
make n = 1 to give us 22 = 4t1 + 2.
This can be rearranged to give:
t1 = a = 22 – 2 = 20 = 5
4        4
Solutions to Exercise 10B
1                                             3
a   d = 10 – 5 = 5.                           a   Geometric sequence, r = 12 = 2.
Thus, tn+1 = tn + 5, where t1 = 5.                                        6
Thus, tn+1 = 2tn, where t1 = 6.
b   d = 8 – 12 = –4.
Thus, tn+1 = tn – 4, where t1 = 12.       b   Arithmetic sequence, d = 12 – 10 = 2.
Thus, tn+1 = tn + 2, where t1 = 10.
c   d = 0.11 – 0.1 = 0.01.
Thus, tn+1 = tn + 0.01, where t1 = 0.1.   c   Arithmetic sequence, d =90 –100= –10.
Thus, tn+1 = tn – 10, where t1 = 100.
d   d = 30 – 15 = 15.
Thus, tn+1 = tn + 15, where t1 = 15.      d   Geometric sequence, r = 1.2 = 1.2.
1
Thus, tn+1 = 1.2tn, where t1 = 1.
e   d = 0 – –9 = 9.
Thus, tn+1 = tn + 9, where t1 = –9.
e   Geometric sequence, r = 1.62 = 0.9.
1.8
f   d = 1.2 – 1.1 = 0.1.                          Thus, tn+1 = 0.9tn, where t1 = 1.8.
Thus, tn+1 = tn + 0.1, where t1 = 1.1.

2                                             f   Geometric sequence, r = -16 = –2.
8
a   r = 4 = 2.                                    Thus, tn+1 = –2tn, where t1 = 8.
2
Thus, tn+1 = 2tn, where t1 = 2.

b   r = 1.1 = 1.1.
1
Thus, tn+1 = 1.1tn, where t1 = 1.

c   r = 80 = 0.8.
100
Thus, tn+1 = 0.8tn, where t1 = 100.

d   r = 1 = 10.
0.1
Thus, tn+1 = 10tn, where t1 = 0.1.

e   r = 144 = 12.
12
Thus, tn+1 = 12tn, where t1 = 12.

f   r = -6 = 1.5.
-4
Thus, tn+1 = 1.5tn, where t1 = –4.
Solutions to Exercise 10C
1                                              3
a   Yes, the equation links two terms one      a   r = 3, d = –4 ⇒ neither
step apart.
b   r = 1, d = 2 ⇒ arithmetic
b   Yes, the equation links two terms one
step apart.                                c   r = 5, d = 0 ⇒ geometric

c   No, the equation links a term to the       d   r = 1, d = –3 ⇒ arithmetic
value of n and not the value of the
preceding or next term.                    e   r = 5, d = –3 ⇒ neither

d   No, the equation links two terms two       f   r = –0.5, d = 0 ⇒ geometric
steps apart.
g   r = –1, d = 2 ⇒ neither
e   No, the equation links two terms two
steps apart.                               h   The equation is not a first-order
equation since it links two terms that
f   No, the equation only contains one             are two steps apart.
term and doesn’t link it to the value of
the preceding or next term.                i   tn+1 + 2tn = –2, tn+1 = –2tn – 2,
thus r = –2, d = –2 ⇒ neither
g   Yes, the equation links two terms one
step apart.

h   Yes, the equation links two terms one
step apart.

2   Note that tn+1 = r × tn + d, and a = t1.
The values of r, d and a can thus be
a   r = 3, d = –4, a = 3

b   r = 1, d = 2, a = 4

c   r = 5, d = 0, a = 1

d   r = 1, d = 2, a = 0

e   r = 5, d = –3, a = –1

f   r = –0.5, d = 0, a = 4

g   r = –1, d = 2, a = 3

h   2tn+1 + tn = 4, 2tn+1 = 4 – tn,
tn+1 = 2 – 0.5tn,
thus r = –0.5, d = 2, a = 3

i   tn+1 + 2tn = –2, tn+1 = –2tn – 2,
thus r = –2, d = –2, a = 0
Solutions to Exercise 10D
1   An arithmetic sequence has r = 1.   2   The general arithmetic sequence
a   r = 5, hence not arithmetic             equation is tn = a + (n – 1)d.
If we know a and d, we can solve the
b   r = 1, hence arithmetic                 difference equation to give us the
general equation for the nth term of
c   r = 5, hence not arithmetic             each sequence.

d   r = 1, hence arithmetic             a   a = 5, d = 1,
tn = 5 + (n – 1)(1)
e   r = –1, hence not arithmetic               =5+n–1=4+n

f   r = 1, hence arithmetic             b   a = 25, d = –4,
tn = 25 + (n – 1)(–4)
g   r = 1, hence arithmetic                    = 25 – 4n + 4 = 29 – 4n

h   r = 1, hence arithmetic             c   a = 5, d = 3,
tn = 5 + (n – 1)(3)
= 5 + 3n – 3 = 2 + 3n

d   a = 10, d = –2,
tn = 10 + (n – 1)(–2)
= 10 – 2n + 2 = 12 – 2n

e   a = 4, d = –1,
tn = 4 + (n – 1)(–1)
=4–n+1=5–n

f   a = 10, d = –5,
tn = 10 + (n – 1)(–5)
= 10 – 5n + 5 = 15 – 5n
Solutions to Exercise 10E
1   A geometric sequence has d = 0.   2   The general geometric sequence
a   d = –4, hence not geometric           equation is tn = a × rn–1.
If we know a and r, we can solve the
b   d = –4, hence not geometric           difference equation to give us the
general equation for the nth term of each
c   d = 0, hence geometric                sequence.

d   d = 0, hence geometric            a   a = 10, r = 2,
tn = a × rn–1 = 10 × 2n–1
e   d = 6, hence not geometric
b   a = 8, r = 1.5,
f   d = 0, hence geometric                tn = a × rn–1 = 8 × 1.5n–1

g   d = 0, hence geometric            c   a = 12, r = –2,
tn = a × rn–1 = 12 ×(–2)n–1
h   d = 2, hence not geometric
d   a = 10, r = 0.9,
tn = a × rn–1 = 10 ×0.9n–1

e   a = 4, r = –1,
tn = a × rn–1 = 4 × (–1)n–1

f   a = 10, r = 5,
tn = a × rn–1 = 10 × 5n–1
Solutions to Exercise 10F
1   The general first–order sequence
equation is
tn = a × rn–1 + d(rn–1 – 1)/(r– 1).
If we know a, d and r, we can solve
the difference equation to give us the
general equation for the nth term of
each sequence.

a   a = 32, d = –8, r = 1.5,
tn = a × rn–1 + d(rn–1 – 1) / (r – 1)
n–1
= 32 × 1.5n–1 + -8(1.5      – 1)
1.5 – 1
n–1
= 32 × 1.5n–1 – -8(1.5     – 1)
0.5
= 32 × 1.5n–1 – 16(1.5n–1 – 1)
= 32 × 1.5n–1 – 16 × 1.5n–1 + 16
= 16 × 1.5n–1 + 16

b   a = 20, d = 14, r = 0.5,
tn = a × rn–1 + d(rn–1 – 1) / (r – 1)
n–1
= 20 × 0.5n–1 + 14(0.5     – 1)
0.5 – 1
n–1
= 20 × 0.5n–1 + 14(0.5     – 1)
-0.5
= 20 × 0.5n–1 – 28(0.5n–1 – 1)
= 20 × 0.5n–1 – 28 × 0.5n–1 + 28
= 28 – 8 × 0.5n–1

c   a = 20, d = –10, r = 0.5,
tn = a × rn–1 + d(rn–1 – 1) / (r – 1)
n–1
= 20 × rn–1 + -10(0.5    – 1)
0.5 – 1
n–1
= 20 × 0.5n–1 – 10(0.5     – 1)
-0.5
= 20 × 0.5n–1 + 20(0.5n–1 – 1)
= 20 × 0.5n–1 + 20 × 0.5n–1 – 20
= 40 × 0.5n–1 – 20
Solutions to Exercise 10G
1                                          D   a = 100, r = 1.1, d = 0
A   a = 35, r = 1, d = –5                      a d = 0, hence geometric sequence
a r = 1, hence arithmetic sequence         b First 5 terms = 100, 110, 121, 133.1,
b First 5 terms = 35, 30, 25, 20, 15          146.41
c Using the StatPlot function of the       c Using the StatPlot function of the
graphics calculator, we get the            graphics calculator, we get the
following graph:                           following graph:

d tn = a + (n – 1)d
= 35 + (n – 1)(–5)
= 35 – 5n + 5 = 40 – 5n                  d tn = a × rn–1
= 100 × 1.1n–1
B   a = 64, r = 0.25, d = 0
a d = 0, hence geometric sequence      E   a = 0, r = 1, d = 0.5
b First 5 terms = 64, 16, 4, 1, 0.25       a r = 1, hence arithmetic sequence
c Using the StatPlot function of the       b First 5 terms = 0, 0.5, 1.0, 1.5, 2.0
graphics calculator, we get the         c Using the StatPlot function of the
following graph:                           graphics calculator, we get the
following graph:

d tn = a × rn–1
= 64 × 0.25n–1                           d tn = a + (n – 1)d
= 0 + (n – 1)(0.5) = 0.5n – 0.5
C   a = 0, r = 1, d = 5
a r = 1, hence arithmetic sequence     F   a = 32, r = 1.5, d = –8
b First 5 terms = 0, 5, 10, 15, 20         a Neither arithmetic or geometric
c Using the StatPlot function of the       b First 5 terms = 32, 40, 52, 70, 97
graphics calculator, we get the            c Using the StatPlot function of the
following graph:                              graphics calculator, we get the
following graph:

d tn = a + (n – 1)d
= 0 + (n – 1)(5) = 5n – 5
G   a = 20, r = 0.5, d = 10                  J   a = 1, r = 0.5, d = 0.5
a Neither arithmetic or geometric            a Neither arithmetic or geometric
b First 5 terms = 20, 20, 20, 20, 20         b First 5 terms = 1, 1, 1, 1, 1
c Using the StatPlot function of the         c Using the StatPlot function of the
graphics calculator, we get the              graphics calculator, we get the
following graph:                             following graph:

H   a = 20, r = 0.5, d = 14                  K   a = 0.1, r = 10, d = 0
a Neither arithmetic or geometric            a d = 0, hence geometric sequence
b First 5 terms = 20, 24, 26, 27, 27.5       b First 5 terms = 0.1, 1, 10, 100, 1000
c Using the StatPlot function of the         c Using the StatPlot function of the
graphics calculator, we get the           graphics calculator, we get the
following graph:                          following graph:

I   a = 20, r = 0.5, d = –10                     d tn = a × rn–1
a Neither arithmetic or geometric              = 0.1 × 10n–1
b First 5 terms = 20, 0, –10, –15,
–17.5                                 L   a = 1, r = 2, d = –2
c Using the StatPlot function of the         a Neither arithmetic or geometric
graphics calculator, we get the           b First 5 terms = 1, 0, –2, –6, –14
following graph:                          c Using the StatPlot function of the
graphics calculator, we get the
following graph:
Solutions to Exercise 10I
1   Since the Fibonacci difference                  4
equation links terms that are two steps         a   i t17 = t15 + t16 = 610 + 987 = 1597
apart, it is a second–order difference              ii t14 = t16 – t15 = 987 – 610 = 377
equation.
b   i t 2 + t 3 = t4
2                                                       ii t200 – t198 = t199
a   Starting with 1, 1, the first 10 terms in           iii t3 + 2t4 + t5 = (t3 + t4) + (t4 + t5)
the sequence can be seen to be 1, 1, 2,                 = t 5 + t6
3, 5, 8, 13, 21, 34, 55.
5
b   Using the Seq function of the graphics          a   Starting with 1, 3, the first 10 terms in
calculator, we can obtain the following             the sequence can be seen to be 1, 3, 4,
graph:                                              7, 11, 18, 29, 47, 76, 123.

b   Reading off the Table function of the
graphics calculator, we see that the
first term to exceed 10 000 has a value
of 15 127, which is the 20th term.

c   Reading off the Table function of the
c   Reading off the Table function of the               graphics calculator, we see that
graphics calculator, we see that the                t20 = 15 127, t21 = 24 476, t22 = 39 603.
first term to exceed 50 000 has a value             Since 15 127 + 24 476 =39 603, this
of 75 052, which is the 25th term.                  shows that t20 + t21 = t22.
d   Reading off the Table function of the           d   Reading off the Table function of the
graphics calculator, we see that                    graphics calculator, we see that
t20 = 6765, t21 = 10 946, t22 = 17 711.             t10 = 123, t11 = 199, t12 = 322.
Since 6765 + 10 946 = 17 711, this                  Since 322 – 123 = 199, this shows that
shows that t20 + t21 = t22.                         t12 – t10 = t11.
e   Reading off the Table function of the
⎡            8          8
⎤
graphics calculator, we see that                6   t8= 1 ⎢⎛ 1 + 5 ⎞ − ⎛ 1 −
⎜      ⎟ ⎜          5⎞
⎟     ⎥ =21.04
5 ⎢⎜ 2 ⎟ ⎜ 2            ⎟     ⎥
⎣⎝        ⎠      ⎝    ⎠
t10 = 55, t11 = 89, t12 = 144.                                                         ⎦
Since 144 – 55 = 89, this shows that                This shows t8 of the Fibonacci
t12 – t10 = t11.                                    sequence to be 21.
3                                                                       4                  4
a   i t11 = t9 + t10 = 34 + 55 = 89                          ⎛1 + 5 ⎞     ⎛1 − 5 ⎞
t4 = ⎜      ⎟     ⎜      ⎟ = 7.01
⎜ 2 ⎟ −⎜ 2 ⎟
7
ii t8 = t9 – t7 = 34 – 13 = 21
⎝      ⎠     ⎝      ⎠
iii t5 + t6 = t7 = 13
This shows t4 of the Lucas sequence to
be 7.
b   i t27 + t28 = t29
ii t102 – t100 = t101
iii t31 + t32 + (t35 – t33) = t33 + t34 = t35
iv t11 + 2t12 + t13 = (t11+t12) + (t12 + t13)
= t13 + t14 = t15
Solutions to Multiple–choice questions
1   Since tn+1 = n doesn’t link two terms         9    Since d = 0 and r < 1, a decreasing
that are one step apart, it is not a first–        geometric sequence is created. ⇒ B
order difference equation. ⇒ C
10   a = 55, r = 1, d = –5, hence arithmetic
2   Since tn+1 = –1 × tn + 3, where t1 = 5,            sequence.
r = –1, d = 3, a = 5. ⇒ A                          tn = 55 + (n – 1)(–5)
= 55 – 5n + 5 = 60 – 5n ⇒ E
3   a = 4; the next term will be 2 × 4 – 5 = 3;
the term after will be 2 × 3 – 5 = 1.         11   a = 10, r = 1.2, d = 0, hence geometric
⇒B                                                 sequence.
tn = a × rn–1 = 10 × 1.2n–1 ⇒ B
4   a = 1; the next term will be –4 × 1 + 2
= –2, followed by –4 × –2 + 2 = 10,           12   The house value increases by 10%
followed by –4 × 10 + 2 = –38. ⇒ A                 each year, so is 110% or 1.10 times the
value of the house in the previous year.
5   a = 5, r = –2, d = 10.                             The initial value (H1) = \$320000.
The general first–order sequence                   Thus, Hn+1 = 1.10 × Hn, where
equation is:                                       H1 = \$320 000. ⇒ C
n–1
tn = a × rn–1 + d r     –1
r–1                      13   The yabbie number increases by 35%
6                         each year, so is 135% or 1.35 times the
Thus, t7 = 5 × 26 + 10 2 – 1                       number of yabbies in the previous
-2 – 1
63                            year.
= 5 × 64 + 10 ×                                 200 yabbies are taken each year and
-3
= 320 – 210 = 110 ⇒ D                            the initial value = 900 yabbies.
Thus Yn+1 = 1.35 × Yn – 200, where
6   t2 of equation A = –14                             Y1 = 900. ⇒ B
t2 of equation B =2, t3 of equation B = 4
t2 of equation C = 4                          14   Adding Fibonacci sequence terms
t2 of equation D = –20.                            together until we get the 19th term, we
t2 of equation E =2, t3 of equation E = 1.         find that t19 = 4181. ⇒ D
⇒B
15   By the formula tn = tn–2 + tn–1, we can
7   If t1 = 8 and t2 = 20, then 20 = a × 8 + 4,        see that t20 = t18 + t19. ⇒ D
8a = 16, a = 2. ⇒ D

8   If t1 = 8 and t2 = 14, then 14 = 2 × 8 + d,
d = 14 – 16 = –2. ⇒ A
Chapter 11 – Revision: Number
patterns and applications
Solutions to Multiple-choice questions
1   Sequence B does not have a common      9    a = –5.4; r = 1.8 = – 1
difference between successive terms,                      -5.4     3
so is not arithmetic. ⇒ B                   S= a
1–r
-5.4
2   d = 12 – 15 = –3 ⇒ A                          =
1– –1
3
3   r = -15 = –3 ⇒ A                                 -5.4 = –4.05 ⇒ B
5                                        =
4/ 3
4   d = 33 – 36 = –3; a = 36.              10   a = 12; d = –2.
t8 = a + 7 × d                              S20 = 20 (2 × 12 + (20 – 1)(–2))
2
= 36 + 7 × –3
= 10(24 + –38)
= 36 – 21 = 15 ⇒ C
= 10 × –14 = –140 ⇒ A
5   t3 – t1 = 2d = 10 – 28 = –18.
11   a = 3; r = 2.
Thus, d = –9; a = 28.
Multiplying each successive term by 2,
4
S4 = (2 × 28 + (4 – 1)(–9))                 the value of the first term to exceed
2
3000 is 3072, which is the 11th term.
= 2(56 – 27)
⇒C
= 2 × 29 = 58 ⇒ C
12   Since t3 and t5 are positive and we are
6   a = 100; r = 50 = 0.5.                      told that r is negative, every odd-
100
n                               numbered term will be positive and
S5 = a 1 – r                                every even-numbered will be negative.
1–r
5                          Thus, we can be certain that the second
= 100 1 – 0.5                             term (which will be a negative
1 – 0.5                            number), will definitely be less than
0.96875
= 100 ×           = 193.75 ⇒ B            the first term (which will be a positive
0.5                          number). ⇒ D
7   a = 10; r = 1.5.                       13   Sequence C could be an arithmetic
t6 = a × rn–1                               sequence with a common difference of
= 10 × 1.55 = 75.9 ⇒ C                   100. ⇒ C
8   a = 10; d = 2.5.
14   Multiplying 3 by 0.6 five times give us
S8 = 8 (2 × 10 + (8 – 1)2.5)                0.23328, which to two decimal places
2
= 4(20 +17.5)                             is 0.23. ⇒ B
= 4 × 37.5 = 150 ⇒ B
15   162 = 81; 5 – 1 = 4; 81 = r4 = 34.
2
r = 3; a = 2.
t3 = 2 × 32
= 2 × 9 = 18 ⇒ D
16   a = 0.5; r = -1/ 4 = –0.5.                 24   a = 8; r = 0.5.
1/ 2                              S= a
a                                            1–r
S=
1–r                                       =    8    = 8 = 16
0.5                                         1 – 0.5   0.5
=          = 0.5                              But we must add the original starting
1 – -0.5   1.5
1 ⇒D                                        height of 20 to give us 36 cm. ⇒ B
=
3
25   a = 15 000; r = 1.05.
17   a = 40; r = 20 = 0.5 ⇒ D                                n
40                                  S5 = a r – 1
r–1
18   The arithmetic sequence’s first 6                                 5
15 000(1.05 – 1)
values are –192, –128, –64, 0, 64, 128.            =
1.05 – 1
The tn = (–2)n sequence’s first 6 values
= 15 000 × 0.2763
are –2, 4, –8, 16, –32, 64.                                       0.05
Thus, 64 is the lowest positive term              = \$82 844 ⇒ D
common to both sequences. ⇒ B
26   a = 15 000; r = 0.98.
19   The number of goats increases by 8%             Vn = 15 000 × (0.98)n–1
each year.                                         = 15 000 × (1 – 0.02)n–1 ⇒ A
Thus, in 2007 there will 1.08 × 600
= 648 goats, and in 2008 there will be     27   a = 30 000; d = 800.
1.08 × 648 = 700 goats. ⇒ D
S15 = 15 (2 × 30 000 + (15 – 1)800)
2
20   a = 25 000; r = 0.8.                               = 7.5(60 000 + 14 × 800)
t6 = 25 000 × 0.85                                 = 7.5 × 71 200 = 534 000 ⇒ C
= 25 000 × 0.32768
= 8192, is closest to 8000 ⇒ B          28   a = 2000; r = 1.035.
n

21   a = 1000; r = 1.08.                             S5 = a r – 1
r–1
Multiplying 1000 by 1.08 successive                              5
2000(1.035 – 1)
times, we find the value of the first             =
1.035 – 1
term to exceed 3000 to be 3172.17,
which is the 16th term.                           = 2000 × 0.1877 = 10 725 ⇒ D
0.035
This represents 15 years since the
money was originally invested. ⇒ B         29   Multiplying each term by 3 and
subtracting 1, starting with 4, gives us
22   a = 18 000; d = –120.                           4, 11, 32. ⇒ A
t181 = 18 000 + (181 – 1)(–120)
= 18 000 – 120 × 180                    30   Multiplying each term by –2 and
= 18 000 – 21 600 = –3600                    adding 100, starting with 5, gives us
This means that all the water has flown         5, 90, –80. ⇒ B
out of the tank, meaning 0 litres are
left. ⇒ A                                  31   t2 of equation A is 0.
t2 of equation B is 0.
23   a = 20 million; r = 1.04.                       t2 of equation C is 10, t3 of equation C
Multiplying 20 million by 1.04                  is 5.
successive times, we find the value of          t2 of equation D is 10, t3 of equation D
the first term to exceed 40 million to          is 0.
be 40.52, which is the 19th term.               t2 of equation E is 10, t3 of equation E
This represents 18 years since the              is 20. ⇒ E
population was 20 million. ⇒ B
32   Multiplying 4 by 3 and adding 6 gives      40   Multiplying 6 by 2 and subtracting 6
us 18 as T2.                                    will give us 6 again.
Multiplying 18 by 3 and adding 6                Thus, the sequence will be 6, 6, 6, 6, 6.
gives us 60 as T3. ⇒ D                          ⇒B

33   Since d = 0.5, the sequence is             41   The sequence required a multiplication
arithmetic and is increasing, since d is        by 0.5 and adding 2.
positive. ⇒ B                                   If we subtract 2 from 12, then divide it
by 0.5 we get 20 as f2.
34   a = 15; d = –5.                                 If we subtract 2 from 20, then divide it
tn = 15 + (n – 1)(–5)                           by 0.5 we get 36 as f1. ⇒ E
= 15 – 5n + 5
= 20 – 5n ⇒ E                           42   a = 3, r = 2.
n
S4 = a r – 1
35   a = 3; d = 4.                                           r–1
tn = 3 + (n – 1)4                                        4
= 3 + 4n – 4                                   =  32 –1
2–1
= 4n – 1 ⇒ B
= 3 × 15 = 45 ⇒ C
1
36   Since T1 = 2 and T2 = 10, we can
substitute these into the equation to      43   a = 10 000, r = 1.065, d = –1000.
give us 10 = 2 × 2 + h;                         Pn = 1.065 × Pn–1 – 1000 ⇒ A
h = 10 – 4 = 6 ⇒ B
44   a = 7500, r = 1.00 – 0.08 = 0.92.
37   a = 3; d = –7.                                  Tn = 0.92 × Tn–1, where T1 = 7500 ⇒ B
tn = 3 + (n – 1)(–7)
= 3 – 7n + 7                            45   a = 1500, r = 1.10, d = –150.
= 10 – 7n ⇒ A                                tn = 1.10 × tn–1 – 150
t2 = 1.10 × 1500 – 150 = 1500
38   Multiplying 19 by 4 and subtracting 5           t3 = 1.10 × 1500 – 150 = 1500 ⇒ E
gives us 71 as X3. ⇒ B
46   a = 1500, r = 1.10, d = –150.
39   Since 0.95 is less than 1, the geometric        tn = 1.10 × tn–1 – 150
sequence will be decreasing. ⇒ C                tn+1 = 1.10 × tn – 150, where t1 = 1500
⇒A
Chapter 12 – Geometry
Solutions to Exercise 12A
1   d and b can be seen to be vertically        c   y is supplementary to the 80 degree
opposite each other. ⇒ E                        angle and so is 180 – 80 = 100
degrees.
2   d and a can be seen to be                       2z is corresponding to the 80 degree
supplementary. ⇒ D                              angle and so is 80 degrees, meaning z
is 40 degrees.
3   c and h can be seen to be cointerior.
⇒B                                          d   x is corresponding to the 50 degree
angle and so is 50 degrees.
4   b and f can be seen to be                       y is supplementary to the 50 degree
corresponding. ⇒ C                              angle and so is 180 – 50 = 130 degrees.
z is corresponding to y and so is 130
5   c and e can be seen to be alternate.            degrees.
⇒A
e   x is cointerior to the 120 degree angle
6                                                   and so is 180 – 120 = 60 degrees.
a   x is corresponding to the 70 degree             y is alternate to the 120 degree angle
angle and so is 70 degrees.                     and so is 120 degrees.
y is supplementary to the 70 degree
and so is 180 – 70 = 110 degrees.           f   2x – 40 and x + 40 are corresponding
z is alternate to the 70 degree angle and       to each other.
so is 70 degrees.                               Thus, 2x – 40 = x + 40,
x = 40 + 40 = 80 degrees.
b   x is alternate to the 40 degree angle
and so is 40 degrees.
y is cointerior to the 40 degree angle
and so is 180 – 40 = 140 degrees.
z is corresponding to the 40 degree
angle and so is 40 degrees.
Solutions to Exercise 12B
1                                              e   x = 40 since it corresponds to the 40
a   x = 180 – 80 – 50 = 50                         degree angle.
y = 80 + 50 = 130                              y = 40 since it corresponds to the other
base angle of the isosceles triangle.
b   2x = 180 – 30 = 150;                           z = 180 – 40 – 40 = 100
x = 75                                        w = 180 – 40 – 40 = 100

c   x = 70 + 30 = 100                          f   a = 75 since it corresponds to the other
y = 180 – 70 = 110                             base angle of the isosceles triangle.
b = 180 – 75 – 75 = 30
d   x = 50 since it corresponds to the other       c = 180 – 40 – 75 – 30 = 35
base angle of the isosceles triangle.
y = 180 – 50 – 50 = 80                     g   2y = 180 – 90 = 90;
y = 45
Angle ECD = 180 – 45 = 135.
2x = 180 – 135 = 45;
x = 22.5
Solutions to Exercise 12C
1                                            5
a   3 sides = triangle                       a   S = 180n – 360
= 180 × 7 – 360
b   4 sides = square                               = 1260 – 360 = 900

c   5 sides = pentagon                       b   S = 180n – 360
= 180 × 6 – 360
d   6 sides = hexagon                              = 1080 – 360 = 720

e   8 sides = octagon                        c   S = 180n – 360
= 180 × 8 – 360
2                                                  = 1440 – 360 = 1080
a   The 4 angles at O are all equal and
sum to 360 degrees. They are therefore   6   S = 180n – 360.
90 degrees each.                             1260 = 180n – 360.
180n = 1620; n = 9 sides.
b   i BAD is a right–angled isosceles
triangle.                             7
ii AOB is a right–angled isosceles       a   r = 2 cm, area = π × r2
triangle.                                   = π × 4 = 12.6 cm2

3                                            b   Area = 12.6 = 2.1 cm2
a   i x = 360 = 72                                           6
5
ii 2y = 180 – 72 = 108;                  8   a = 180 × 6 – 360 = 720 = 120
y = 54                                            6          6
b= 180 × 3 – 360 = 180 = 60
b   S = 180n – 360                                        3          3
= 180 × 5 – 360                            Since a and b are supplementary
= 900 – 360 = 540                          angles – that is, a + b = 180 – at any
vertex the sum of angles will be 360
4                                                degrees, meaning the pattern
tessellates perfectly.
a   x = 360 = 60
6
180n – 360
9   x = 135 =
b   2y = 180 – 60 = 120;                                         n
y = 60                                      135n = 180n – 360
360 = 45n; n = 8 sides.
Solutions to Exercise 12D
1
4    VW = 2.4 2 + 4.6 2 = 5.76 + 21.16
a   x = 10 2 + 6 2 = 100 + 36 = 11.7cm                = 5.19 cm

b   x=   5 2 + 112 = 25 + 121 = 12.1 cm       5    AD = 32 2 − 10 2 = 1024 − 100
= 30.40 cm
c   x = 10 2 + 3 2 = 100 + 9 = 10.4 cm
6    Height = 18 2 − 7 2 = 324 − 49
d   x=   9 2 − 7 2 = 81 − 49 = 5.7 cm                    = 16.58 m

7    Diagonal = 40 2 + 9 2 = 1600 + 81
e   x = 33 2 + 44 2 = 1089 + 1936
= 41 m
= 55 cm
8    CB = 2 × 14 2 − 8 2 = 196 − 64
f   x = 15 − 12 = 225 − 144 = 9 cm
2     2
= 2 × 11.49 = 23.0
2
9    Distance = 14 2 − 2 2 = 196 − 4
a   x = 4.8 2 + 3.2 2 = 23.04 + 10.24                       = 13.9 cm
= 5.77 cm
10   2x2 = 182 = 324;
b   x = 2.8 2 + 6.2 2 = 7.84 + 38.44                x2 = 162, x = 12.73 cm
= 6.80 cm
11   XY = 170 2 − 90 2 = 28900 − 8100
c   x = 9.8 2 − 5.2 2 = 96.04 − 27.04                 = 144.2 m
= 8.31 cm
12   Area = 169 cm2, so side length = 13 cm.
d   Let us define the hypotenuse of the            Diagonal = 13 2 + 13 2 = 169 + 169
triangle with smaller sides of length 3
= 18.38 cm
and 4 as y.
y=    4 2 + 3 2 = 16 + 9 = 5 cm           13
x = 5 2 + 3.5 2 = 25 + 12.25              a    2x2 = (8 2 )2 = 128; x2 = 64
= 6.10cm                                     Area = x2 = 64 cm2

3                                             b    2x2 = 82 = 64; x2 = 32.
a   x2 = 36 + 16 = 52;                             Area = x2 = 32 cm2
x = 7.21
14   AB = 20 2 − 4 2 = 400 − 16
2
b   x = 81 – 52 = 29;                                 = 19.6 cm
x = 5.39

c   x2 = 4.62 + 37.21 = 41.83;                15   CA = CE = 2 2 + 2 2 = 4 + 4 =2.8
x = 6.47                                      DE = CE – CD
= 2.8 – 2 = 0.8 cm

16   Side length = 12 + 12 = 1 + 1 = 1.41
Area = 1.412 = 2 cm2
Solutions to Exercise 12E
1                           3    x = 0.3 × 33 = 41.25 m
0.24
a   x = 5 × 9 = 11.25
4
4    Height = 20 × 15 = 7.5 m
40
b   x= 6 ×5=3
10
5    Height = 1 × 300 = 15 m
20
c   x = 13 × 8 = 26
12        3
6    CY = 15 × 45 = 22.5 m
30
d   x = 14 × 10 = 11.7
12
7    h = 2 × 32 = 10.32 m
6.2
e   x = 6 × 10 = 7.5
8
x    20 – x
8      =
f   x= 2 ×6=3                    4      8
4                         8x = 20 – x
4
g   x = 12 × 24 – 12             3x = 20; x = 6.7 cm
16
= 18 – 12 = 6
9    Height = 12 × 30 + 80
100
h   x= 2 ×5–2                               = 3.6 + 80 = 83.6 cm
3
= 10 – 2 = 4           10      x = 1.5
3       3                1.3 – x 0.8
0.8x = 1.5(1.3 – x),
x      2                       0.8x = 1.95 – 1.5x,
i         =
x+8 8                            2.3x = 1.95; x = 0.85 m
8x = 2(x + 8)
6x = 16; x = 8          11   Height = 103.5 × (3 – 1.7) + 1.7
3                             3.5
= 38.44 + 1.7 = 40.14 m
x
j           = 10
x + 1.5    12
12   x = 9 × 8 = 7.2 m
12x = 10(x + 1.5)                  10
2x = 15; x = 7.5
13   Height = 1.1 – 0.6 × 8 + 0.6
3
2   AC = 15 × 14 = 17.5
12                          = 1.33 + 0.6 = 1.93 m
AE = 12 ,
AE + 4    15
15AE = 12(AE + 4),
3AE = 48; AE = 16
AB = AE + EB
= 16 + 4 = 20
Solutions to Exercise 12 F
1                                                5
a   V = π r2h                                    a    V = 2 × π r3
= π × 6.32 × 2.1                                     3
= 261.85 cm3                                      = 2×    π × 123 = 3619.11 cm3
3
b   V = 2.1 × 8.3 × 12.2
= 212.65 cm3                               b    V = 2 × π r3
3

c   V = 2.8 × 6.2
= 2×    π × 163 = 8578.64 mm3
3
= 17.36 cm3
c    V = 2 × π r3
d   V = π r2h                                             3
= π × 2.32 × 4.8                                  = 2×    π × 163 = 8578.64 mm3
3
= 79.77 cm3

2   Surface area                                 d    V = 2 × π r3
3
= bh + bl + hl + l b 2 + h 2                      = 2×    π × 153 = 7068.58 cm3
3
= 4 × 4 + 2 × 4 ×12 +12 × 4 2 + 4 2
= 16 + 96 + 12 × 5.66                       6    V = 1 × 18 × 15 × 20 = 1800 m3
3
= 112 + 67.88 = 179.88 cm2
Volume = 4 × 4 × 12
2                            7    V = 1 × 275 × 275 × 175 = 4 411 458 m3
2                         3
= 8 × 12 = 96.00 cm
8    VX2 = 122 + 52 = 144 + 25
3
= 169 = 132; VX = 13
a   Surface area = 2(13 × 4 + 13 × 3 + 3 × 4)
= 2(52 + 39 + 12)                    Surface area = 4 × 10 × 13 + 102
2
= 206 cm2                                       = 2 × 130 + 100
= 360 cm2
b   Volume = 3 × 4 × 13 = 156 cm3
Volume = 1 × 10 × 10 × 12 = 400 cm3
3
4
a   V = 4 × π r3                                 9    Surface area = 4 π r2 + 2 π rh
3                                                          = 4× π ×4+2× π ×2 ×5
= 4×   π × 43 = 268.08 mm3                                   = 113.10 cm2
3
Volume = 4 × π r3 + π r2h
3
b   V = 4 × π r3                                                = 4 ×π × 8 +       π ×4×5
3
3
= 4×   π × 11.5 = 6370.63 cm
3                3
= 96.34 cm     3
3

10
c   V = 4 × π r3                                 a    Surface area
3
= 4×   π × 3.8 = 229.85 m
3            3                  = 2(4 × 3 ) + 2(5 × 2) + 10(4 + 3 + 2 + 5 + 2)
3                                                        2
= 12 + 20 + 160 = 192 m2
d   V = 4 × π r3
3                                        b    Volume = 10(4 × 3 ) + 10(5 × 2)
= 4×   π × 7.5 = 1767.15 cm
3                3                                 2
3                                                      = 10(6 + 10) = 160 m3
11                                                      13
a    Triangle slope height                              a    Surface area
=    24 + 7 = 576 + 49 = 25
2           2
= π × 52 + 1 × 2 × π × 5 × 20 +
2
Surface area = 4(25 × 14 ) + 5 × 142                        20(5 + 10 + 5) + 2(10 × 5)
2
= 25 π + 100 π + 400 + 100
= 700 + 980 = 1680 cm2
= 125 π + 500
= 892.70 cm2
b    Volume = 1 × 142 × 24 + 143
3
= 1568 + 2744 = 4312 cm3                  b    Volume = 1 × π × 52 × 20 + 10 × 5 × 20
2
12                                                                   = 250 π + 1000
a    Triangle hypotenuse                                             = 1785.40 cm3
= 3 2 + 4 2 = 9 + 16 = 5
Surface area
= 2(3 × 4 ) + 2(4 × 4) + 10(5 + 3 + 4 + 4 + 4)
2
= 12 + 32 + 200 = 244 cm2

b    Volume = 10(3 × 4 + 42)
2
= 10(6 + 16) = 220 cm3
Solutions to Exercise 12G
1   Area = 6 × 1.22 = 8.64 cm2                 10   Volume ratio = (Length ratio)3
= 53:103 – 53
2
Area = 20 × ⎛ 5 ⎞ = 55.6 cm2
= 125:875 = 1:7
2               ⎜ 3⎟
⎝ ⎠
11
3
2
Area = 30 / ⎛ 3 ⎞ = 13.33 = 40 cm2         a    Area ratio = (Length ratio)2
⎜ 2⎟
⎝ ⎠              3                             = 12:42 = 1:16

20           2                 b    Volume ratio = (Length ratio)3
4   Area =        2 = 4.54 cm
2.1                                               = 12:43 = 1:64

5                                              12   Volume ratio = (Length ratio)3
= 13:23 – 13 = 1:7
a   BF =       2 2 − 12 = 4 − 1 = 1.73 cm           Thus, water fills one eighth of the
container.
b   2 = 1.73 × a                                    Volume = 100 = 12.5 m3
2                                                     8
Thus, a = 2 × 2
1.73
4 = 2.31 cm                  13   i Area ratio = (Length ratio)2
=                                                 = 22:52 = 4:25
1.73
ii Volume ratio = (Length ratio)3
c   Length scale factor (determined by                              = 23:53 = 8:125
ratio of vertical heights of the
14
triangles) = 2/ 3 .
a    Length ratio = 15:45 = 1:3
4
Area ratio = (2/ 3 )2 =
3
b    Area ratio = (Length ratio)2
= 12:32 = 1:9
6   Area ratio = 16:25
= 42:52 = (Length ratio)2
c    Volume ratio = (Length ratio)3
Length ratio = 4:5
= 13:33 = 1:27
7   Area ratio = 81:144
15
= 92:122 = (Length ratio)2
a    Volume ratio = 1:27
Length ratio = 9:12
= 13/2:93/2 = (Area ratio)3/2
Side length = 9 × 30 = 22.5 cm                  Area ratio = 1:9
12

8   Area = 20 × 1.83 = 116.64 cm3              b    Volume ratio = 1:27
= 13:33
9   Volume ratio = 8:125                                         = (Length ratio)3 = 1:3
= 23:53 = (Length ratio)3
i Length ratio = 2:5
ii Length ratio = 2:5
iii Area ratio = (Length ratio)2
= 22:52
= 4:25
Solutions to Multiple-choice questions
1   PRS is supplementary to 120 degrees.       9    XY = 3 × 12
PRS = 180 – 120 = 60 ⇒ D                             5
= 7.2 cm ⇒ E
2   RPS = 180 – 60 – 70 = 50 ⇒ C
10   Area ratio = (Length ratio)2
3   PSQ = 180 – 70 = 110, due to being                         = 12:22 = 1:4
supplementary to the PSR.                       Area = 60 = 15 ⇒ C
4
QPS = 50 due to PS bisecting QPR.
Thus, PQS = 180 – 110 – 50 = 20.
⇒A                                         11   x = 18 × 1.2
1.8
= 12 ⇒ A
4   x = 180 – 125 = 55, due to being
supplementary to an angle
12   x = 180 × 12 – 360
corresponding with the 125 degree                         12
angle. ⇒ E                                        =  2160 – 360
12
5   y = 55, due to being vertically opposite          = 1800 = 150 ⇒ D
12
angle x. ⇒ E

6   z = 125, due to corresponding with the     13   XY = 4 × 10 = 5.7 ⇒ B
7
125 degree angle. ⇒ B
14   Length ratio = 10:40 = 1:4
7   BC2 = 62 + 82                                   Volume ratio = (Length ratio)3
= 36 + 64                                                = 13:43
= 100 = 102                                              = 1:64 ⇒ C
BC = 10 ⇒ A
15   135 = 180n – 360
8   BC2 = 72 + 92                                                 n
= 49 + 81                                   135n = 180n – 360
= 130 = 11.402                              45n = 360
BC = 11 (to the nearest cm) ⇒ E                 n = 8 ⇒C

16   Diagonal length
= 10 2 + 10 2 = 2 × 100
= 10 2 ⇒ C
Chapter 13 – Trigonometry
Solutions to Exercise 13A
1    x = sin(31)                                 h = tan(49)
10
6
20
x = sin(31) × 10 = 5.1504                   h = tan(49) × 20 = 23 m
y = cos(31)
10                                     7
y = cos(31) × 10 = 8.5717
a    sin(ACB) = 1
6
2   tan(35) = x                                   ACB = sin–1( 1 ) = 9.6 degrees
10                                                   6
x = 10 × tan(35) = 7.0021
b     BC = cos(9.6)
3                                                 6
a   x = cos(35)                                   BC = cos(9.6) × 6 = 5.9 m
5
x = cos(35) × 5 = 4.10               8
x = sin(5)                            a    cos( θ ) = 10
b                                                         20
10
x = sin(5) × 10 = 0.87                    θ = cos–1( 1 ) = 60 degrees
2

c   x = tan(20.16)                               h = tan(60)
8                                      b
10
x = 8 × tan(20.16) = 2.94                   h = tan(60) × 10 = 17.3 m

d   x = tan(30.25)                         9
7                                           3 = sin(26)
x = 7 × tan(30.25) = 4.08            a
l
l=      3    = 6.8 m
e   tan(x) = 10                                        sin(26)
15
x = tan–1( 2 ) = 33.69               b    tan(26) = 3
3                                       h
h=    3    = 6.2 m
f   10 = tan(40)                                     tan(26)
x
x=      10 = 11.92
tan(40)                         10   sin(x) = 13
60
x = sin–1( 13 ) = 12.51 degrees
4   sin(60) = 20                                              60
1
l=      20 = 23.09 cm                      h
sin(60)                         11       = sin(66),
200
h = sin(66) × 200 = 183 m
5   Let the two equal angles = x and the
other angle = y.                            400
12       = sin(16)
cos(x) = 6                                   d
15
x = cos–1( 2 ) = 66.42                     d = 400 = 1451 m
5                                  sin(16)
y = 180 – 2(66.42) = 47.16
13   Since the lengths of the diagonals are    16    d = tan(32),
equal, ABCD must be a square (which            50
is a special type of rhombus).                   d = tan(32) × 50 = 31 m.
Thus, angle OBC is 45 degrees.
17   Let x be the angle the ladder makes
with the ground.
a    sin(45) = 5
l                                  cos(x) = 1.7
5                                            4.7
l=         = 7.07 cm
sin(45)                                  x = cos ( 1.7 ) = 68.8 degrees.
–1
4.7
b    Since ABCD is a square, angle ABC               h
= tan(68.8),
1.7
= 90 degrees.
h = tan(68.8) × 1.7 = 4.4 m.
14   Let h be the pendulum’s highest point
18   The angle between the path and the
height.
h = cos(15),                                  route straight across the river is
90                                             90 – 60 = 30 degrees.
h = cos(15) × 90 = 86.93                     50 = cos(30),
Since the string is 90 cm long, the             d
lowest point must be 90 cm.                      d=      50 = 57.7 m
x = 90 – 86.93 = 3.1 cm.                              cos(30)

15   Let l be the length of half the string.
15 = sin(52.5),
l
l=  15      = 18.91.
sin(52.5)
Thus, length of string = 18.91 × 2
= 37.8 cm.
Solutions to Exercise 13B
1                                    3
a      x     = 10                    a   A = 180 – 59 – 73 = 48 degrees
sin(40) sin(70)                         b     = 12
x = 10 × sin(40) = 6.84            sin(59)    sin(48)
sin(70)                     b = 12 × sin(59) = 13.84
sin(48)
b   angle ZXY = 180 – 65 – 37               c     = 12
= 78 degrees             sin(73)    sin(48)
x     =      6                      c = 12 × sin(73) = 15.44
sin(78)      sin(65)                            sin(48)
x = 6 × sin(78) = 6.48
sin(65)               b   C = 180 – 75.3 – 48.25 = 56.45 degrees
a      =      5.6
c       x    = 5.6                       sin(75.3)     sin(48.25)
sin(100) sin(28)                       a = 5.6 × sin(75.3) = 7.26
x = 5.6 × sin(100) = 11.75                      sin(48.25)
sin(28)                       c      =      5.6
sin(56.45)     sin(48.25)
d   angle ZXY = 180 – 38 – 92 = 50         c = 5.6 × sin(56.45) = 6.26
sin(48.25)
x    = 12
sin(50) sin(92)
x = 12 × sin(50) = 9.20        c   B = 180 – 123.2 – 37 = 19.8 degrees
sin(92)                       b      =      11.5
sin(19.8)     sin(123.2)
2                                          b = 11.5 × sin(19.8) = 4.66
sin(θ) = sin(72)                                   sin(123.2)
a                                           c            11.5
7         8                                =
sin(37)     sin(123.2)
θ = sin (7 × sin(72) )
–1
8                      c = 11.5 × sin(37) = 8.27
= 56.32 degrees                                 sin(123.2)

sin(θ) = sin(42)                 d   C = 180 – 23 – 40 = 117 degrees
b
8.3       9.4                         b     = 150
sin(40)     sin(23)
θ = sin (8.3 × sin(42) )
–1
9.4                   b = 150 × sin(40) = 246.76
= 36.22 degrees                                sin(23)
c     = 150
sin(117)     sin(23)
c   sin(θ) = sin(108)
c = 150 ×  sin(117) = 342.05
8         10
sin(23)
θ = sin (8 × sin(108) )
–1
10
e   C = 180 – 140 – 10 = 30 degrees
= 49.54 degrees
a     =    20
sin(10)    sin(140)
sin(θ) = sin(38)
d
9         8                         a = 20 × sin(10) = 5.40
sin(140)
θ = sin (9 × sin(38) )
–1
c     =    20
8                    sin(30)    sin(140)
= 43.84 degrees
c = 20 × sin(30) = 15.56
sin(140)
4                                             6
a   sin(B) = sin(48.25)                       a   ACB = 49 – 37 = 12 degrees
17.6         15.3                            CBO = 180 – 90 – 37 = 53 degrees
B = sin (17.6 × sin(48.25) )
–1                                     CBA = 180 – 53 = 127 degrees
15.3
= 59.12 degrees                         b   CAB = 180 – 12 – 127 = 41 degrees
A = 180 – 59.12 – 48.25 = 72.63 degrees         BC = 60
a      =     15.3                        sin(41)   sin(12)
sin(72.63)     sin(48.25)
BC = 60 × sin(41) = 189.3 m
a = 15.3 × sin(72.63) = 19.57                            sin(12)
sin(48.25)
c     OB = 189.3
b   sin(C) = sin(129)                             sin(37)   sin(90)
4.56        7.89
OB = 189.3 × sin(37) = 113.9 m
C = sin–1(4.56 × sin(129) )                                  sin(90)
7.89
= 26.69 degrees                         7   ABP = 180 – 46.2 = 133.8 degrees
A = 180 – 26.69 – 129 = 24.31 degrees         APB = 180 – 133.8 – 27.6 = 18.6
a      = 7.89                               PB     =     34
sin(24.31)     sin(129)                       sin(27.6)   sin(18.6)
sin(24.31) = 4.18
a = 7.89 ×                                    PB = 34 × sin(27.6) = 49.39 m
sin(129)                                    sin(18.6)
OP     = 49.39
c   sin(B) = sin(28.25)                           sin(46.2)   sin(90)
14.8          8.5
OP = 49.39 × sin(46.2) = 35.6 m
B = sin (14.8 × sin(28.25) )
–1                                                      sin(90)
8.5
= 55.50 degrees                         8   angle ACB = 180 – 74 – 69
C = 180 – 55.50 – 28.25 = 96.25 degrees                 = 37 degrees
c      =      8.5                          CA = 1070
sin(96.25)     sin(28.25)                     sin(69)   sin(37)
c = 8.5 × sin(96.25) = 17.85                  CA = 1070 × sin(69) = 1659.9 m
sin(28.25)                                      sin(37)

5   angle BAC = 180 – 68 – 72                 9   Let the point O be the centre
= 40 degrees                        intersection, i.e. the intersection of the
AC = 400                                    lines AY and BX.
sin(68)   sin(40)
AC = 400 × sin(68) = 577 m              a   XAB = 88 + 32 = 120
sin(40)                          AXB = 180 – 120 – 20 = 40
AX = 50
sin(20)    sin(40)
AX = 50 × sin(20) = 26.60 m
sin(40)

b   YAB = 89 + 20 = 109
BYA = 180 – 109 – 32 = 39
AY    = 50
sin(109)    sin(39)
AY = 50 × sin(109) = 75.12 m
sin(39)
Solutions to Exercise 13C
1   BC2 = 102 + 152 – 2 × 10 × 15 × cos(15)       6
= 35.16; BC = 5.93 cm                     a    angle ABC = 180 – 48 = 132
AC2 = 42 + 52 – 2 × 4 × 5 × cos(132)
2       2           2
= 67.7; AC = 8.23 cm
2   cos(ABC) = 8 + 5 – 10 = –0.137
2×8×5
ABC = cos–1(–0.137)                           b    BD2 = 42 + 52 – 2 × 4 × 5 × cos(48)
= 97.90 degrees                                    = 14.21; BD = 3.77 cm
2       2       2
cos(ACB) = 10 + 5 – 8 = 1.89
2×10×5                  7    d2 = 542 + 422 – 2 × 54 × 42 × cos(70)
ACB = cos–1(1.89)                                     = 3136; d = 56 cm
= 52.41 degrees
8
3                                                 a    BD2 = 42 + 62 – 2 × 4 × 6 × cos(92)
2     2      2
a   a = 16 + 30 – 2 × 16 × 30 × cos(60)                    = 53.73; BD = 7.33 cm
= 676; a = 26
b    sin(BDC) = sin(88)
b   b2 = 142 + 122 – 2 × 14 × 12 × cos(53)                  5        7.33
= 137.83; b = 11.74                             BDC = sin (5 × sin(88) ) = 42.98 degrees
–1
7.33
2           2       2        DBC = 180 – 88 – 42.98 = 49.02 degrees
c   cos(ABC) = 27 + 46 – 35 = 0.652
2×27×46                          CD      = 7.33
ABC = cos–1(0.652)                                 sin(49.02)   sin(88)
= 49.29                                          CD = 7.33 × sin(49.02) = 5.53 cm
sin(88)
d   b2 = 172 + 632 – 2 × 17 × 63 × cos(120)
= 5329; b = 73                             9
2       2       2
a    cos(AO’B) = 6 + 6 – 8 = 0.111
e   c2 = 312 + 422 – 2 × 31 × 42 × cos(140)                               2×6×6
= 4719.70; c = 68.70                              AO’B = cos–1(0.111) = 83.62
2       2       2
f   cos(BCA) = 102 + 122 – 92 = 0.679             b    cos(AOB) = 7.5 + 7.5 – 8 = 0.431
2×10×12                                      2×7.5×7.5
BCA = cos–1(0.679) = 47.22                           AOB = cos–1(0.431) = 64.46

g   c2 = 112 + 92 – 2 × 11 × 9 × cos(43.2)        10
= 57.61; c = 7.59                          a    AB2 = 902 + 702 – 2 × 90 × 70 × cos(65)
= 7673.76; AB = 87.6 m
2           2           2
h   cos(CBA) = 8 + 15 – 10 = 0.787
2×8×15                  b     sin(OAB) = sin(65)
CBA = cos–1(0.787) = 38.05                              70         87.6
OAB = sin (70 × sin(65) ) = 46.40 degrees
–1
4   AB2 = 42 + 62 – 2 × 4 × 6 × cos(20)                                    87.6
= 6.76; AB = 2.6 km                            OC2 = 902 + 43.82 – 2 × 90 × 43.8 × cos(46.40)
= 4581.5; OC = 67.7 m
5   AB2 = 42 + 62 – 2 × 4 × 6 × cos(30)
= 10.24; AB = 3.2 km
Solutions to Exercise 13D
1   Note that area = 0.5 × b × c × sin(A).   5   Note that s = 0.5(a + b + c) and that
a   Area = 0.5 × 6 × 4 × sin(70)                 A = s ( s − a )( s − b)( s − c )
= 11.28 cm2
a   s = 0.5(5 + 8 + 10) = 11.5
b   Area = 0.5 × 6.2 × 5.1 × sin(72.8)
= 15.10 cm2                             A = 11.5(11.5 − 5)(11.5 − 8)(11.5 − 10)
= 19.81 cm2
c   Area = 0.5 × 3.5 × 82 × sin(130)
= 10.99 cm2                         b   s = 0.5(2 + 11 + 10) = 11.5
A = 11.5(11.5 − 2)(11.5 − 11)(11.5 − 10)
d   ACB = 180 – 25 – 25 = 180 – 50 = 130
= 9.05 cm2
Area = 0.5 × 5 × 5 × sin(130)
= 9.58 cm2
c   s = 0.5(8 + 8 + 10) = 13
2                                                A = 13(13 − 8)(13 − 8)(13 − 10)
a   Area = 0.5 × 6.2 × 6.2 × sin(60)               = 31.22 cm2
= 16.65 cm2
6
b   Area = 0.5 × 3.7 × 3.7 × sin(60)         a   6 = Area = 0.5 × 3 × 5 × sin(x)
= 5.93 cm2                              sin(x) = 0.8;
x = sin–1(0.8) = 53.13 degrees.
3   31 = Area = 0.5 × 9 × 13 × sin(x)            Also, x = 180 – 53.13 = 126.87
sin(x) = 0.5299;                             degrees, due to symmetry of the sin
x = sin–1(0.5299) = 32 degrees.         trigonometric identity (refer to the unit
Also, x = 180 – 32 = 148 degrees, due        circle).
to symmetry of the sin trigonometric
identity (refer to the unit circle).     b   BC2 = 32 + 52 – 2 × 3 × 5 × cos(53.13)
= 16.00; BC = 4.00 cm
4   Area = 0.5 × 8 × 9 × sin(108.6)              Or, BC2 = 32 + 52 – 2 × 3 × 5 × cos(126.87)
= 34.12 cm2                                     = 52.00; BC = 7.21 cm.
Solutions to Multiple-choice questions
1   10 = sin(35)                               7    sin(XYZ) = sin(62)
XZ                                                 18          21
XZ =      10 ⇒ C                                XYZ = sin (18 × sin(62) )
–1
sin(35)                                                     21
= 49.2 degrees ⇒ D
2      AC    = 11
sin(102)    sin(28)                        8    angle ABC = 180 – 52 = 128 degrees.
AC = 11 × sin(102)                            AC2 = 82 + 122 – 2 × 8 × 12 × cos(128)
sin(28)                              = 326.20
= 22.9 ⇒ B                                 AC = 18.06 = 18 ⇒ B
2     2        2
3   cos(ACB) = 5.2 + 6.8 – 7.3 = 0.283         9    Area = 0.5 × 5 × 3 × sin(30)
2 × 5.2 × 6.8                          = 3.75 cm2 ⇒ B
ACB = cos–1(0.283)
= 74 degrees ⇒ C                       10      AC    = 11
sin(109)   sin(32)
4   c2 = 302 + 212 – 2 × 30 × 21 × 51                 AC = 11 × sin(109) = 19.6 cm ⇒ C
53                         sin(32)
= 128.55
c = 11.34 = 11 ⇒ C                         11   BC2 = 72 + 62 – 2 × 7 × 6 × cos(60)
= 43 ⇒ E
5   AC2 = 102 + 122 – 2 × 10 × 12 × cos(110)
= 326.08                               12   AC2 = 252 – 82 = 561
AC = 18.06 = 18 ⇒ B                             AC = 23.69
2   2     2

6   Let the two smaller angles = x.                 cos(ABC) = 25 + 8 – 23.69
2 × 25 × 8
2x = 180 – 130 = 50,
= 0.32 = 8 ⇒ A
x = 25 degrees.                                                   25
r    =      10
sin(25)     sin(130)                       13   AC2 = a2 + b2;
r = 10 ×  sin(25) = 5.52 cm ⇒ A               AC =     a2 + b2
sin(130)
cos(x) = AB
AC
b
=           ⇒B
2    2
a +b
Chapter 14 – Applications of
geometry and trigonometry
Solutions to Multiple-choice questions
1   tan(x) = 5 ;, x = tan–1( 5 ) = 36          7    AB2 = 242 + 162 + 102 = 932
7                  7                   AB = 30.53 = 31 cm ⇒ C
Bearing = 360 – 36
= 324 degrees ⇒ C                  8    AB2 = 202 – 122 = 256; AB = 16
CB2 = 182 – 122 = 180; CB = 13.4.
2   215 – 180 = 35                                  AC2 = 162 + 13.42 = 435.56;
Bearing of return = 035 degrees ⇒ A              AC = 21 cm ⇒ B

3   tan(20) = 500                              9    angle ABC = 45 + 60 = 105
d
AC2 = 1002 + 602 – 2 × 100 × 60 × cos(105)
d=     500 = 1374 m ⇒ D
= 16705.8
tan(20)
AC = 129.3 = 129 km ⇒ D

4   tan(x) = 80                                10   Let a diagonal from the centre of the
1300
–1 80                                    base to one of the 4 base vertices = d.
x = tan (      )
1300                                  Let the slope length = s.
= 3.5 = 4 ⇒ B                                 d2 = 42 + 42 = 32; d = 5.66
s2 = 162 + 5.662 = 288
5   35 – 10 = 25 degrees                              s = 288 ⇒ A
d2 = 1002 + 502 – 2 × 100 × 50 × cos(10)
= 2651.92                               11   angle ABC = 60 + 60 = 120 degrees.
AB = 51.497 = 51 km ⇒ A                         AC2 = 42 + 62 – 2 × 4 × 6 × cos(120),

6   Let d = distance to the start from the            AC =    6 2 + 4 2 − 48 cos(120) ⇒ B
finish point.
Let x be the angle between the 7km         12   Let us call the distance Q is east of P,
path and the path that would return the         q, and the distance R is east of Q, r.
hiker to the start.                              q = sin(50); q = 35sin(50)
20 + 35 = 55 degrees.                           35
d2 = 5.22 + 72 – 2 × 5.2 × 7 × cos(55)           r = sin(20); r = 20sin(20)
= 34.28; d = 5.86 km                        20
2         2        2               Thus, R is east of P by
cos(x) = 7 + 5.86 – 5.2 = 0.686                 35sin(50) + 20sin(20) km ⇒ A
2 × 7 × 5.86
x = cos–1(0.686) = 47 degrees.
Bearing = 47 + 35 + 180                    13   The steepest part is where the contours
= 262 ⇒ E                               are closest together = between 250 m
and 300 m contour lines. ⇒ D
Chapter 15 – Revision: Geometry
and trigonometry
Solutions to Multiple-choice questions
1   PQ = cos(40)                            10   AC2 = 102 – x2 and AC2 = 62 + x2.
16.5                                         Thus, 102 – x2 = 62 + x2
PQ = 16.5 × cos(40) = 12.6 ⇒ C               2x2 = 102 – 62 = 100 – 36 = 64
x2 = 64 = 32
2   angle BCD = 42 + 62 = 104 ⇒ A                           2
x=       32 = 4 2 ⇒ D
3   2BAC = 180 – 30 = 150
BAC = 75 ⇒ C                               EF = 5
11
9    6
4   Area = 0.5 × 10 × 14 × sin(70)                 EF = 5 × 9 = 7.5
= 66 ⇒ C                                                   6
AG = 6
15 5
5   sin(p) = 9
11                                    AG = 6 × 15 = 18
5
p = sin–1( 9 ) = 54.9 ⇒ E                  AE = 18 – 9 = 9
11
x 6
=
6   angle ACB = 180 – 42 – 63 = 75                 6 9
AB = 35                                     x = 6 × 6 = 4 ⇒B
sin(75)    sin(42)                                          9
AB = 35 × sin(75) = 51 mm ⇒ E                                 2       2   2
sin(42)                    12   cos( θ ) = 4 + 6 – 8 = –0.25 ⇒ A
2×4×6
7   Largest angle (x) is between the 7 cm
and 3cm sides.                          13   Volume ratio = 4:9 = (Area ratio)3/2
2        2   2                     Area ratio = (Volume ratio)2/3
cos(x) = 7 + 3 – 8 = –0.143,                            = 42/3:92/3
2×7×3
x = cos–1(–0.143) = 98 degrees ⇒ D                    = 3 16 : 3 81 ⇒ E

8   2ACB = 180 – 80 = 100                   14   Side AE is proportional to AB since
ACB = 50                                   both are opposite angles of the same
AB = 10 = 7.779                            size (marked by the ×).
sin(50)     sin(80)
AD is 4 × AE, but BC is 2 × AB, so AD
Of the available options,                           3
5     = 7.779 ⇒ C                         and BC are not in proportion.
cos(50)                                      Thus, DE must be in proportion to BC.
This gives us:
9   Since SQR is an isosceles triangle,          DE = 12
angle RSQ = 180 – 2x.                         3     4+2
Since PS is a transversal,                     DE = 12 × 3 = 6 ⇒ A
angle PSR = 180 – 42 = 138.                                         6
Thus, 180 – 2x + y = 138
2x = y + 42
2x – y = 42 ⇒ E
15   BC2 = 32 + 62 = 9 + 36 = 45               23   CA2 = 402 – 202 = 1200
BC = 45                                        CA = 1200
DB2 = 32 + ( 45 )2 = 9 + 45 = 54               AB2 = 302 – 202 = 500
DB =       54 = 3 6 ⇒ D                       AB = 500
CB2 = CA2 + AB2 = 1200 + 500 = 1700
16   Volume ratio = (Length ratio)3                  CB = 1700 ⇒ A
= 403:103
= 64 000:1000 = 64:1       24   OB2 = 12 + 12 = 2
However, we must account for the fact          OC2 = OB2 + 12 = 2 + 1 = 3
that this ratio (1 part air to 64 parts        OD2 = OC2 + 12 = 3 + 1 = 4
water), includes the 1 part air in both         OB = 2 ⇒ D
the air and water categories.
25   Volume ratio = (Length ratio)3
Thus, the ratio = (64 – 1):1
= 13:23 = 1:8
= 63:1 ⇒ D
Volume = 80 = 10 ⇒ C
8
2
17   Area AXY = ⎛ 2 ⎞ × 108
⎜ ⎟
⎝ 3⎠                                 sin(β) = sin(α)
108 = 48 cm2 ⇒ B
26
=4×                                             8         10
9                                                    2 × 8 = 8 ⇒B
sin( β ) =
3    10    15
18   x2 = 72 + 82 – 2 × 7 × 8 × cos(130)
= 49 + 64 – 2(7)(8)cos(130).           27   Area = 0.5 × 5 × 6 × sin(40)
But since cos(130) = –cos(180 – 130)                = 15sin(40) ⇒ C
= –cos(50),
we can say that                           28   b2 = 32 + 42 – 2 × 3 × 4 × cos(120) = 37
x2 = 49 + 64 + 2(7)(8)cos(50). ⇒ A               b = 37 ⇒ C

19   Let x be the line from the 25 degree      29   Let x be half one of the base diagonals.
angle point to the top of the tower.           x2 = 42 + 42 = 32; x = 5.66
180 – 25 = 155; 180 – 155 – 15 = 10            s2 = 32 + 5.662
= 9 + 32 = 41
x     = 100
sin(15)      sin(10)                           s = 41 ⇒ A
x = 100 × sin(15)                       30   angle ACB = 180 – 100 – 30 = 50
sin(10)
h           x                                AB =        12
=                                    sin(50)    sin(100)
sin(25)      sin(90)
h = sin(25) × x                                AB = 12sin(50) ⇒ D
sin(100)
= sin(25) × 100 × sin(15)
sin(10)           31   Let QT = 1. Thus, RT = PU = SU = 1,
= 100 sin(15) sin(25) ⇒ B
since QRT is a right isosceles triangle.
sin(10)
QR2 = 12 + 12 = 2; QR = SP = 2
20   tan( θ ) =    400 = 400 ⇒ E                       SQ =       2
300/ 2 150                        sin(90) sin(60)
21   AC2 = 82 + 62 – 2 × 8 × 6 × cos(120)           SQ =     2 [since sin(60) = 3 ]
3 /2                     2
AC = 100 − 96 cos(120) ⇒ B
= 2 2
22   Let x be half the length of one of the                  3
diagonals.                                                      1
x2 = 32 + 42 = 9 + 16 = 25; x = 5              sin(θ) = SU =         = 3
2   2   2
SQ (2 2 )/ 3 2 2
cos( θ ) = 5 + 5 – 6 = 0.28
2×5×5                             Multiplying by 2 gives us 6 ⇒ E
cos–1(0.28) = 73.7 ⇒ D                                      2          4
32   Area ratio = (Length ratio)2                40   The hiker travelled 5 km at 030
= (1.4)2:12                           degrees but then 10 km at 330 degrees.
= 1.96:1 = 49:25 ⇒ A                  Since 030 is 30 degrees east of north
and 330 is 30 degrees west of north,
33   Area = 0.5 × 6 × 8 × sin(40)                     the first 5 km of her path from Q to R
= 24sin(40) ⇒ E                             returned her to the north midline – that
is, when the hiker had travelled a total
34   Area = 0.5 × 15 × 10 × sin(30)                   distance of 10 km, she was directly
= 75 × sin(30) = 37.5 ⇒ B                   north of the start point P.
The remaining 5 km to R at 330
35   Area ratio = (Length ratio)2                     degrees took her further north and a
= 92:62                               little west of the start point.
= 81:36 = 9:4
Thus, area = 9 × 20 = 45 ⇒ D                     Let the distance she is west of the start
4                                   point be given the letter w.
Since sin(30) = w ,
36   Volume ratio = (Length ratio)3                                     5
= 103:303                              w = 5 × sin(30) = 2.5 ⇒ A
= 1000:27 000
= 1:27 ⇒ C                      41   ZX 2 = 82 + 2.52
= 64 + 6.25 = 70.25
37   Let d be the length of half of one of the        ZX = 8.4 cm ⇒ D
base diagonals.
d2 = 52 + 52 = 50; d = 7.07                 42   If the line on the map is 4 cm long, it
h2 = 102 – d2                                    represents 4 × 20 000 = 80 000 cm
= 100 – 50 = 50                               = 800 m in reality.
h = 50 ⇒ B                                       It starts at the 100 m contour and ends
at the 250 m contour. This represents a
sin(y) = sin(x)                                  250 – 100 = 150 m rise over 800 m.
38
6        8                                    The average slope = 150
800
sin(y) = 6 × sin(x)                                                    = 0.1875 ⇒ C
8
=6×   3/ 7                                              2       2    2
8                             43   cos(PQR) = 5 + 7 – 8 = 0.143
6×3                                                   2×5×7
=
7×8                                       cos–1(0.143) = 81.8 ⇒ D
18
=     = 9 ⇒B                         44   Volume ratio = (Length ratio)3
56 28
= 43:13 = 64:1 ⇒ E
39   Largest angle is MNP.
2    2     2                  45   If PQ is twice QS, then PS must be 1.5
cos(MNP) = 6 + 5 – 10 = –0.65                    times PQ.
2×6×5
cos–1(–0.65) = 73.7                              Area ratio = (Length ratio)2
= 130.5 degrees ⇒ D                                       = 12:1.52 = 1:2.25
Thus, area of PQR = 1 × 18
2.25
= 8 ⇒C
Chapter 16 – Constructing and
interpreting linear graphs
Solutions to Exercise 16A
y2 – y1                   4   Note that (x2, y2) is the coordinate with
1   Note that m =
x2 – x1                       the lower x value.
a   Coordinates are (–2,0) and (0,6)         a   m = 6 – 8 = -2 = – 1
12 – 4     8             4
m= 6–0 = 6 =3
0 – -2   2
b   m = -12 – 8 = -20 = – 5
b   Coordinates are (0,–4) and (3,0)                  2 – -6         8             2
m = 0 – -4 = 4
3–0     3                           c   m = -1.5 – 3.5 = -5 = –2
5.5 – 3           2.5
c   Coordinates are (0,6) and (5,0)
m = 0 – 6 = -6 = – 6                     d   m = 0 – 16 = -16 = –8
5–0      5      5                             12 – 10        2

d   Coordinates are (–3,0) and (0,–4)        e   m= 0–0 = 0 =0
3 – -3     6
m = -4 – 0 = -4 = – 4
0 – -3      3   3
f   m = -3 – 0 = -3 = –1
e   Coordinates are (–5,0) and (0,–5)                 0 – -3     3

m = -5 – 0 = -5 = –1
0 – -5      5                        g   m = 16 – 9 = 7 = 7
4–3        1
f   Coordinates are (–2,0) and (0,6)
m= 6–0 = 6 =3                            h   m = 25 – 64 = -39 = –13
0 – -2   2                                    -5 – -8        3

2   Any increasing line (i.e. a line going
upwards towards the right) which
makes an angle of 45 degrees with the
positive x-axis is possible.

3   Any line with no slope (i.e. any
horizontal line) is possible.
Solutions to Exercise 16B
1                                    e   Gradient = –2, x-intercept = 4
a   Gradient = 1, x-intercept = 1

f   Gradient = –1, x-intercept = 6

b   Gradient = 2, x-intercept = 1

2   For a graph to pass through the origin,
the y-intercept = c = 0.
a   y = –x + 1, y–-intercept = 1,
doesn’t pass through origin.

c   Gradient = 3, x-intercept = –6   b   y = –x, y-intercept = 0,
passes through the origin.

c   y = 3x , y-intercept = 0,
2
passes through the origin.

d   y = 3x – 3x + 3 = 3, y-intercept = 3,
doesn’t pass through the origin.

e   y = x – 1, y-intercept = –1,
doesn’t pass through the origin.

d   Gradient = –1, x-intercept = 4   3
a   y = –x + 1, m = gradient = –1

b   y = –x, m = gradient = –1

c   y = 3x , m = gradient = 3
2                       2

d   y = 3, m = gradient = 0

e   y = x – 1, m = gradient = 1
4                   d   x + y = 3;
a   4x + 2y = 12;       y=3–x
2y = 12 – 4x
y = 6 – 2x

e   y – 2x = 6;
y = 2x + 6

b   6x + 3y = 12;
3y = 12 – 6x
y = 4 – 2x

f   2y + 3x = 6;
2y = 6 – 3x
y = 3 – 3x
2
c   6y – 4x = 24;
6y = 4x + 24
y = 2x + 4
3
g   x + 2y = 1;   i   2x – 3y = 6;
2y = 1 – x        3y = 2x – 6
y = 1 – 1x        y = 2x – 2
2   2             3

h   y – 2x = 1;
y = 2x + 1
Solutions to Exercise 16C
1   Note that y = mx + c                       3
a   m = 2 – 0 = 2 = 2,                         a   m = 3 , c = 1, y = 3 x + 1
0 – -1      1                                  2              2
c = 2, y = 2x + 2
b   m = 1 , c = 6, y = 1 x + 6
2              2
b   m = 0 – 6 = -6 = –3,
2–0     2
c = 6, y = –3x + 6                         c   m = –3. y = –3x + c
Substituting in (1,4): 4 = –3(1) + c
c   m = 0 – -3 = 3 = 3,                            c = 7, y = –3x + 7
1–0         1
c = –3, y = 3x – 3                         d   m = –1, c = 2, y = –x + 2

d   m = infinite,                              e   m = –2, y = –2x + c
c = non–existent., x = 2                       Substituting in (–1,4): 4 = –2(–1) + c
c = 2, y = –2x + 2
e   m = 0, c = –3, y = –3
f   m = –1, y = –x + c
f   m = 0 – 2 = -2 = –1,                           Substituting in (3,7): 7 = –(3) + c
0 – -2         2
c = 10, y = –x + 10
c = 0, y = –x.
4
g   m = 3 – 0 = 3 = 3,
1–0        1                          a   m = 0 – 3 = -3 = –1,
3–0      3
c = 0, y = 3x
c = 3, y = –x + 3

h   m = 0 – 2 = -2 = – 1 ,
8–0     8      4                      b   m = -6 – 0 = -6 = –2,
0 – -3   3
c = 2, y = – 1x + 2
4                                 c = –6, y = –2x – 6

2   Note that since c is the y–intercept, if   c   m = 2 – 4 = -2 = – 1 ,
one of the given points is the                      4–0     4      2
y–intercept, we know c without having          c = 4, y = – 1x + 4
2
to substitute into the y equation.

a   m = 1 , c = 1, y = 1 x + 1                 d   m = 1 – 0 = 1 = 1,
2                  2                           0 – -1   1
c = 1, y = x + 1
b   m = 2, y = 2x + c
Substituting in (1,4): 4 = 2(1) + c        e   m = 6 – 4 = 2 = 1,
c = 2, y = 2x + 2                                  5–1 4           2
y= 1x + c
c   m = 0, c = 6, y = 6                               2
Substituting in (1, 4): 4 = 1 1 + c
d   m = 1, y = x + c                                                            2
Substituting in (2,–4): –4 = 2 + c             c = 7 , y = 1x + 7
c = –6, y = x – 6                                  2       2    2

e   m = –2, y = –2x + c
Substituting in (1,3): 3 = –2(1) + c,
c = 5, y = –2x + 5
Solutions to Exercise 16D
1                           e   –3x + 4y = 15;
a   2x – 3y = 12;               4y = 3x + 15
3y = 2x – 12                 y = 3 x + 15
4           4
y = 2x – 4
3

f   7x – 2y = 14;
b   4x – y = 8;                 2y = 7x – 14
y = 4x – 8                   y = 7x – 7
2

2
c   3x – 4y = 24;           a   2x – y = 6,
4y = 3x – 24                y = 2x – 6, m = 2
y = 3x – 6
4                  b   x + 4y = 12,
4y = 12 – x,
y = 3 – 1 x, m = – 1
4           4

c   –x – 2y = 6,
2y = –x – 6,
y = – 1 x – 3, m = – 1
2                   2

d   2x – 5y = 20;           d   5x – 2y = 10,
5y = 2x – 20                2y = 5x – 10,
y = 2x – 4                  y = 5 x – 5, m = 5
5                           2               2

e   x – 5y = 10,
5y = x – 10,
y = 1 x – 2, m = 1
5               5

f   –x + 2y = 8,
2y = x + 8,
y = 1 x + 4, m = 1
2               2
Solutions to Exercise 16F
1   Where elimination is used, the two   j   2x + 3y = 12, 10x + 15y = 60 (A)
equations are denoted A and B.           5x + 4y = 23, 10x + 8y = 46 (B)
a   x + 2y = 6 (A)                           A – B: 7y = 14, y = 2
x + 3y = 4 (B)                           2x + (3 × 2) = 12,
B – A: y = –2                              2x = 6, x = 3
x + (2 × –2) = 6,
x = 6 + 4 = 10                     k   5x + 4y = 21, 15x + 12y = 63 (A)
3x + 6y = 27, 6x + 12y = 54 (B)
b   7x + 6y = 0 (A)                          A – B: 9x = 9, x = 1
5x – 6y = 48 (B)                         (6 × 1) + 12y = 54,
A + B: 12x = 48, x = 4                     12y = 48, y = 4
(7 × 4) + 6y = 0,
l   9x + 8y = 17, 18x + 16y = 34 (A)
6y = –28, y = – 14
3                    2x – 6y = –4, 18x – 54y = –36 (B)
A – B: 70y = 70, y = 1
c   3x + 2y = 12 (A)                         9x + (8 × 1) = 17,
x + 2y = 8 (B)                             9x = 9, x = 1
A – B: 2x = 4, x = 2
2 + 2y = 8,                          m   y=6–x
2y = 6, y = 3                          2x + y = 8,
2x + (6 – x) = 8, x = 2
d   x – 2y = 6 (A)                           y=6–2=4
4x + 2y = 14 (B)
A + B: 5x = 20, x = 4                n   9 + x = y, x = y – 9
4 – 2y = 6,                              x + 2y = 12,
2y = –2, y = –1                          (y – 9) + 2y = 12,
3y = 21, y = 7
e   x + 3y = 12 (A)                          x = 7 – 9 = –2
x + y = 8 (B)
A – B: 2y = 4, y = 2                 o   2y = 4 + x, x = 2y – 4
x + 2 = 8, x = 6                       y = x + 8, x = y – 8
x = 2y – 4 = y – 8,
f   9x + 2y = 48 (A)                           2y – 4 = y – 8, y = –4
x – 2y = 2 (B)                           x = –4 – 8, x = –12
A + B: 10x = 50, x = 5
5 – 2y = 2,                          p   x+4=y
2y = 3, y = 3                          y = 10 – 2x
2                           y = x + 4 = 10 – 2x,
x + 4 = 10 – 2x,
g   2x + 3y = 13 (A)
3x = 6, x = 2
2x + 5y = 21 (B)
y = 10 – (2 × 2), y = 6
B – A: 2y = 8, y = 4
2x + (3 × 4) = 13,                   q   y=4–x
2x = 1, x = 1                          y=x+6
2                          y = 4 – x = x + 6,
h   3x – y = 10 (A)                            4 – x = x + 6,
x + y = –2 (B)                             2x = –2, x = –1
A + B: 4x = 8, x = 2                     y = 4 – –1 = 5
2 + y = –2, y = –4                   r   y = 4 + 2x
i   3p + 5q = 17 (A)                         y – 2x = 6, y = 6 + 2x
4p + 5q = 16 (B)                         y = 4 + 2x = 6 + 2x,
B – A: p = –1                              4 + 2x = 6 + 2x, 2 = 0
3 × –1 + 5q = 17,                        Since we have arrived at a nonsense
5q = 20, q = 4                         equation (2 = 0), this means that this
question has no solutions.
Solutions to Exercise 16H
1   Break–even point is at S = C.         4
Thus, 0.75x = 0.25x + 100,            a   C = R, 7x = 6x + 800,
0.5x = 100, x = 200                        x = 800

2   Note that S = n and thus break–even   b
point will be at C = nx.
a   2x + 16 = 10x,
8x = 16, x = 2

b   12x + 84 = 40x,
28x = 84, x = 3

c   90x + 60 = 100x,
10x = 60, x = 6

d   3x + 39 = 16x,                        c   C = 1.05(6x + 800) = 6.3x + 840
13x = 39, x = 3                         C = R; 6.3x + 840 = 7x,
0.7x = 840, x = 1200.
e   30x + 350 = 100x,
70x = 350, x = 5                    5   Note that break–even point is where C = R.
a   2x + 4500 = 3x, x = 4500
f   400x + 800 = 500x,
100x = 800, x = 8

3
a   C = 22x + 45 000; S = 72x
Thus, 22x + 450 00 = 72x,
50x = 45 000, x = 900

b   Profit = P = S – C
= 72x – 22x – 45000                 22 x + 1200 = 14x
= 50x – 45 000                  b
3
Since x = 1800,                              20 x = 1200
P = 50 × 1800 – 45 000 = \$45 000.             3
20x = 3600, x = 180
c   Loss = L = C – S
= 22x + 45 000 – 72x
= 45 000 – 50x
Since x = 450,
L = 45 000 – 50 × 450 = \$22 500.

d   P = 50x – 45 000 = 90 000
50x = 135 000, x = 2700.
c   0.85x + 600 = 1.05x
0.2x = 600, x = 3000

d   0.16x + 360 = 0.25x
0.09x = 360, x = 4000
Solutions to Multiple-choice questions
1   Points are (0,1) and (1,–1)                10   m = 3, y = 3x + c
m = -1 – 1 = -2 = –2 ⇒ B                        Substituting in (5,9): 9 = (3 × 5) + c
1–0      1                                   c = –6, y = 3x – 6 ⇒ C

2   Points are (0,–3) and (1,–2)               11   320 = 1.5x + 20,
-2 – -3                                       1.5x = 300, x = 200 ⇒ C
m = 1 – 0 = 1 = 1,
1
c = –3, y = x – 3                          12   3x – 2y = –6;
3 = x – y ⇒B                                  2y = 3x + 6
y = 1.5x + 3
3   m = 0, c = 1, y = 1 ⇒ E                         The only graph with a y–intercept of 3
and a positive gradient is A. ⇒ A
4   y = 1 and y = x – 3 intersect at
1 = x – 3, x = 4                           13   4x + 2y = 8, 2y = 8 – 4x, y = 4 – 2x and
We know y = 1, since that is the only           y = 3x – 1 intersect at:
allowable value of line C.                        4 – 2x = 3x – 1
Therefore, intersection is at (4,1). ⇒ B              5x = 5, x = 1
y = (3 × 1) – 1 = 2
5   2x – y = 10                                     Intersection is at (1,2) ⇒ D
x + 2y = 0, x = –2y
2 (–2y) – y = 10                           14   2x + 4y – 6 = 0
–5y = 10, y = –2                           4y = 6 – 2x,
x = –2y                                            y = 3 – 1x
= –2 × –2 = 4 ⇒ C                                   2 2
m= – 1 ⇒A
2
6   m = 3, y = 3x + c
Substituting in (1,9): 9 = (3 × 1) + c
15   Competitors = x + y = 219
c = 6, y = 3x + 6 ⇒ C
Entry fees = 2x + y = 361 ⇒ B
7   5x – y + 7 = 0, y = 5x + 7
16   Points are (–2,5) and (2,–5)
ax + 2y – 11 = 0
2y = 11 – ax                                  m = -5 – 5 = -10 = –2.5
2 – -2       4
y = 11 – a x                                 y = –2.5x + c
2    2
Substituting in (–2,5): 5 = (–2.5 × –2) + c
Since the lines are parallel, the
c=5–5=0
y = –2.5x,
5 = – a , a = –10 ⇒ C                           2y = –5x,
2
2y + 5x = 0 ⇒ A
8   2.5x + 65 = 750
2.5x = 685                            17   m=      Y-int = 4 = – 4
x = 274 ⇒ E                                          – X-int    -3  3
c = 4, y = –  4x + 4
3
9   Let chips = C and coke = K.
y –x
3C + 2K = 13.20, 6C + 4K = 26.40 (A)                =   + 1,
4   3
2C + 3K = 11.80, 6C + 9K = 35.40 (B)              y x
B – A: 5K = 9.00, K = 1.80                          +  = 1 ⇒A
4 3
2C + (3 × 1.80) = 11.80
2C = 6.40, C = 3.20 ⇒ C
18   m = 0, c = 3, y = 3 ⇒ B

19   2x + y = 1
y + 14 = 3x, y = 3x – 14
2x + (3x – 14) = 1
5x = 15, x = 3.
y = (3 ×3) – 14
= 9 – 14 = –5 ⇒ B

20   2x + 3y – 6 = 0
3y = 6 – 2x
y = 2 – 2 x.
3
Since the gradient is – 2 , C is
3
incorrect ⇒ C
Chapter 17 – Graphs
Solutions to Exercise 17A
1                                                   Added to the 5 hours of cycling and
a   Reading the distance off the graph for          1.5 hours of resting the cyclist had
the time given:                                 done prior to this segment, the cyclist
i 30 km                                         was away from home for 10 hours.
ii 20 km                                    3
iii 70 km                                   a   Reading the times off the horizontal
axis of the various marked cities:
b   Reading the time off the graph for the          i 5:00
distance given:                                 ii 12:00
i 30 minutes                                    iii 9:00
ii 2 hours and 45 minutes                       iv 6:00
iii 1 hour and 30 minutes
b   He arrived in Dover at 5:00 and left at
c                                                   6:00 = 1 hour.
i Speed = distance / time
c   He travelled to Paris at constant speed
= 10 – 0 = 10 km/h
1–0                                 from Calais and it took 3 hours, so he
ii Speed = distance / time                      would have been halfway 1.5 hours
= 30 – 10 = 20 km/h                    after leaving Calais at 9:00 = 10:30.
2–1
iii Speed = distance / time                 4
= 70 – 30 = 40 km/h               a   Noting the increase in rate at 5
3–2                              minutes, we can draw the following
iv Speed = distance / time                      graph:
= 80 – 70 = 10 km/h
4–3

2
a   Noting that there are 5 segments to the
cyclist’s journey, namely a 15 km/h
segment, a rest, a 12 km/h segment, a
rest, and a 20 km/h return segment, we
can draw the following graph:

b   Let V = amount of water in tank and
t = time in minutes.
i Since 10L are being poured in per
minute and we are starting at the
origin (0,0), we get V = 10t.
ii 15L are being poured in per minute,
giving us V = 15t.
However, this assumes that the rate
b   Distance = speed × time.                           was 15 L/min for the first 5 minutes
15 km/h × 3 h = 45 km                            as well. We must this correct for the
12 km/h × 2 h = 24 km                            difference in volumes during the
Time = distance / speed                            first 5 minutes.
= 45 + 24 = 69 = 3.45 = 3.5 h                Since 15 L/min × 5 min – 10 L/min
20         20
× 5 min = 75 – 50 = 25, we must
Thus, the 20 km/h segment took 3.5 hours.          subtract the difference of 25 from
V = 15t, giving us V = 15t – 25.
c   V = 15t – 25, so at t = 15,              7
V = 15 × 15 – 25                         a   The bath first reached 200 L at 4
= 225 – 25 = 200 L                         minutes.

5                                            b   The bath was filled to 50 L in 2
a   1000 – 600 = 400 L flows out in the          minutes.
first 40 minutes.                            50 = 25 L/min.
400 = 10 L/min                               2
40
c   200 L was emptied in 10 – 9 = 1
b   m = –10, c = 1000, y = –10x + 1000           minute.
Since V is the y-axis and t is the           200 = 200 L/min.
x-axis, we get V = 1000 – 10t.                1

c   600 L flows out in 80 – 50 = 30
minutes.
600 = 20 L/min.
30

6
a   Speed = distance / time
= 150 = 50 km/h.
3

b   She reaches town A after 5 hours.

c   The point where the paths cross is at
time = 2 hours.

d   The paths cross at a point 100 km from
town A.

e   She stops for 4 – 3.5 = 0.5 hours.

f   She travels 60 km in 1 hour = 60 km/h.
Solutions to Exercise 17B
1                                              b   Sketching the 3 steps on the one graph
a   Sketching the 11 steps on the one              gives us the following graph:
graph gives us the following graph:

Note that the open circles mean that
4
y-value is not the cost for that weight,
a   Putting the information from the
whereas the closed circles mean that
paragraph into equation form we get:
y-value is the cost for that weight.
C = \$0.30 for 0 ≤ d ≤ 25
b   Reading the prices of the graph for the          = \$0.40 for 25 < d ≤ 50
given steps:                                     = \$0.70 for 50 < d ≤ 85
i \$1.38                                          = \$1.05 for 85 < d ≤ 165
ii \$0.90                                         = \$1.22 for 165 < d ≤ 745
iii \$0.76                                        = \$1.77 for 745 < d.

2   Sketching the 4 steps on the one graph     b   Sketching the 6 steps on the one graph
gives us the following graph:                  gives us the following graph:

3
a   Putting the information from the
paragraph into equation form we get:
C = \$1.20 for 0 ≤ M ≤ 20
= \$2.00 for 20 < M ≤ 50                  5
= \$3.00 for 50 < M ≤ 150.                a   Since 40 minutes is in the first step,
parking for 40 minutes is free.

b   Since 2 hours is in the second step,
parking for 2 hours costs \$3.00.

c   Since 3 hours is in the third step,
parking for 3 hours costs \$5.00.
Solutions to Exercise 17C
1                                           3
a   Drawing the curve to include all the
known data points gives us the
following graph:

a   Interpolating the y-value for x = 25
from the graph gives us around
b   Interpolating the y-value for 4 years       y = 2800.
from the graph gives us around 70 cm.
b   Extrapolating the y-value for x = 60
c   Interpolating the x-value for 80 cm         from the graph gives us around
from the graph gives us around 5.25         y = 1250.
years.
4
2
a   It rose from 9:00 to reach a maximum
at 12:00 ⇒ 3 hours.

b   Maximum temperature is at 12:00
⇒ 40 degrees.

c   The temperature is 39 degrees at
4 x-values, which are 10:00, 14:00,
16:00 and 18:50.
a   Interpolating the y-value for x = 30
from the graph gives us around
d   Reading off the graph, temperature at
y = 176.
18:00 = 39.5 degrees.
b   Extrapolating the y-value for x = 90
e   37 degrees could be expected to be
from the graph gives us around y = 30.
reached by about 20:15, assuming the
graph continues the same downwards
trend.
Solutions to Exercise 17D
1   The following tables are calculated by                         e
substituting each integer x–value                                  x   –3     –2    –1     0       1    2     3
between –3 and 3 inclusive into the                                y    2     1     2      –       –2   –1   –2
relevant equation for y:                                                3                                      3
a
x   –3     –2    –1        0        1    2       3
y   4.5    2     0.5       0       0.5   2      4.5

b                                                                  f
x    –3     –2     –1          0      1         2      3
x   –3     –2    –1         0   1    2    3
y   –4.5    –2    –0.5         0     –0.5       –2    –4.5
y   –27    –8    –1         0   1    8    27

c
x   –3     –2    –1    0       1     2      3
y    4     1     4     –       4     1      4
9                                      9
g
x    –3     –2    –1        0     1       2       3
y   13.5    4     0.5       0    –0.5     –4    –13.5

d
x   –3     –2    –1    0       1         2       3
y   –4     –1    –4    –       –4        –1     –4
9                                            9
h                                           2   Substituting in (1,5):
x   –3   –2   –1   0   1     2     3          5 = k(1)3, k = 5
y   9    6    3    0   –3    –6    –9
3   Substituting in (2,30):
30 = k(2)2,
4k = 30, k = 7.5

4   Substituting in (16,4):
4= k ,
16
k = 16 × 4 = 64

5   Substituting in (2,1):
k
1 = 2,
i                                                     2
x   –3   –2   –1   0    1     2    3          k = 1, k = 4
4    2                              4
y   2              0   –2    –4    –2
3    3         3     3
6   Substituting in (3,–10):
–10 = k(3), k = – 10
3
Solutions to Exercise 17E
1                                                      b
a   Substituting in (2,20):
20 = k(2)3,
8k = 20, k = 2.5
b
x   0    2         3       6       7
y   0    20       67.5    540    857.5

2
a   Substituting in (2,6):                             5
k                                           a
6 = 2,
2
k = 6, k = 24
4

b
x   1         2     4        8       12
y   24        6    1.5     0.375    0.167

3
b   Substituting in (5,75):
a   Substituting in (2,4):
75 = k(5)2,
4 = k(2)3,
25k = 75, k = 3
8k = 4, k = 0.5
Thus, P = 3d2.
b
6
x   2     6        4       3      8
a
y   4    108       32     13.5   256

c

b   Substituting in (1,100):
100 = k/(1), k = 100.
Thus, P = 100 .
V
4
a   Substituting in (0.5,4):                           7
4 = k(0.5)2,                                     a   Substituting in (2,39.2):
k = 4, k = 16                                          39.2 = k(2)2,
4                                                      4k = 39.2, k = 9.8
x   0    0.354      0.5     1      2     3.536
y   0      2         4      16     64     200    b   If t = 5, d = 9.8(5)2
= 9.8 × 25 = 245 m
8                                  10   The graphs are all linear so are of the
a   y is multiplied by (2) = 1
–2            form y = mx + c. However, the x-axes
4          are not simply x and are of the form xn.
Thus, the equations will be of the form
b   y is multiplied by (2)–1 = 1        y = mxn + c.
2

c   y is multiplied by (2)1 = 2    a    m = 14.7 = 1.2, c = 0,
12.25
2
y = 1.2x2
d   y is multiplied by (2) = 4
b    m = 0.25 = 1 , c = 0,
e   y is multiplied by (2)3 = 8                 2       8
y = 1(1) = 1
9   Substituting in (10,0.125):             8 x    8x
0.125 = k(10)3,
1000k = 0.125                c    m = 6 = 6, c = 0,
k = 0.000125                     1
y = 6x3
a   V = 0.000125(12)3
= 0.216 L                    d    Points are (0,1) and (4,3)
m = 3 – 1 = 2 = 1 , c = 1,
4–0 4           2
b   1 = 0.000125(h)3,                      1 x2 + 1.
h3 = 8000,                        y=
2
h = 20 cm
e    Points are (0,6) and (4,4).
m = 4 – 6 = -2 = – 1 , c = 6,
4–0      4           2
y= – 1 x2 + 6.
2

f    m = 2 = 1 , c = 0,
4       2
1     1
y = 1( 2) = 2
2 x    2x
Solutions to Multiple-choice questions
1   Substituting (0.5, 10) into option C      7    m = 9 = 3 , c = 0,
6 2
gives us 10 = 5 ,                                 3 x2 ⇒ C
0.5                          y=
10 = 5 × 2 = 10                                 2
This is the only option that includes
the point (0.5,10). ⇒ C                   8    m = 2.5, c = 15
C = 2.5t + 15 ⇒ D
2   The greatest increase is from 10 to 19
billion during the 1960–1969 period.      9    Maximum height of graph is at 35.
⇒E                                             ⇒B
(Note the change of scale in the y-axis
between 15 and 25 billion.)               10   m = 13.5 = 9 , c = 0,
3   2
y= 9 x3 ⇒ A
3   Substituting in (1,4.2):                          2
4.2 = k(1)n, k = 4.2
Substituting in (2,33.6):                 11   Total number of punnets sold
33.6 = 4.2(2)n,                              = x + y = 150,
2n = 8 = 23, n = 3 ⇒ E                       Total cost = 3x + 2.5y = 410
Multiplying by 2 gives 6x + 5y = 820.
4   3 kg is in the third step = \$4.50. ⇒ A         ⇒C
5   Points are (0,30) and (20,60)             12   The period of greatest decrease in the
m = 60 – 30 = 30 = 3 , c = 30                  graph, i.e. where the gradient is most
20 – 0   20 2
3 x + 30 ⇒ C                              negative, is between 7 and 10 minutes.
C=
2                                         ⇒E

6   Segment 1 for x between 1 and 4.          13   The 3 kg parcel is in the third step, so
Points are (0,5) and (4,3).                    will cost \$5.00.
The 1.5 kg parcel is in the second step,
m = 3 – 5 = -2 = – 1 , c = 5,
4–0     4           2                      so will cost \$4.00.
y= – 1x + 5                                    Total cost = \$9.00 ⇒ D
2
14   200a + b = 430 (A)
Segment 2 for x between 4 and 5.               315a + b = 660 (B)
Points are (4,3) and (5,0).                    B – A: 115a = 230, a = 2
m = 0 – 3 = -3 = –3,                           (200 × 2) + b = 430,
5–4      1
b = 430 – 400 = 30 ⇒ B
y = –3x + c

Substituting in (4,3):
3 = –3(4) + c,
c = 3 + 12 = 15
y = –3x + 15 ⇒ E
Chapter 18 – Linear Programming
Solutions to Exercise 18A
1                                             b
a   Substituting in (2,2), we get
5(2) + 6(2) = 10 + 12 = 22.
This is thus on the line 5x + 6y = 22.
⇒C

b   Substituting in (3,1), we get
5(3) + 6(1) = 15 + 6 = 21.
This is thus less than the line
5x + 6y = 22. ⇒ B

c   Substituting in (4,2), we get
5(4) + 6(2) = 20 + 12 = 32.               c
This is thus greater than the line
5x + 6y = 22. ⇒ A

d   Substituting in (0,0), we get
5(0) + 6(0) = 0 + 0 = 0.
This is thus less than the line
5x + 6y = 22. ⇒ B

2   Note that x is positive to the right of
the graph and negative to the left,
while y is positive to the top of the
graph and negative to the bottom.         d
Note that ≤ and ≥ are denoted by a
solid line, with < and > denoted by a
dotted line.
The following graphs can thus be
drawn.
a

e
f                                                d   x – y ≥ 10,
y ≤ x – 10

3   Once each of the inequations are in the
form y (<,>, ≤ , ≥ ) mx + c, if the
inequality sign is ‘greater than’, the       e   2x – 6y ≤ 3,
shaded area will be above the line.                   6y ≥ 2x – 3,
If the inequality sign is ‘less than’, the             y ≥ 1x – 1
shaded area will be below the line.                        3      2

a   y ≥ 2x + 5

f   y ≤ 2x
b   3x – 2y < 6
2y > 3x – 6
y > 3x – 3
2

g   x – y > –3,
y<x+3
c   x + y ≤ 10,
y ≤ 10 – x
h   x – 2y > –3,       j   3x – 4y < 12,
2y < x + 3,             4y > 3x – 12,
y < 1x + 3              y > 3x – 3
2    2                   4

i   2x + y ≥ 12,
y ≥ 12 – 2x
Solutions to Exercise 18B
1   Note that to find the point of            c   For point of intersection:
intersection, treat all inequalities as       2x + 3y = –6 (A);
equals signs and use simultaneous             3x – y = 6,
equations accordingly.                        9x – 3y = 18 (B)
A + B: 11x = 12, x = 12
a   For point of intersection:                                            11
y = –2x + 10; y = x – 2                       3×  12 – y = 6,
11
–2x + 10 = x – 2
3x = 12, x = 4                                y = 36 – 66 = – 30
11    11       11
y=4–2=2                                                      12 , – 30 )
Intersection = (4,2).                         Intersection = (
11     11

d   For point of intersection:
b   For point of intersection:                    2x + 3y = 6; x ≥ 0
3x – 2y = 6; x – 3y = 9                       3y = 6 – 2x,
2y = 3x – 6,                                    y = 2 – 2x
3
y = 1.5x – 3
3y = x – 9,                                   x = 0; y = 2 – 2 (0) = 2
3
y = 1x – 3                                 Intersection = (0,2)
3
1.5x – 3 = 1 x – 3,
3
1.5x = 1 x, x = 0
3
y = 1.5(0) – 3 = –3
Intersection = (0,–3)

e   For point of intersection:
2y – 3x = 6; x < 0
2y = 3x + 6,
y = 1.5x + 3
x = 0; y = 1.5(0) + 3 = 3
Intersection = (0,3)
f   For point of intersection:               h   For point of intersection:
y – 3x = 6; y = 3x + 6                       2x – 2 = y; 2x = y + 2
x – y = –3,                                  2x = 3 – 2y
y=x+3                                     y + 2 = 3 – 2y,
3x + 6 = x + 3,                                3y = 1, y = 1
3
2x = –3, x = – 3                                1 +2= 7
2                               2x =
3 +3= 3                                    3     3
y= –                                             7
2     2                                 x=
6
Intersection = ( – 3 , 3 )
2 2                      Intersection = ( 7 , 1 )
6 3

i   For point of intersection:
2y = 4x + 2; y = 2x + 1
g   For point of intersection:                    y = 2x + 1
x – y = 1; y = x – 1                         Since both equations are exactly the
y – x = 1, y = x + 1                         same, there is an infinite number of
Since both line have the same gradient       points of intersection.
but different y-intercepts, they are
parallel and thus do not intersect.
Solutions to Exercise 18C
1   The following graphs are found by         c   3x + 4y ≤ 60,
shading the included area of the 4 or 5            4y ≤ 60 – 3x,
inequations given.                                  y ≤ 15 – 3 x
Equations that are not already in the y                               4
(<,>, ≤ , ≥ ) mx + c form are given in
that form below.

a   x + y ≤ 7,
y ≤ 7–x
5x – 3y ≤ 15,
3y ≥ 5x – 15,
y ≥ 5x – 5
3

d   x + 2y ≥ 10,
2y ≥ 10 – x,
y ≥ 5 – 1x
2

b   –x + 2y ≤ 12,
2y ≤ x + 12,
y ≤ 1x + 6
2

e   –4x + 5y ≤ 20,
5y ≤ 4x + 20,
y ≤ 4x + 4
5
2x + 3y ≥ 6,
3y ≥ 6 – 2x,
2
y ≥ 2 – 3x
f   x + y ≤ 9,               h   3x + y ≥ 15,
y ≤ 9–x                       y ≥ 15 – 3x
24x + 30y ≥ 360,             2x + 3y ≥ 36,
30y ≥ 360 – 24x,            3y ≥ 36 – 2x,
y ≥ 12 – 4 x               y ≥ 12 – 2 x
5                      3

g   x + y ≥ 6,
y ≥ 6–x              i   x + y ≤ 4,
2x + 5y > 20,                    y ≤ 4–x
5y > 20 – 2x,
y > 4 – 2x
5
Solutions to Multiple-choice questions
1   Substituting in the corner points:      6    The red line is given by y ≥ x.
(0,40) gives us 2(0) + 3(40) = 120.          The blue line is y ≤ 6 – x,
(16,40) gives us 2(16) + 3(40) = 152.         x + y ≤ 6.
(48,20) gives us 2(48) + 3(20) = 156.        The yellow line is y ≥ 2 – 2x,
Thus, (48,20) gives us the maximum P.         2x + y ≥ 2. ⇒ C
⇒D
7    The red line is given by y ≤ x,
2   5x + 3y ≤ 30,                                 x ≥ y.
3y ≤ 30 – 5x,                          The blue line is given by x + y ≤ 3.
y ≤ 10 – 5x                           ⇒D
Graph A is the only option with a
y-intercept of 10. ⇒ A                  8    Substituting in the corner points:
(0,0) gives us 3(0) – 4(0) + 25 = 25.
3   The linear graph given has                   (0,4) gives us 3(0) – 4(4) + 25 = 9.
m = -1 = – 1 and c = 1.                      (3,6) gives us 3(3) – 4(6) + 25 = 10.
3     3                                 (7,0) gives us 3(7) – 4(0) + 25 = 46.
y≥ – 1 x + 1,                                Thus 9 is the lowest value. ⇒ B
3
3y ≥ –x + 3,
9    For the blue line:
3y + x ≥ 3 ⇒ D
m = -6 = – 6 , c = 6,
7     7
4   Substituting in the corner points:                6 x + 6,
y≤ –
(0,60) gives us 2(0) + 3(60) = 180.               7
(20,40) gives us 2(20) + 3(40) = 160.          7y ≤ 42 – 6x,
(30,0) gives us 2(30) + 3(0) = 60.             6x + 7y ≤ 42
Thus, (0,60) gives us the maximum T.         For the red line:
⇒A                                           m = -8 = – 8 , c = 8,
5     5
5   Substituting in the corner points:           y≤ – 8 x + 8,
5
(0,5) gives us 2(0) + 3(5) = 15.
5y ≤ 40 – 8x,
(3,2) gives us 2(3) + 3(2) = 12.
8x + 5y ≤ 40 ⇒ D
(4,0) gives us 2(4) + 3(0) = 8.
Thus 15 is the highest P. ⇒ D
10   Substituting in corner points:
(0,30) gives us 2(0) + (30) = 30.
(20,60) gives us 2(20) + (60) = 100.
(40,80) gives us 2(40) + (80) = 160.
Thus, highest P is at (40,80). ⇒ B
Chapter 19 – Revision: Graphs and
relations
Solutions to Multiple-choice questions
1   90 is in the third step, so costs \$1.00.   7    Substituting in corner points:
30 is in the first step, so costs \$0.50.        (2,9) gives us 2(2) +(9) = 13.
Total cost = \$1.50 ⇒ D                          (4,11) gives us 2(4) + (11) = 19.
(6,10) gives us 2(6) + (10) = 22.
2   Substituting in (2,1):                          (6,1) gives us 2(6) + (1) = 13.
1 = 3(2) + c,                                   Highest z value is thus 22. ⇒ B
c = 1 – 6 = –5 ⇒ B
8    6x – 2y = 5 (A)
3   y + 8 = 0,                                      2x – y = 2, 4x – 2y = 4 (B)
y = –8                                      A – B: 2x = 1, x = 0.5.
x – 12 = 0,                                     2(0.5) – y = 2,
x = 12                                       y = 1 – 2 = –1.
Intersection = (12,–8) ⇒ E                      y
= -1 = –1 × 2 = –2 ⇒ A
x    0.5
4   There are 3 segments in which the car      9    2x – y + 3 = 0,
is moving and 2 segments in which the                y = 2x – 3
car is at rest.                                 ax + 3y – 1 = 0,
First moving segment:                                3y = 1 – ax,
speed = distance / time
y = 1 – ax
= 350 = 70 km/h.                                 3       3
5                                   Since the lines are parallel, the
Second moving segment:                          gradients must be equal.
speed = distance / time
– a = 2, a = –6 ⇒ A
= 250 = 83.33 km/h.                        3
3
Third moving segment:                      10   x + 2y = –2,
speed = distance / time                             x = –2 – 2y.
= 100 = 50 km/h.                         x – 6y = 18.
2
Substituting the first equation into
Thus, second moving segment (9 to 12
second gives us:
hours) has the fastest speed. ⇒ C               (–2 – 2y) – 6y = 18,
–8y = 20,
5   After 12 hours, the car has travelled                        y = –2.5
350 km away from the origin to its              x = –2 – 2(–2.5)
destination, rested there for 4 hours,            = –2 + 5 = 3 ⇒ A
and travelled 250 km back to the
origin, giving a total distance of 600     11   First segment: points are (0,1) and (2,3)
km. ⇒ D                                         m= 3–1 = 2 =1
2–0       2
6   Points are (4,320) and (6,450).                 c = 1, y = x + 1
m = a = 450 – 320 = 130 =65                     Second segment: points are (2,3) and (2.5,5)
6–4           2
C = 65x + b                                     m= 5–3 = 2 =4
Substituting in (4,320):                             2.5 – 5       0.5
320 = 65(4) + b,                                y = 4x + c
b = 320 – 260 = 60                            Substituting in (2,3): 3 = 4(2) + c,
C = 65x + 60,                                     c = 3 – 8 = –5.
If x = 1, C = 65(1) + 60 = 125. ⇒ E           y = 4x – 5 ⇒ A
19   To be parallel to the y-axis, the line
12   Substituting in (–1,–1):                       must have an infinite gradient, i.e. the
2(–1) + 3(–1) + c = 0,                         line must be vertical.
c = 2 + 3 = 5 ⇒E                             Of the given options, only x = –3 is
vertical, with all the other lines have a
13   x + y = 2,                                     gradient between –1 and 1. ⇒ D
y = 2 – x.
x + 3y = –4.                              20   The line is parallel to the x-axis. It is
Substituting first equation into second        thus horizontal, meaning that every
gives us:                                      y-value on the line will be the same.
x + 3(2 – x) = –4,                             The point Q is on the y–axis, so will
x + 6 – 3x = –4,                             have x–coordinate 0 and must have
2x = 10, x = 5 ⇒ E                           y-coordinate 6.
Q = (0,6) ⇒ D
14   Points are (–4,8) and (6,10).
m = 10 – 8 = 2 = 1                        21   2x – 5y = 7,
6 – -4         10       5                       5y = 2x – 7,
1x + c
y=
5                                                 y = 2x – 7
5     5
Substituting in (6,10):
Thus, the gradient = 2 . ⇒ B
10 = 1 (6) + c,                                                       5
5
c = 10 – 6 = 44                         22   y = kxn.
5            5
Substituting in (1.0, 4.50):
y= 1 x + 44 ,
5      5                                    4.5 = k(1)n,
5y = x + 44,                                   k = 4.5; y = 4.5xn.
5y – x = 44 ⇒ D                              Substituting in (1.2,7.78):
7.78 = 4.5(1.2)n,
15   The x-intercept is at y = 0.                     1.729 = 1.2n
3(0) = 3x – 4,                                 Since 1.23 = 1.728, n = 3.
y = 4.5x3 ⇒ D
3x = 4, x = 4 ⇒ A
3
23   20x – 14y = 63 (A)
16   3y = 3x – 4,                                   5x + 6y = 68,
y = x – 4,                                     20x + 24y = 272 (B)
3                                    B – A: 38y = 209,
m = 1 ⇒C                                                 y = 5.5.
5x + 6(5.5) = 68,
17   2y = 3x – 4,                                            5x = 35, x = 7 ⇒ D
y = 3 x – 2.
2
3 , so the correct given line will    24   m = 5 , c = 0.
m=                                                  2
2
The x–axis is x2 and not x, so y = 5 x2.
have a gradient of 3 as well.                                                       2
2                          ⇒E
Taking option D: 2y = 3x + 1,
y = 3x + 1                               25   Break-even is when R = C,
2       2
None of the other given lines have a           7n = 500 + 5n,
2n = 500, n = 250. ⇒ D
gradient of 3 . ⇒ D
2
26   The line is above y = 2200 between
18   Points are (1,2) and (3,–4).                   x = 4 and x = 6, and then between
m = -4 – 2 = -6 = –3 ⇒ A                       x = 19 and x = 29, giving us a total of
3–1           2                           12 seconds above 2200 rpm. ⇒ C
27   Substituting in (3,–2):                   31   3x = 5 – y,
k                                         y = 5 – 3x.
–2 = 2 ,
3                                        2x – 3y = 7.
k = 9 × –2 = –18 ⇒ A                         Substituting first equation into second:
2x – 3(5 – 3x) = 7,
28   Points are (0,8) and (3,6).                      2x – 15 + 9x = 7,
11x = 22, x = 2.
m = 6 – 8 = – 2 , c = 8.
3–0        3                                y = 5 – 3(2) = –1 ⇒ E
y= – 2 x + 8.
3                                    32   3x – y ≥ 3,
Substituting in y = 0:                              y ≤ 3x – 3.
0 = – 2 c + 8,                                 This represents the yellow line (the red
3
line in option C), so the shaded region
2 c = 8,
3                                            should be below the yellow line.
2c = 24, c = 12 ⇒ C
x + y ≤ 3,
29   Revenue from 1000 copies = 25(1000)                y ≤ 3 – x.
= 25 000.                                      This represents the red line (the yellow
Therefore the incorrect statement is D.        line in option C), so the shaded region
should be below the red line.
30   P=R–C
= 25x – 15 000 – 15x                          The only option where both of these
= 10x – 15 000 ⇒ B                            conditions are met is in option A.
Chapter 20 – Principles of financial
mathematics
Solutions to Exercise 20A
1                                           3   Note that percentage discount
a   i Mark–down = 0.1 × 20 = \$2.00              = 100% – final price / initial price × 100%.
ii New price = 0.9 × 20 = \$18.00
a   Percentage discount
b   i Mark–down = 0.25 × 3.6 = \$0.90            = 100% – 15 × 100%
ii New price = 0.75 × 3.6 = \$2.70                      20
= 100% – 75% = 25%
c   i Mark–down = 0.3 × 75 = \$22.50
ii New price = 0.7 × 75 = \$52.50        b   Percentage discount
= 100% – 27 × 100%
30
d   i Mark–down = 0.5 × 40 = \$20.00
= 100% – 90% = 10%
ii New price = 0.5 × 40 = \$20.00
c   Percentage discount
e   i Mark–down = 0.2 × 2.99 = \$0.60
ii New price = 0.8 × 2.99 = \$2.39           = 100% – 19.00 × 100%
39.99
= 100% – 50% = 50%
f   i Mark–down = 0.15 × 14.5 = \$2.18
ii New price = 0.85 × 14.5 = \$12.33     d   Percentage discount
g   i Mark–down = 0.125 × 42.99 = \$5.37         = 100% – 19082.50 × 100%
22450
ii New price = 0.875 × 42.99 = \$37.62         = 100% – 85% = 15%

2                                           4   Note that percentage increase =
a   i Mark–up = 0.1 × 20 = \$2.00                final price / initial price × 100%.
ii New price = 1.9 × 20 = \$22.00
a   Percentage increase
b   i Mark–up = 0.25 × 3.6 = \$0.90              = 20 × 100% – 100%
ii New price = 1.25 × 3.6 = \$4.50              15
= 133.33% – 100% = 33.33%
c   i Mark–up = 0.3 × 75 = \$22.50
ii New price = 1.3 × 75 = \$97.50        b   Percentage increase
= 11.94 × 100% – 100%
d   i Mark–up = 0.5 × 40 = \$20.00                  9.95
ii New price = 1.5 × 40 = \$60.00              = 120% – 100% = 20%

e   i Mark–up = 0.2 × 2.99 = \$0.60          c   Percentage increase
ii New price = 1.2 × 2.99 = \$3.59           = 1050 × 100% – 100%
1000
= 105% – 100% = 5%
f   i Mark–up = 0.15 × 14.50 = \$2.18
ii New price = 1.15 × 14.50 = \$16.68
d   Percentage increase
g   i Mark–up = 0.125 × 42.99 = \$5.37           = 2250 × 100% – 100%
2000
ii New price = 1.125 × 42.99 = \$48.36         = 112.5% – 100% = 12.5%
5   Let x be the original price of the item.   9    Note that percentage discount =
a   20 = 0.85 × x,                                  100% – final price / initial price × 100%
x = 20 = \$23.53
0.85                                     Racquets:
Sporties: 100% – 78.99 × 100%
b   44.96 = 0.9 × x,                                                     94.99
= 100% – 83.16% = 16.8%.
x = 44.96 = \$49.96
0.9                                     Goodalls: 100% – 78.99 × 100%
92.99
c   50 = 0.75 × x,                                            = 100% – 84.94% = 15.1%.
x = 50 = \$66.67                               Shoes:
0.75
Sporties: 100% – 45.98 × 100%
64.50
d   18 880 = 0.92 × x,
= 100% – 71.29% = 28.7%.
x = 18 880 = \$20 521.74
0.92                                   Goodalls: 100% – 45.98 × 100%
72.99
= 100% – 63.99% = 37.0%.
6   Let x be the original price of the item.
a   20 = 1.15 × x,                                  Skirts:
x = 20 = \$17.39                               Sporties: 100% – 23.50 × 100%
1.15                                                          33.98
= 100% – 69.16% = 30.8%.
b   24 = 1.125 × x,
Goodalls: 100% – 23.50 × 100%
x = 24 = \$21.33                                                    32.99
1.125                                              = 100% – 71.23% = 28.8%.

c   65.95 = 1.05 × x,                               T–shirts:
x = 65.95 = \$62.81                            Sporties: 100% – 24.99 × 100%
1.05                                                          33.75
= 100% – 74.04% = 26.0%.
d   125 00 = 1.025 × x,                             Goodalls: 100% – 24.99 × 100%
33.99
x = 12 500 = \$12 195.12                                 = 100% – 73.5% = 26.5%.
1.025

7   Petrol = 20 × 0.978 × 0.9                       Costless’ claim is incorrect for the
= \$17.604                                racquets but correct for all the other
Food and drink = (1.70 + 3.05) × 0.85           items.
= \$4.0375
Total = \$21.64                             10
a    Reduced price 1 = 240 × 0.85 = \$204
8   New airfare = \$1850 × 1.025                     Reduced price 2 = 204 × 0.85 = \$173.40
= \$1896.25
New accommodation = \$550 × 1.07            b    Percentage discount
= \$588.5                      = 100% – 173.4 × 100%
240
New total = \$2484.75                              = 100% – 72.25% = 27.75%.
Price increase
= \$2484.75 – (\$1850 + \$550)             11
= \$2484.75 – \$2400                      a    Increased price 1 = 24 950 × 1.03
= \$84.75                                        = \$25 698.5
Increased price 2 = 25 698.5 × 1.035
= \$26 597.95

b    Percentage increase
= 26597.95 × 100% – 100%
24950
= 6.61%.
12                                        13   Let x be the value of the goods.
a    Cost = 50 × 1.15 × 0.9 = \$51.75
a    The discounted value of the goods he
b    New price per litre = 1.15 × 1.10         will be able to buy is \$100, meaning
= \$1.265                                \$100 is 80% of the value of the goods.
New price per litre after discount        100 = 0.8 × x,
= 1.265 × 0.9                             x = 100 = \$125
= \$1.14 / L                                    0.8

c    Percentage discount                  b    The discounted value of the goods he
will be able to buy is \$100, meaning
= 100% – 1.14 × 100%
1.15                            \$100 is 70% of the value of the goods.
= 100% – 99%                            100 = 0.7 × x,
= 1% discount                             x = 100 = \$142.86
0.7

14   Let x be the original price.
153.75 = x × 0.75,
x = 153.75 = \$205.00
0.75

15   Let x be the original price.
2.35 = 1.125 × x,
x = 2.35 = \$2.09
1.125
Solutions to Exercise 20C
1   Note that I = P × (1 + r )n – P
100                 6
6 )4 – 2000
a   I = 2000 × (1 +
100                       a     I = 7000 × (1 + 8 )4 – 7000
100
= 2000 × 1.26 – 2000 = \$524.95                   = 7000 × 1.36 – 7000 = \$2523.42

b   I = 10 000 × (1 + 12 )5 – 10 000          b     I = 2995 × (1 + 7.2 ÷ 12 )240 – 2995
100                                             100
= 10 000 × 1.76 – 10 000                         = 2995 × 4.203 – 2995 = \$9591.71
= \$7623.42
7
c   I = 8000 × (1 + 12.5 )3 – 8000            a     A = 850 × (1 + 13.25 ÷ 52 )26
100                                                100
= 8000 × 1.42 – 8000 = \$3390.63                  = 850 × 1.068 = \$908.14

2   A = 3000 × (1 + 10 )5                     b     Typing the equation into the
100
Y= function of the graphics calculator
= 3000 × 1.61 = \$4831.53
gives us the following graph:
3   I = 3300 × (1 + 7.5 )10 – 3300
100
= 3300 × 2.06103 – 3300
= \$3501.40

4   Simple interest = 4500 × 0.11 × 6
= \$2970
8
Compound interest
= 4500 × (1 + 11 )6 – 4500              a     A = 1500 × (1 + 11 ÷ 52 )260
100
100
= 4500 × 1.87 – 4500                            = 1500 × 1.732 = \$2598.37
= \$3916.87
Thus, compound interest earns             b     A = 1500 × (1 + 11.75 ÷ 4 )20
100
3916.87 – 2970 = \$946.87 more than
= 1500 × 1.784 = \$2676.48
simple interest.
c     A = 1500 × (0.125) × 5 + 1500
5                                                     = \$2437.50
a   i A = 1000 × (1 + 7 )1
100
= 1000 × 1.07 = \$1070                 Thus, b is the best option for Philip.
ii A = 1000 × (1 + 7 ÷ 4 )4
100                9     A = 300 × (1 + 18 ÷ 12 )6
= 1000 × 1.07186 = \$1071.86                                   100
= 300 × 1.093 = \$328.03
iii A = 1000 × (1 + 7 ÷ 12 )12
100
= 1000 × 1.07229 = \$1072.29           10
iv A = 1000 × (1 + 7 ÷ 52 )52
a     Price = 0.8 × 1299 = 1039.2
100                     A = 1039.2 × (1 + 18 ÷ 12 )4
= 1000 × 1.07246 = \$1072.46                                         100
v A = 1000 × (1 + 7 ÷ 356 )356
= 1039.2 × 1.061 = \$1102.97
100
= 1000 × 1.07250 = \$1072.50           b     No, since this way he saves
1299 – 1102. 97 = \$196.03
b   Interest is greater the more frequently
it is compounded.
11                                            17   7774.69 = 6000 × (1 + 5.25 ÷ 2 )2n
A = 3000 × (1 + 5.65 ÷ 52 )260
100
a
100                          1.026252n = 1.2958 = 1.0262510
= 3000 × 1.326 = \$3978.72                     2n = 10, n = 5 years

b    Typing the equation into the             18   20 000 = 200 × (1 + 4.75 )n
Y= function of the graphics calculator                             100
gives us the following graph.                   1.0475n = 100 ≈ 1.0475100
n = 100 years

19   Using the equation solver function of
the graphics calculator:
a    P = 1000, r = 10, n = 4, A = 2000,
gives t = 8 years

12   12 000 = x × (1 + 6.8 )4                 b    P = 1000, r = 12, n = 4, A = 2000,
100                        gives t = 6 years
12 000 = x × 1.301
x = 12 000 = \$9223.51                  20   Using the equation solver function of
1.301
the graphics calculator:
13                                            a    P = 1000, r = 5, n = 12, A = 2000,
gives t = 14 years
a    15 000 = x × (1 + 6.25 )5
100
15 000 = x × 1.354                     b    P = 1000, r = 6, n = 12, A = 2000,
gives t = 12 years
x = 15 000 = \$11 077.62
1.354
21   Using the equation solver function of
b    15 000 = x × (1 + 6.25 ÷ 365 )1825            the graphics calculator:
100                      P = 1000, n = 12, A = 1051.16, t = 1,
15 000 = x × 1.367                          gives r = 5%
x = 15 000 = \$10 974.53
1.367
22   Using the equation solver function of
11 077.62 – 10 974.53 = \$103.09 less.         the graphics calculator:
P = 18 000, n = 4, A = 19 299.27, t = 2,
14   30 000 = x × (1 + 9.7 ÷ 4 )10                 gives r = 3.5%
100
30 000 = x × 1.271
23
x = 30 000 = \$23 608.13                a    Using the equation solver function of
1.271
the graphics calculator:
2695.55 = x × (1 + 11.65 ÷ 2 )42
P = 30 000, n = 12, A = 60 000, t = 3,
15
100                  gives r = 23%
2695.55 = x × 10.782
x = 2695.55 = \$250.00                  b    No; the time taken to double an invest
10.782                                 depends on the rate, time and
compounding frequency but is
16   5241.61 = 3500 × (1 + 6.75 ÷ 12 )12n          independent of the principal.
100
1.00562512n = 1.4976 = 1.00562572
12n = 72, n = 6 years
Solutions to Exercise 20D
1                                                       4
End of year    Interest    Repayment     Balance    a    i Amount owing = \$24551.42
1          1920.00       2700       11 220.00          interest paid = \$1551.43
2          1795.20       2700       10 315.20
3          1650.43       2700         9265.63    End of month    Interest   Repayment     Balance
4          1482.50       2700         8048.13          1          312.50       400       24 912.50
Total        6848.13                                     2          311.41       400       24 823.91
3          310.30       400       24 734.21
a      After 4 years, \$8048.13 is still owing.                 4          309.18       400       24 643.38
5          308.04       400       24 551.42
b      After 4 years, \$6848.13 in interest has              Total        1551.42
been paid.
ii Amount owing = \$23013.45
2                                                               interest paid = \$1513.45
End of month    Interest    Repayment    Balance     End of month    Interest   Repayment     Balance
1           96.75        225       6321.75           1          312.50       700       24 612.50
2           94.83        225       6191.58           2          307.66       700       24 220.16
3           92.87        225       6059.45           3          302.75       700       23 822.91
4           90.89        225       5925.34           4          297.79       700       23 420.69
5           88.88        225       5789.22           5          292.76       700       23 013.45
Total         464.22                                 Total        1513.45
a      After 5 months, \$5789.22 is still
owing.                                           b    The \$700 per month repayment plan is
probably better, since the loan will be
b      After 5 months, \$464.22 in interest has               paid off faster and the total interest
been paid.                                            paid will be lower, meaning Jack saves
time and money.
3
End of month       Interest    Repayment     Balance
1             312.50        595       24 717.50
5
End of month    Interest    Repayment    Balance
2             308.97        595       24 431.47
1          687.50        840      10 9847.50
3             305.39        595       24 141.86
2          686.55        840      10 9694.05
4             301.77        595       23 848.64
3          685.59        840      10 9539.63
5             298.11        595       23 551.74
4          684.62        840      10 9384.26
Total           1526.74
Total        2744.26
a      After 5 months, \$23 551.74 is still
a    After 4 months, \$109 384.26 is still
owing.
owing.
b      After 5 months, \$1526.74 in interest
has been paid.                                   b    After 4 months, \$2744.26 in interest
has been paid.
6                                                  7
a                                                  a      A = 149.99 × (1 + 18 ÷ 12 )3
End of month   Interest   Repayment    Balance                                   100
1         687.50       750      109 937.50             = 149.99 × 1.0457 = \$156.84.
2         687.11       750      109 874.61
3         686.72       750      109 811.33
b
4         686.32       750      109 747.65
End of month   Interest   Repayment   Balance
Total       2747.65
1          2.25        50        102.24
Amount owing = \$109 747.65,                             2          1.53        50         53.77
interest paid = \$2747.65                                3          0.81         0         54.58
Total         4.59
b                                                         Final payment at end of third month
End of month   Interest   Repayment    Balance            would have been \$54.58.
1         687.50       900      109 787.50
2         686.17       900      109 573.67
Saving would be 156.84 – (50 + 50 +
3         684.84       900      109 358.51
4         683.49       900      109 142.00          54.58) = \$2.26.
Total       2742.00
Amount owing = \$109 142.00,
interest paid = \$2742.00
Solutions to Multiple-choice questions
1   Percentage discount                         10   Reduced price = 0.8 × 250 = \$200.
= 100% – 485 × 100%                            Discounted price = 0.875 × 200
598                                = 1\$75. ⇒ E
= 100% – 81.10% = 19% ⇒ A
11   175 × 100% = 70%.
2   Let x be the original price.                     250
375 = 0.75 × x,                                  Thus, discount = 100% – 70% = 30%.
x = 375 = 500 ⇒ B                              ⇒B
0.75
12   Using the equation solver function of
3   4687.50 = 25000 × r × 5,                         the graphics calculator:
100
P = 4500, r = 5.75, n = 12, A = 10 000,
4687.5 = 1250 × r = 4687.50,
gives t = 14 years ⇒ D
r = 3.75% ⇒ A
13   We must first find the total money in
4   I = 3000 × 0.049 × 3
the account at the end of 5 years and
= 441 ⇒ E
then calculate the interest in the 6th
year.
5   Before price rise, 2 hours = \$110.
New price = 110 × 1.15 = \$126.50.                A = 10 000 × (1 + 5.3 ÷ 12 )60
100
Price increase = 126.50 – 110                      = 10 000 × 1.3027 = \$13 026.71.
= \$16.50 ⇒ D
Interest in 6th year
6   A = 6000 × (1 + 4.75 ÷ 12 )24                      = 13 026.71 × (1 + 5.3 ÷ 12 )12 –13 026.71
100                                                         100
= 6000 × 1.099                                   = 13 026.71 × 1.054 – 13 026.71
= 6596.72 = \$6597 ⇒ C                            = \$707.43 ⇒ A

7   20 000 = P × 0.05 × 3 + P,                  14
1.15 × P = 20 000,                        End of month   Interest     Repayment     Balance
P = 20 000 = 17391.30 ⇒ A                       1         131.55         225       17 446.55
1.15                                      2         130.85         225       17 352.40
3         130.14         225       17 257.54
8   20 000 = P × (1 + 5 ÷ 4 )12,                     a = \$130.14 ⇒ A
100
20 000 = P × 1.161,                       15   From the above table, b = \$17 257.54.
P = 20 000 = \$17230 ⇒ C                        ⇒C
1.161

9   Since there is an initial amount
invested, the y-intercept will be greater
than 0.
The graph will tend upwards since the
amount of money in the investment is
increasing; the graph will curve
upwards since the compound interest
means the interest at each compound
will be greater each time. ⇒ B
Chapter 21 – Applications of
financial mathematics
Solutions to Exercise 21A
1                                            3
a   Tax bracket = 1                          a   Monthly salary = 45 732 = \$3811
Tax = \$0                                                        12

b   Tax bracket = 3                          b   Tax bracket = 3.
Tax = \$0 for bracket 1,                      Tax = \$0 for bracket 1,
0.17 × 15 600 = \$2652 for bracket 2,         0.17 × 15 600 = \$2652 for bracket 2,
0.30 × 2270 = \$681 for bracket 3.            0.30 ×24 132=\$7239.60 for bracket 3.
Tax = 2652 + 681 = \$3333                     Tax = 2652 + 7239.60 = \$9891.60.
Net income = 45 732 – 9891.60
c   Tax bracket = 3                                          = \$35 840.40 per year.
Tax = \$0 for bracket 1,                        35 840.40 = \$2986.70 per month.
12
0.17 × 15 600 = \$2652 for bracket 2,
0.30 × 37 090 =\$11 127 for bracket 3.
4
Tax = 2652 + 11 127 = \$13 779
a   Income = 18 × 52 × 22.50 = \$2 1060
d   Tax bracket = 5
b   Tax bracket = 2.
Tax = \$0 for bracket 1,
Tax = \$0 for bracket 1,
0.17 × 15600 = \$2652 for bracket 2,
0.17 ×15 060=\$2560.20 for bracket 2.
0.30 × 48400 =\$14 520 for bracket 3,
Tax = \$2560.20
0.42 × 55000 =\$23 100 for bracket 4,
0.47 × 3950 = \$1856.50
c   Net income = 21 060 – 2560.20
Tax = 2652 + 14 520 + 23 100 + 1856.50
= \$18 499.80.
= \$42 128.50
Let net hourly rate = h.
18 499.80 = 18 × 52 × h,
2
h = \$19.76 / hour
a   Tax bracket = 3
Tax = \$0 for bracket 1,
5
0.17 × 15600 = \$2652 for bracket 2,
a   Profit = 23 000 – 13 500 = \$9500.
0.30× 27325=\$8197.50 for bracket 3.
Tax = 2652 + 8197.50 = \$10 849.50
b   Tax bracket = 4.
Tax = 0.42 × 9500 = \$3990
b   This money will be taxed at the
bracket 3 rate, so tax = 0.30 × 4500
6   Tax bracket = 4.
= 1350.
Tax = 0.42 × 34 345 = \$14 424.90
Thus, the money she saves is
4500 – 1350 = \$3150.
7
a   Tax bracket = 5.
Tax = 0.47 × 62 000 = \$29 140

b   Tax bracket = 3.
Tax = 0.30 × 62 000 = \$18 600
8                                   11   39 990 = 1.1 × x,
a    GST = 0.1 × 121.30 = \$12.13           x = 39 990 = \$36 354.55.
1.1
b    GST = 0.1 × 67.55 = \$6.76           GST = 39 990 – 36 354.55
= \$3635.45
c    GST = 0.1 × 985.50 = \$98.55
12
d    GST = 0.1 × 395 = \$39.50       a    318.97 = 1.1 × x,
x = 318.97 = \$289.97
1.1
9
a    Price after GST = 1.1 × 139
b    GST = 318.97 – 289.97 = \$29.00
= \$152.90
13   Duty bracket = 3.
b    Price after GST = 1.1 × 2678
Duty = 2560 + 0.06 × (660 000 – 115 000)
= \$2945.80
= 2560 + 37 200 = \$35 260
c    Price after GST = 1.1 × 9850
14   Duty bracket = 4.
= \$10 835
Duty = 0.055 × 1 280 000
= \$70 400
d    Price after GST = 1.1 × 1395
= \$1534.50
15   Duty bracket = 2.
Duty = 280 + 0.024 × (114 600 – 20 000)
10   2399 = 1.1 × x,
= 280 + 2270.40 = \$2550.40
x = 2399 = \$2180.91
1.1
16   Duty bracket = 3.
Duty = 2560 + 0.06 × (523 000 – 115 000)
= 2560 + 24 480 = \$27 040
Solutions to Exercise 21B
1                                             6
a   Balance = 5595 – 1500 – 950 = \$3145       a   Minimum daily balance is 73 226 for 9
days, 71 226 for 20 days, 75 226 for 63
b   Minimum monthly balance was \$3145.            days.
Interest = 3145 × 0.0075 = \$23.59             Interest = 5.3 ÷ 365 × (9 × 73226 + 20
100
2   Total interest = 0.0425 × (130 + 460.50       × 71226 + 63 × 75226)
12                           = 0.053 × 6 822 792 = \$990.71
+ 545.63 + 391.49)                                 365
= 0.00354 × 1527.62 = \$5.41
b   Minimum monthly balance is 71 226
3   Minimum monthly balance in April              for 1 month, 75226 for 2 months.
= 5525 – 500 = \$5025.                       Interest = 5.4 ÷ 12 × (71 226 + 75 226
Minimum monthly balance in May                            100
= \$5025.                                    + 75 226)
Minimum monthly balance in June                 = 0.054 × 221 678 = \$997.55
12
= 5025 + 175 = \$5200.
Interest = 0.06 × (5025 + 5025 + 5200)    7   Minimum daily balance is 500 for 39
12
= 0.005 × 15 250 = \$76.25                   days, 900 for 124 days, 1250 for 21
days.
4                                                 Interest = 3 ÷ 365 × (39 × 500 + 124 ×
100
a   Minimum daily balance is 650.72 for 5
900 + 21 × 1250)
days, 445.82 for 14 days, 1241.37 for
12 days.                                        = 0.03 × 157 350 = \$12.93
365
Interest = 3 ÷ 365 × (5 × 650.72 + 14 ×
100
8   Minimum daily balance is 10 000 for
445.82 + 12 × 1241.37)
12 days, 14 350.98 for 28 days,
= 0.06 × 24 391.52 = \$2.00                  12 073.54 for 21 days.
365
Interest = 4 ÷ 365 × (12 × 10 000 + 28
100
b   Minimum monthly balance is 445.82.
× 14 350.98 + 21 × 12 073.54)
Interest = 3 ÷ 12 × 445.82 = \$1.11
100                                 = 0.04 × 775 371.78 = \$84.97
365
5
a   Minimum daily balance is 650.72 for 8
days, 900.72 for 13 days, 1150.72 for
10 days.
Interest = 3.75 ÷ 365 × (8 × 650.72 +
100
13 × 900.72 + 10 × 1150.72)
= 0.0375 × 28 422.32 = \$2.92
365

b   Minimum monthly balance is 650.72.
Interest = 3.75 ÷ 12 × 650.72 = \$2.03
100
Solutions to Exercise 21C
1   Note that flat rate = 100 × I .    3
P×t       a   I = 3000 × 0.185 = \$555
a   I = 30 × 90 – 2400 = \$300              Total cost = \$3555
Flat rate = 100 × 300 = 5%             Monthly cost = 3555 = \$296.25
2400 × 2.5                                12

b   I = 15 × 115 – 1500 = \$225.        b   I = 5500 × 0.125 × 1.5 = \$1031.25
Flat rate = 100 × 225 = 12%
Total cost = \$6531.25
1500 × 1.25                Monthly cost = 6531.25 = \$362.85
18
c   I = 24 × 350 – 8000 = \$400
c   I = 15 000 × 0.14 × 6 = \$12 600.
Flat rate = 100 × 400 = 2.5%           Total cost = \$27 600
8000 × 2
Monthly cost = 27600 = \$383.33
72
d   I = 6 × 135 – 750 = \$60
Flat rate = 100 × 60 = 16%         d   I = 2250 × 0.0975 × 0.5 = \$109.69
750 × 0.5
Total cost = \$2359.69
e   I = 18 × 420 – 7250 = \$310             Monthly cost = 2359.69 = \$393.28
6
Flat rate = 100 × 310 = 2.85%
7250 × 1.5             e   I = 4200 × 0.1125 × 1.25 = \$590.63.
Total cost = \$4790.625
2
a   I = 52 × 42.50 – 2000 = \$210           Monthly cost = 4790.625 = \$319.38
15
Flat rate = 100 × 210 = 10.5%
2000 × 1               4
a   I = 1800 × 0.113 × 0.75 = \$152.55
b   I = 26 × 29.90 – 750 = \$27.40          Total cost = \$1952.55
Flat rate = 100 × 27.40 = 7.31%        Weekly cost = 1952.55 = \$50.07
750 × 0.5                                  39

c   I = 39 × 150 – 5500 = \$350         b   I = 14950 × 0.098 × 2 = \$2930.20
Flat rate = 100 × 350 = 8.48%          Total cost = \$17 880.20
5500 × 0.75                Weekly cost = 17 880.20 = \$171.93
104
d   I = 65 × 97.50 – 6000 = \$337.50
Flat rate = 100 × 337.50 = 4.5%
c   I = 15000 × 0.144 × 5 = \$10 800
6000 × 1.25                Total cost = \$25 800
Weekly cost = 25 800 = \$99.23
e   I = 156 × 22.70 – 3000 = \$541.20                       260

Flat rate = 100 × 541.20 = 6.01%   d   I = 22 250 × 0.1175 × 0.5 = \$1307.19
3000 × 3
Total cost = \$23 557.19
Weekly cost = 23 557.19 = \$906.05
26

e   I = 1250 × 0.1075 × 0.25 = \$33.59
Total cost = \$1283.59
Weekly cost = 1283.59 = \$98.74
13
5   Note that interest to be paid            10
= (price – deposit) × rate × years.      a    325 × 6 × 12 = \$23 400

a   I = (350 – 35) × 0.1                     b    Cash payment = 41 999 – 23 200
= \$31.50                                                  = \$18 799
Saving = 23 400 – 18 799 = \$4601.
b   I = (790 – 100) × 0.12 × 1.25
= \$103.50                              c    Flat rate = 100 × 4601 = 4.08%
18799 × 6

c   I = (3550 – 450) × 0.145 × 2.5           d    Effective rate = 4.08% × 2(72)
= \$1123.75                                                           72 + 1
= 4.08 × 144 = 8.05%
6                                                                         73
a   Cost = 25 × 52 × 3 + 50 = \$3950          11
a    Interest = 48 × 24 + 150 – 1100 =\$202
b   Flat rate = 100 × 3061 = 121.6%
839 × 3
b    Flat rate = 100 × 202 = 10.63%
950 × 2
c   Effective rate = 121.6% × 2(156)
156 + 1       c    Effective rate =10.63% × 2(24)
= 121.6 × 312 = 241.65%                                 24 + 1
157                                = 10.63 × 48 = 20.41%
25
7
12
a   Flat rate = 100 × 4000 = 26.67%          a    Interest = 2000 × 0.18 × 2.25 = \$810
5000 × 3

b    Total price = \$2810.
b   Effective rate = 26.67% × 2(36)
36 + 1              Monthly repayment = 2810 = \$104.07
72 = 51.89%                                    27
= 26.67 ×
37
c    Flat rate = 100 × 810 = 18%
8                                                            2000 × 2.25
a   Flat rate = 100 × 4000 = 10%             i    Exact effective rate = 18% × 2(27)
8000 × 5                                                         27 + 1
= 18 × 54 = 34.71%
28
b   Effective rate = 10% × 2(60)             ii   Approximate effective rate
60 + 1
120 = 19.67%              = 18% × 2 = 36%
= 10 ×
61
13
9                                            a    Interest = (4560 – 320) × 0.15 × 2
a   Interest = 35 × 24 + 85 – 775 = \$150.                  = \$1272

Flat rate = 100 × 150 = 10.87%
b    Total price = 5832 – 320 = \$5512
b
690 × 2                           Monthly repayment = 5512 = \$229.67
24
c   Effective rate = 10.87% × 2(24)
24 + 1         c    Flat rate = 100 × 1272 = 15%
48 = 20.87%                       4240 × 2
= 10.87 ×
25                    i Exact effective rate = 15% × 2(24)
24 + 1
= 15 × 48 = 28.8%
25
ii Approximate effective rate
= 15% × 2 = 30%
Solutions to Exercise 21D
1                                            9
a   Price = 3.50 × 1.027 = \$3.59             a    200 000 = x × (1 + 0.03)10,
x = 200 000 = \$148 818.78
b   Price = 3.59 × 1.035 = \$3.72                         1.344

2                                            b    200 000 = x × (1 + 0.13)10,
a   Price = 825 × 1.032 = \$851.40                   x = 200 000 = \$58 917.67
3.395
b   Price = 851.40 × 1.053 = \$896.52
10
a    1000 = x × (1 + 0.026)20,
3   Price in 2007 = 4.95 × 1.045 × 1.021 ×
1.033 = \$5.46                                   x = 1000 = \$598.48
1.671

4   Price in 2007 = 295 × 1.041 × 1.031 ×    b    1000 = x × (1 + 0.069)20,
1.021 = \$323.26
x = 1000 = \$263.40
3.798
5
a   2.40 × 1.02310 = \$3.01                   c    1000 = x × (1 + 0.143)20,
x = 1000 = \$69.04
b   2.40 × 1.06710 = \$4.59                              14.485

6                                            11
a               20
1.20 × 1.019 = \$1.75                     a    5000 = x × (1 + 0.021)5,
x = 5000 = \$4506.52
b   1.20 × 1.07120 = \$4.73                              1.110

7                                            b    5000 = x × (1 + 0.057)5,
a   500 000 × 1.02612 = \$680 359.31                 x = 5000 = \$3789.61
1.319
b   500 000 × 1.06912 = \$1 113 531.40
c    5000 = x × (1 + 0.128)5,
8                                                 x = 5000 = \$2737.94
1.826
a   52 500 × 1.035 = \$60 861.89

b   525 00 × 1.145 = \$101 084.27
Solutions to Exercise 21E
1                                          6
a   Unit cost = 37 000 – 5000              a    Book value = 65 000 – 65 000 × 0.1 × t
100 000                                 = 65 000 – 6500t
=   32 000 = \$0.32
100 000
b    Graphing y = 65 000 – 6500x gives us:
b   Production = 5234 + 6286 + 3987
= \$15 507
Book value = 37000 – 0.32 × 15507
= \$32 037.76

c   Time in use = 37 000 – 5000
5169 × 0.32           c    Yearly depreciation = 6500(1) = \$6500
= 19.3 years
d    Years = 65 000 – 13 000
6500
2   Drop in value = 29 000 – 5000                       52 000 = 8 years
= \$24 000                           =
6500
x × 0.25 = 24 000,
x = 24 000 = 96 000 km               7
0.25                            a    Book value = 5600 – 5600 × 0.225 × 3
= \$1820
3   Drop in value = 35 400 – 25 700
= \$9700                  b       5600      = 4.44 years
x × 25 000 = 9700,                          5600 × 0.225

x = 9700 = \$0.388 / km.              8
25 000
a    Book value = 7000 – 7000 × 0.175 × 2
4                                                          = \$4550
a   i Unit cost = 11 0000 – 2500           b    7000 – 875 = 6125,
4 000 000
6125     = 5 years
= \$0.026875                     7000 × 0.175
ii Book value = 110 000 – 0.026875 ×
1 500 000 = \$69 687.50              9
iii Depreciation = 0.026875 × 750000   a    Book value = 1200 × 0.887 = \$490.41
= \$20 156.25
b    Depreciation = 1200 – 490.41
b   Book value = 110 000 – 0.026875 ×                        = \$709.59
(750 000 × 5) = \$9218.75
c    1200 × 0.88n = 215,
0.88n = 0.1792
c   70 000 = 110 000 – 0.026875 × x,
n = 13.45, meaning the value is
0.026875x = 40 000,
reached during the 14th year.
x = 1 488 372 pages
10
5                                          a    Book value = 38 500 × 0.9055
a   Depreciation after 3 years = 1700 ×                    = \$23 372
0.125 × 3 = \$637.50
b    Depreciation = 38 500 – 23 372
b   Book value = 1700 – 637.50                               = \$15 128
= \$1062.50
c    38 500 × 0.905n = 10 000,
0.905n = 0.2597
n = 13.51, meaning the value is
reached during the 14th year.
11   x × 0.91810 = 13 770,
x = 13 770 = \$32 400
0.425

12   x × 0.8756 = 56 100,
x = 56 100 = \$125 000
0.449

13   8000 × r3 = 6645,
r3 = 0.8306, r = 0.94 = 6%
Solutions to Multiple-choice questions
1    Tax bracket = 3.                           11   3400 × 0.854 = \$1775 ⇒ E
Tax = \$0 for bracket 1,
0.17 × 15 600 = \$2652 for bracket 2,      12   Using the TVM Solver function of the
0.30 × 16 350 = \$4905 for bracket 3.           graphics calculator, the repayment
Tax = 2652 + 4905 = \$7557 ⇒ A                   must be \$585 to pay off the loan of
\$6000 in 12 equal quarterly
2    Total income = \$39 346                          instalments. ⇒ A
Tax bracket = 3.
Tax = \$0 for bracket 1,                    13   Using the TVM Solver function of the
0.17 × 15 600 = \$2652 for bracket 2,           graphics calculator, the balance of the
0.30 ×17 746 = \$5323.80 for bracket 3.         account (initially \$35 300) after 12
Tax = 2652 + 5323.80 = \$7976 ⇒ E                months of monthly \$220 withdrawals
and interest compounding will be
3    399 = 1.1 × x,                                  closest to \$35 125. ⇒ A
x = 399 = \$362.73 ⇒ C
1.1                                  14   Using the TVM Solver function of the
graphics calculator, the balance of the
4    Duty bracket = 3.                               account (initially \$5000) after 8
Duty = 2560 + 0.06 × (452 500 – 115 000)        quarters of monthly \$430 deposits and
= \$22 810 ⇒ C                              quarterly interest compounding will be
closest to \$16 600. ⇒ C
5    Minimum monthly balance = \$1112.00
Interest = 6 ÷ 12 × 1112 = \$5.56 ⇒ A       15   Using the TVM Solver function of the
100
graphics calculator, the amount still
owed at the end of the second year on
6    The effective interest rate is always           a \$12 000 loan at 12% monthly
greater than the flat rate of 11% due to        interest, compounded monthly, is
the fact that the amount owing is
closest to \$8040. ⇒ C
continually being reduced, whereas the
amount repaid at each repayment is
16   Using the TVM Solver function of the
always the same. ⇒ C
graphics calculator, repayment must be
\$159 to pay off the loan of \$25 000 in
7    Interest paid = 20 × 52 + 100 – 995
60 equal monthly instalments. ⇒ B
= \$145
Flat rate = 100 × 145 = 16.2% ⇒ B          17   Using the TVM Solver function of the
895 × 1
graphics calculator, the principal
required for an investment with yearly
8    Effective rate = 11% × 2(50)                    \$3000 withdrawals and interest at 3.5%
50 + 1
100                       should be closest to \$85 700. ⇒ B
= 11 ×
51
= 21.6% ⇒ E

9    1.20 × 1.02520 = \$1.97 ⇒ D

10   62 000 = x × (1 + 4.1 )15,
100
x= 62 000 = \$33 935 ⇒ D
1.827
related mathematics
Solutions to Multiple-choice questions
1   A = 15 000 × (1 + 6.5 )3                   9    A = 3000 × (1 + 6 ÷ 12 )12n
100                                               100
= \$18 119 ⇒ B                                   = 3000 × 1.00512n ⇒ B

2   Interest = 600 × 12 × 3 + 3000 – 20 250    10   2400 = x × 0.025 × 3,
= \$4350 ⇒ C                              x = 2400 = \$32 000 ⇒ E
0.075
3   Interest = (20 250 – 3000) × 0.1 × 3
= \$5175 ⇒ D                       11   Using the TVM Solver function of the
graphics calculator, the value of an
4   Minimum monthly balance = 2270 for              investment after 2 years that was
1 month, 6326 for 2 months.                     initially \$5000, with monthly \$500
deposits at 6% per annum, will be
Interest = 3 ÷ 12 × (2270 + 6326 + 6326)        \$18 351.78. ⇒ C
100
= 0.0025 × 14 922
= \$37.31 ⇒ A                      12   Value of investment at end of 4th year
= 6350 × (1 + 5.9 ÷ 4 )16
5   Using the TVM Solver function of the                                100
graphics calculator, the monthly                  = 6350 × 1.264 = \$8026.36
repayment on a 20-year, \$80 000 loan            Interest in 5th year
at 7.2% interest, compounded monthly,             = 8026.36 × (1 + 5.9 ÷ 4 )4 – 8026.36
100
will be closest to \$629.88. ⇒ C
= 8026.36 × 1.0603 – 8026.36
= \$484.14 ⇒ C
6   For a reducing balance loan, the
quarterly repayments are always the
same. Since the outstanding balance is     13   1400 = x × (1 + 3.5 )1 – x,
100
being reduced each quarter, so is the             1400 = 0.035 × x,
payable interest, meaning less of the             x = \$40 000 ⇒ E
repayment is going towards interest
and more is going towards repaying         14   Total interest paid = 48 × 480 – 18 000
the principal. ⇒ B                                                  = \$5040
7   Depreciation amount = 5000 × 0.2                Flat rate = 100 × 5040 = 7%.
18 000 × 4
= \$1000 per year.
5000 – 1500 = 3500,                            Effective rate = 7% × 2(48)
48 + 1
3500 = 3.5 years ⇒ D                                                96
1000                                                         =7×
49
= 13.71% ⇒ B
8   Flat rate depreciation means the graph
should be a negative linear one and        15   Depreciation per year = 24 000 × 0.16
should not curve either upwards or                                    = \$3840
downwards.                                      Depreciation after 6 years = 3840 × 6
At time = 0 years, the value of the                                        = \$23 040
machine is 20 000. At time = 5 years,           Machine value after 6 years
the value of the machine is 4000.                = 36 000 – 23 040
The only graph to satisfy the gradient           = \$12 960 ⇒ C
and end–point criteria is A. ⇒ A
16   Using the TVM Solver function of the       23   Annual depreciation = 6500 – 2000
graphics calculator, the pay-off period                                        5
for \$1200 monthly repayments, for a               = 4500 = \$900 ⇒ B
5
\$135 000 loan at 7% interest
compounded monthly, is 15 years.
24   After first month, balance
This is 5 years less than the 20 years
taken normally. ⇒ C                             = 12 200 × (1 + 3 ÷ 12 ) – 5
100
= 122 00 × 1.0025 – 5
17   A = P × (1 + r )n                                 = \$12 225.50.
100
According to the graph, P = \$2000 and           After second month, balance
r = 6%.                                         = 12 225.5 × 1.0025 – 5
= \$12 251.06 ⇒ A
Thus, A = 2000 × (1 + 6 )n
100
= 2000 × (1.06)n ⇒ A                25   Balance after 5 years
= P × (1 + r ÷ 4 )4t
18   Let x be the original price.                                 100
Increasing the price by 5% and then               = 8000 × (1 + r )4×5
400
another 5% is given by x × 1.05 × 1.05
= x × 1.1025.                                   = 8000 × (1 + r )20 ⇒ D
400
This represents a 10.25% increase in x.
⇒D                                         26   Using the TVM Solver function of the
graphics calculator, with an initial loan
19   Tax bracket = 3.                                balance of \$80 000 for 10 years (with
Tax = \$0 for bracket 1,                         5.6% interest), and comparing monthly
0.17 × 15 600 = \$2652 for bracket 2,           compounding repayments of \$555 to
0.30 × 32 400 = \$9720 for bracket 3.           weekly compounding repayments of
Tax = 2652 + 9720 = \$12 372 ⇒ C                 \$132, we can see that the weekly
option will result in more money being
20   Tax bracket = 4.                                repaid faster, which will reduce the
Tax = 0.30 × 3345 = \$1003.50 ⇒ C                period of the loan. ⇒ D

21   92.80 = 1.1 × x,
x = 92.80 = \$84.36 ⇒ D
1.1

22   Using the TVM Solver function of the
graphics calculator, the balance of the
loan after 5 years with initial borrowed
amount of \$2200 00 at 6.1% interest,
compounded monthly with monthly
repayments of \$1800, will be \$172 000.
⇒B
Chapter 23 – Undirected graphs
Solutions to Exercise 23A
1                                               e
a   i A has 3 edges, degree = 3.                          A   B   C   D
ii B has 2 edges, degree = 2.                     A   0   1   1   1
B   1   0   1   1
iii H has 1 edge, degree = 1.
C   1   1   0   1
D   1   1   1   0
b   Counting edges in the figure, the
following table can be drawn:               f
A   B   C   D   H                              A   B   C   D   E   F
A    0   1   1   1   0                          A   0   1   1   0   0   0
B    1   0   1   0   0                          B   1   0   0   1   0   0
C    1   1   0   2   0                          C   1   0   0   1   0   0
D    1   0   2   0   1                          D   0   1   1   0   0   0
H    0   0   0   1   0                          E   0   0   0   0   0   1
F   0   0   0   0   1   0
c   It is not simple since there are multiple
edges (between C and D).                    g
A   B   C   D
2   Counting edges in the figure, the                 A   0   0   0   0
B   0   0   0   1
following tables can be drawn:                    C   0   0   0   2
a                                                     D   0   1   2   0
A   B   C   D
A    0   1   1   0
B    1   0   1   1                        h
C    1   1   0   0                                  A   B   C   D
D    0   1   0   0                              A   0   1   1   1
B   1   0   1   0
C   1   1   0   1
b                                                     D   1   0   1   0
A   B   C   D
A    0   1   1   0
B    1   0   0   1                        3
C    1   0   0   1                        a   The simple graphs are a, b, c, e, f and h;
D    0   1   1   0                            d and g are not simple ,as they contain
loops and multiple edges respectively.
c
A   B   C   D                        b   The connected graphs are a, b, d, e & h;
A    0   1   0   0                            c, f and g are not connected, as least one
B    1   0   0   0                            other vertex cannot be reached from
C    0   0   0   1
D    0   0   1   0
each of the vertices in these graphs.

d
A
A       1
4   From the tables and matrices given, the   5
following graphs can be drawn:            a   Below are three of the many possible
a                                                 subgraphs:

b

c

b   A has 5 edges from it, degree = 5.
d                                             c   Each vertex has degree 5, there are 6
vertices. Vertex sum = 5 × 6 = 30.

d   There are 60 possible paths. One of the
possible paths is A–B–C–D–A.

e

f
Solutions to Exercise 23B
1   The following answers are determined          3   Note that Euler’s formula is v – e + f = 2.
by counting the vertices, edges and
faces of the graphs given.                    a   8 – 10 + f = 2,
Note that Euler’s formula is v – e + f = 2.         f = 2 + 2 = 4.

a   i 8 vertices, 12 edges, 6 faces.              b   v – 14 + 4 = 2,
ii 8 – 12 + 6 = 2                                   v = 2 + 10 = 12.

b   i 6 vertices, 12 edges, 8 faces.              c   5 – 14 + f = 2,
ii 6 – 12 + 8 = 2                                   f = 2 + 9 = 11.

c   i 7 vertices, 12 edges, 7 faces.              d   10 – e + 11 = 2,
ii 7 – 12 + 7 = 2                                   e = 21 – 2 = 19.

d   i 7 vertices, 9 edges, 4 faces.               4
ii 7 – 9 + 4 = 2                              a   One way of representing a cube as a
planar graph is as below:
2   The following are possible ways in
which each of the given graphs can be
redrawn to show their planar natures.
a

b   v = 8, e = 12, f = 6.
Since v – e + f = 2, 8 – 12 + 6 = 2.

b

c
Solutions to Exercise 23C
1   The graph to be drawn is K6.           2
a   n = 6, edges = 6 × 6 – 1               a   n = 7, edges = 7 × 7 – 1
2                                      2
=6× 5                                       =7× 6 = 21 edges.
2                                           2
= 15 edges = 15 matches.
b
b

3   n = 8, edges = 8 × 8 – 1
c                                                                 2
7 = 28 edges.
⎡0   1 1 1 1 1⎤                                         =8×
2
⎢1                ⎥
⎢    0 1 1 1 1⎥
⎢                 ⎥
⎢1   1 0 1 1     1⎥
⎢                 ⎥
⎢1   1 1 0 1     1⎥
⎢                 ⎥
⎢1   1 1 1 0     1⎥
⎢                 ⎥
⎣1   1 1 1 1     0⎦
Solutions to Exercise 23D
1
a   There is no Euler circuit since A and E   e   There is an Euler circuit, and thus many
are both of odd degree.                       Euler paths, with one possible circuit
One possible Euler path is A–B–C–D–           being A–E–F–D–E–B–D–C–B–A.
B–E–A–D–E.
2   Below are possible Hamilton circuits
b   There is no Euler circuit since all the       for the given graphs:
vertices are of odd degree.
There is no Euler path since more than    a   A–B–C–F–I–H–E–G–D–A
2 vertices are of odd degree.
b   A–B–C–D–E–F–A
c   There is no Euler circuit since A and F
are both of odd degree.                   c   A–B–C–D–E–A
There is an Euler path since A and F
are the only two vertices of odd          d   A–F–E–D–C–B–G–A
degree.
One possible Euler path is                e   No Hamilton circuit exists, since any
A–C–E–C–B–D–E–F. Note that this               path that connects all vertices will pass
path must start and end at A and F, the       through at least one vertex more than
odd degree vertices.                          once.

d   There is an Euler circuit, and thus       3   One possible Hamilton path, starting at
many Euler paths, with one possible           F and finishing at G, is
circuit being A–B–C–D–E–C–A.                  F–A–D–E–H–C–B–G.
Solutions to Exercise 23E
1   Minimum spanning trees for the given   2   The graph is not drawn to scale, and
graphs are shown below.                    the shortest path is in fact A–C–D–E
with a weight of 11.
a   Tree weight = 10
3   The shortest Hamilton path is 24 units
long and is A–F–E–D–C–G–B.

b   Tree weight = 80

c   Tree weight = 47

d   Tree weight = 730
Solutions to Multiple-choice questions
1    The simplest way to connect a number       12   Of the given options, the only one to
of vertices is to put them in a straight        have both Euler and Hamilton circuits
line and join them up, each to the next.        is A. ⇒ A
This results in 6 edges. ⇒ C
13   Of the given options, the only one to
2    C is the only graph to connect all 7 of         have a Euler circuit is E, as it is the
the original vertices, contain no               only one to have no vertices of odd
circuits and only utilise edges that            degree and be connected. ⇒ E
were present in the original
graph. ⇒ C                                 14   Of the given options, the only one to
not contain circuits, use only edges
3    The only vertex with 5 edges, and thus          from the original graph and contain all
degree 5, is Q. ⇒ A                             points is C. ⇒ C

4    v–e+f=2                                    15   Of the give options, the only one to
15 – e + 12 = 2,                               have all vertices of even degree and
e = 27 – 2 = 25 ⇒ D                            thus have an Euler circuit is B. ⇒ B

5    While option A has an Euler path, it       16   A counter-example in this
contains two vertices of odd degree so          circumstance is an example that
does not contain an Euler circuit. ⇒ A          disproves a statement.
To find a counter-example to the
6    8 – 13 + f = 2,                                 statement, we must find a graph with 7
f = 2 + 5 = 7 ⇒C                              vertices, each having a degree of 3 or
greater, which does not contain a
7    Of the given options, the only one              Hamilton circuit.
which traces a Hamilton circuit is              The only 7–vertex graph to have all
A–E–F–D–C–B–A. ⇒ B                              vertices of degree 3 or greater is graph
D, which can be shown to not have a
8    Since each edge will contribute once to         Hamilton circuit. ⇒ D
the degrees of exactly 2 vertices, the
sum of the degrees will be equal to        17   A degree sum of 20 corresponds to a
twice the number of edges.                      graph with 20 = 10 edges. The only
Thus, the graph with degree sum 12                           2
will have 6 edges. ⇒ C                          graph with 10 edges is A. ⇒ A

9    There are 11 edges, so the degree sum      18   Of the given options, the only one
will be 11 × 2 = 22. ⇒ C                        which traces a Hamilton circuit is
P–Q–U–V–R–S–T–P. ⇒ C
10   We currently have 4 vertices (A,C,D,E)
of odd degree. To have an Euler path,      19   Note that the edges of our graph
we should have only 2 vertices of odd           represent routes between towns and
degree. This can be achieved by                 not individual roads.
joining two of the vertices A,C,D and           From the road diagram we can see
E with an arc.                                  there are 2 routes between A and B,
Of the given options, only the arc DE           and 2 routes between C and B. There is
does this ⇒ B                                   1 route between each of B and C, B
and D, A and C, A and D.
11   The minimum length tree utilises the            The graph that most closely represents
edges of length 8, 7, 5, 2 and both the         this is B; however, note that B does not
edges of length 4, giving it a total            have an edge to represent the route
length of 30. ⇒ A                               between A and C. ⇒ B
Chapter 24 – Directed graphs
Solutions to Exercise 24A
1                                            c   All two-step paths in the network are
a   The adjacency matrix can be                  displayed in the matrix below:
calculated straight from the graph as           A B C D E F
below:                                       A ⎡ 0 0 0 0 1 0⎤
⎡0     0 0 0 1⎤                               ⎢
B ⎢ 0 0 0 1 0 0⎥
⎥
⎢1              ⎥
⎢      0 0 0   0⎥                             ⎢                 ⎥
C⎢0               0⎥
⎢               ⎥                                   0 0 0 1
⎢0     1 0 0   0⎥                             ⎢                 ⎥
⎢               ⎥                            D⎢ 0   0 0 0 0    1⎥
⎢0     0 1 0   1⎥                             ⎢                 ⎥
⎢               ⎥                            E⎢0    0 1 0 0    0⎥
⎣0     1 1 0   0⎦                             ⎢                 ⎥
F⎣0    0 0 1 0    0⎦

b   i Vertex C is reachable from A via E.    3
ii Vertex A is reachable from D via      a   The one-step dominance between
either C or E and then B.                teams is displayed in the matrix below:
iii Vertex D is unreachable from A.              A B C D
A ⎡ 0 1 1 0⎤
c   Taking into account one-, two- and            ⎢
B ⎢ 0 0 1 0⎥
⎥
three-step reachability, the following
matrix can be created:                        ⎢            ⎥
C⎢0    0 0   0⎥
A B C D E                                ⎢            ⎥
D⎣ 1   1 1   0⎦
A ⎡ 0 2 1 0 1⎤
⎢
B ⎢ 1 0 1 0 1⎥
⎥                            Thus, the ranking by one-step
dominance is D, A, B, C.
⎢              ⎥
C⎢1    1 0 0   1⎥
⎢              ⎥                        b   The two-step dominance between
D⎢ 3   3 2 0   2⎥                            teams is displayed in the matrix below:
⎢              ⎥
E⎣2    2 1 0   0⎦                                A B C D
A ⎡ 0 0 1 0⎤
⎢
B ⎢ 0 0 0 0⎥
⎥
2
a   One possible directed graph for the           ⎢            ⎥
C⎢0    0 0   0⎥
adjacency matrix given is as follows:         ⎢            ⎥
D⎣ 0   1 2   0⎦
The ranking, taking into account both
one-step and two-step dominance, is
D, A, B, C.
4
a   One possible graph that is described by
the matrix is as below:

b   i E can be reached from B via C and D
ii B cannot be reached from A.
iii C can be reached from A via D, E
and F.

b   Dominance scores for each student are:
A = 4, B = 1, C = 0, D = 2, E = 2.
Thus, the rank of each student is, in
order, A, D and E, B, C.
Note that D and E are equal.
Solutions to Exercise 24B
1   Capacity of C1 is 6 + 8 = 14.        3   There are 6 possible cuts as per the
Capacity of C2 is 4 + 5 + 3 = 12.        diagram below.
Capacity of C3 is 8 + 10 + 3 = 21.       Capacity of C1 = 8 + 7 = 15.
Capacity of C2 = 2 + 4 + 8 = 14.
2                                            Capacity of C3 = 4 + 8 + (0) + 7 = 19.
a   Maximum flow = minimum cut               Capacity of C4 = 4 + (0) + 7 = 11.
= CT + BT                    Capacity of C5 = 2 + 8 + 4 = 14.
= 5 + 4 = 9.                 Capacity of C6 = 2 + 8 + 5 + 8 = 23.

b   Maximum flow = minimum cut
= AC + BC + BD
= 3 + 2 + 6 = 11.

c   Maximum flow = minimum cut
= AC + AB + SB
= 3 + (0) + 5 = 8.

d   Maximum flow = minimum cut
= CT + DT
= 8 + 10 = 18.
Solutions to Exercise 24C
1                                                  2
a   i The critical path can be seen to be          a   The linear diagram to represent the
as follows:                                      repair of an engine component is as
follows:
ii The non-critical activities are A, C,
I, G and D.
The float times are the blue numbers        b   i C is immediately preceded by A.
minus the yellow numbers and are                ii F is immediately preceded by B
A = 1, C = 14, I = 1, G = 1, D = 1.                and D.

b   i The critical path can be seen to be          c
as follows:                                      i L is immediately preceded by I
and K.
ii E is immediately preceded by B
ii The non-critical activities are A, C               and D.
and B.
The float times are the blue numbers        d   The digraph to represent the
minus the yellow numbers and are                production and assembly of two new
A = 1, C = 15, B = 1.                           products is as follows:
c   i The table can be completed as
follows:
Activity   Duration   Preceding activities
A           3               -
B           6               -
C           6              A,B
D           5               B
E          7              C,D
F          1               D
G           3               E
H           3               F
I          2               B
ii The critical path can be seen to be
as follows:

iii The non-critical activities are A, F,
H, I and D.
The float times are the blue numbers
minus the yellow numbers and are
A = 3, F = 7, H = 7, I = 14, D = 1.
Solutions to Exercise 24D
1                                                 We now subtract 1 from every entry:
a   After subtracting the lowest values                    A     B       C      D
from each row and the columns                    W     0     0       1      2
without zeroes, we get:                          X     0     0       0      0
A     B    C     D                      Y     1     0       3      0
W      5    13   10    0                      Z     0     0       0      0
X      0    0    15    0                   Minimum lines to cover the 0s is 4 =
Y     25    1    15    0                   number of rows.
Z      0    3    0     0
Minimum lines to cover the 0s is 3.         One possible allocation is W = A,
X = C, Y = B, Z = D.
We now add the minimum uncovered
number (1) to row X, row Z and            2   After subtracting the lowest values
column D:                                     from each row and the columns
A    B     C     D                   without zeroes, we get:
W     5    13    10    1                          100 m   400 m       800 m   1500 m
X     1    1     16    2                     D      0       4           4       32
Y     25    1    15    1                     J      0       0           4       10
Z     1     4    1     2                     C      0       2           8       21
E      0       3           0       0
We now subtract 1 from every entry:
Minimum lines to cover the zeroes is 3.
A    B     C     D
W     4    12     9    0                   We now add the minimum uncovered
X     0    0     15    1                   number (2) to row J, row E and
Y     24    0    14    0                   column 100 m:
Z     0     3    0     1
Minimum lines to cover the zeroes                    100 m   400 m       800 m   1500 m
D      2       4           4       32
is 4 = number of rows.
J      4       2           6       12
We can see thus see that the only least         C      2       2           8       21
cost allocation possible is W = D,              E      4       5           2       2
X = A, Y = B, Z = C.
We now subtract 2 from every entry:
b   After subtracting the lowest values                  100 m   400 m       800 m   1500 m
from each row and the columns                   D      0       2           2       30
without zeroes, we get:                         J      2       0           4       10
C      0       0           6       19
A     B     C    D
E      2       3           0       0
W     0     1     1    3
X     0     1     0    1                   Minimum lines to cover the zeroes
Y     0     0     2    0                   is 4 = number of rows.
Z     0     1     0    1
Minimum lines to cover the 0s is 3.           Thus, the best possible allocation will
be Dimitri = 800 m, John = 400 m,
We now add the minimum uncovered              Carol = 100 m, Elizabeth = 1500 m.
number (1) to row Y, column A and
column C:
A     B     C    D
W     1     1     2    3
X     1     1     1    1
Y     2     1     4    1
Z     1     1     1    1
3   After subtracting the lowest values     4   After subtracting the lowest values
from each row and the columns               from each row and the columns
without zeroes, we get:                     without zeroes, we get:
A     B     C                               W     X    Y     Z
J     0     0     0                         A     12    6     0    28
M     0     4     12                        B     6     0    6     0
A     0     0     2                         C     21    0     0    12
D      0    3    12    9
Thus, best job allocation is Joe = C,       Minimum lines to cover the zeroes
Meg = A, Ali = B.                           is 4 = number of rows.

Thus, best job allocation is A = Y,
B = Z, C = X, D = W.
Solutions to Multiple-choice questions
1   The only point not reachable from P         8    The only network to follow all 3
and indeed from any other point is T.            conditions is network B. ⇒ B
⇒E
9    Maximum flow = minimum cut
2   Considering both one- and two-step                           = 5 + 7 + (0) + 8
pathways, the correct matrix is option                       = 20 ⇒ A
D. ⇒ D
10   Maximum flow = minimum cut
3   W has one-step dominance over X, Z                           = 4 + 6 = 10 ⇒ A
and Y, so has a one-step dominance
score of 3. ⇒ D                             11   The critical path of 23 days long is
A–D–H–I–K. ⇒ B
4   The matrix that correctly shows one-
and two-step (but not three-step)           12   The network with the greatest possible
dominances is B. ⇒ B                             flow is that with the highest minimum
cut, which is 3 + 3 + 2 = 8, belonging
5   For F to start, the critical path of A, D        to network A. ⇒ B
and E must occur, which takes 9 + 7 +
8 = 24 days. ⇒ E                            13   Time required = critical path
=3+6+2+6
6   Capacity= 3 + 4 + 3 = 10 ⇒ D                                   = 17 ⇒ D

7   Kieran and Miriam can both be seen to       14   Earliest starting time for G
have played 3 sports each. ⇒ C                    = critical path to G
=B+C+E
= 3 + 4 + 1 = 8 ⇒E
Chapter 25 – Revision: Networks
and decision mathematics
Solutions to Multiple-choice questions
1   The only graph to only use edges of            Minimum lines to cover the zeroes is 4.
the original graph is graph B. ⇒ B
We now add the minimum uncovered
2   All of the graphs are equivalent in that       number (1) to row I, row N, column
they contain the same vertices and             and column E:
edges with the same connectivity.                       A     B     C      D     E
In each graph, the four vertices of the           F      1    4     87     1      1
square are in the same position, but the          D      2    1     1      1      2
top vertices of the triangles and the             H     92    3     3      1      3
I      1    1     4      4     90
outlier vertex change position relative           N      1    92    1      3      2
to the square. ⇒ E
We now subtract 1 from every entry:
3   There are 2 connections between F and
A     B     C      D     E
H, 2 between F and G, 2 between F                 F      0    3     86     0      0
and H, 1 between F and I, 1 between I             D      1    0     0      0      1
and H, and a loop from F to F.                    H     91    2     2      0      2
All these are depicted only in graph A.           I      0    0     3      3     89
⇒A                                                N      0    91    0      2      1
Minimum lines to cover the zeroes is 5.
4    After subtracting the lowest values
from each row and the columns                  Herman must do D, since he cannot do
without zeroes, we get:                        any other task most efficiently.
A     B     C     D     E
David must thus be allocated one of B
F     2    5     88    0      1             and C, leaving Indira and Natalie to do
D     3    2     2     0      2             A and the other one of B and C that
H    93    4     4     0      3             David doesn’t do.
I     0    0     3     1     88             This leaves Francis doing E. ⇒ E
N     0    91    0     0      0
Minimum lines to cover the zeroes is 3.    5   Sum of degrees = 2 × edges
= 2 × 8 = 16 ⇒ E
We now add the minimum uncovered
number (1) to row I, row N and             6   We currently have 4 vertices of odd
column D:                                      degree, namely S, Z, U and W. To have
A     B     C     D     E              an Euler path, we must have only 2
F     2    5     88    1      1             vertices of odd degree.
D     3    2     2     1      2             Thus, the added edge must join two of
H    93    4     4     1      3
these vertices. The only one that does
I     1    1     4     3     89
N     1    92    1     2      1             this is SU. ⇒ B

We now subtract 1 from every entry:        7   v – e + f = 2.
A     B     C     D     E              In this case, v = f, so 2v – e = 2.
F     1    4     87    0      0             2v – 20 = 2,
D     2    1     1     0      1               2v = 22, v = 11 ⇒ C
H    92    3     3     0      2
I     0    0     3     2     88
N     0    91    0     1      0
8   A minimal spanning tree will give us
the shortest path. ⇒ B
9    Since critical path tasks cannot be         20   A is a predecessor of C, which
delayed without delaying the entire              precedes E, so E cannot start until A is
project, knowing critical path tasks can         finished. ⇒ E
be used to determine which tasks are
able to be delayed without holding up       21   v–e+f=2
the project. ⇒ A                                 v – 12 + 4 = 2,
v = 2 + 8 = 10 ⇒ B
10   Vertex O cannot be reached from
vertex Q; indeed, no other vertex can       22   The adjacency matrix that correctly
be reached from Q either. ⇒ E                    shows both the single links and the
2 sets of multiple edges is given by B.
11   The 23–hour long critical path is                ⇒B
K–N–Q–T–U. ⇒ D
23   Minimum cost = minimal spanning
12   Shortest path = 2 + 6 + 4                        tree = 70 + 120 + 100 + 110
= 12 ⇒ B                                = 400 ⇒ A

13   Ann and Tom have visited 3 resorts in       24   Capacity = 3 + 2 + (0) + 8
total, but Matt and Maria have visited 4.                 = 13 ⇒ D
⇒E
25   Degree sum = 2 × edges
14   The cheapest network would                                  = 2 × 7 = 14 ⇒ C
incorporate all the vertices with the
least edge length, and thus would be a      26   v – e + f = 2.
minimum-length spanning tree. ⇒ C                9 – 20 + f = 2,
f = 2 + 11 = 13 ⇒ B
15   While Sally and Jon can translate 4
languages between them, Kate and            27   Minimum spanning tree
Greg can translate 5. ⇒ E                         =7+7+6+4+5+3
= 32 ⇒ C
16   The only graph to correctly display all
the given information is graph D. ⇒ D       28   Maximum flow = minimum cut
=1+2+4
17   Minimum spanning tree                                         = 7 ⇒C
=8+7+4+3+5+4
= 31 ⇒ D                                   29   There are 5 possible paths, three of
them via node 2 and two via node 3.
18   A complete graph requires 2 × v edges            ⇒E
= 2 × 5 = 10 edges. ⇒ C
30   The only graph not to contain vertices
19   An Euler circuit requires no vertices of         of odd degree and thus contain an
odd degree, but currently A and D are            Euler circuit is graph C. ⇒ C
of odd degree.
The arc AD would make all vertices of       31   The only graph to correctly display
even degree. ⇒ C                                 all the links shown in the matrix is
graph A. ⇒ A
Chapter 26 – Matrices and
applications I
Solutions to Exercise 26A
1                                           l   a1,4 is 0.
a   The square matrices are C and E (they
have dimensions 2×2 and 3×3             m   b3,1 is 1.
respectively).
n   c1,1 is 0.
b   Matrix B has 3 rows.
o   d4,1 is 4.
c   The row matrix is A (it contains a
single row only).                       p   e2,2 is –1.

d   The column matrix is B (it contains a   q   d3,2 is 3.
single column only).
r   b1,1 is 3.
e   Matrix D has 4 rows and 2 columns.
s   c1,2 is 1.
f   The order of matrix E is 3×3.
t   a1,2 is 1.
g   The order of matrix A is 1×5.
u   e1,3 is 1.
h   The order of matrix B is 3×1.
2   Using the Matrix function of the
i   The order of matrix D is 4×2.               graphics calculator, the given graphs
can be displayed on the Home screen.
j   There are 9 elements in matrix E.

k   There are 5 elements in matrix A.
Solutions to Exercise 26B
1                                            3   The matrix representing the paired
a   Complete matrix:                             digits listed one under the other is as
⎡4      2        1⎤                          follows:
⎢                 ⎥                          ⎡3    5       8        7        0   2   3   6⎤
⎢6      2        3⎥                          ⎢                                            ⎥
⎢                 ⎥                          ⎣4    2       2        9        0   0   0   9⎦
⎣2      1        0⎦
Order = 3×3.                            4   The adjacency matrices for the given
graphs are given below:
b   Row matrix:                                   ⎡0       1       0⎤
[6      2        3]                           ⎢                 ⎥
a
⎢1       0       1⎥
Order = 1×3.                                  ⎢                 ⎥
⎣0       1       0⎦
c   Column matrix:
⎡1 ⎤                                          ⎡0       1       0        1
⎤
⎢ ⎥                                           ⎢1                         ⎥
⎢3 ⎥                                          ⎢        0       1        0⎥
⎢ ⎥                                      b
⎢                          ⎥
⎣0 ⎦                                          ⎢0       1       0        1⎥
⎢                          ⎥
Order = 3×1.                                  ⎣1       0       1        0⎦
The sum of the elements will represent
the total number of computers owned
by members of the three households.           ⎡0       0       1        1
⎤
⎢0                         ⎥
⎢        0       0        1⎥
2                                            c
⎢                          ⎥
a   Complete matrix:                              ⎢1       0       0        1⎥
⎡24         32        11 ⎤                    ⎢                          ⎥
⎢                         ⎥                   ⎣1       1       1        0⎦
⎣32         34        9   ⎦
Order = 2×3.                             5   With Town 1 as the first row and
column, Town 2 as the second and
b   Row matrix:                                  Town 3 as the third, the following
[24     32       11]                         matrix can be created:
⎡0     0      1⎤
Order = 1×3.                                 ⎢              ⎥
⎢0     0      3⎥
c   Column matrix:                               ⎢              ⎥
⎣1     3      0⎦
⎡24 ⎤
⎢ ⎥
⎣32 ⎦                                    6
a   f3,4 = 1, so girls 3 and 4 must be
Order = 2×1.                                 friends.
The sum of the elements will represent
the total number of small cars sold by   b   f2,5 = 0, so girls 2 and 5 must not be
both dealers.                                friends.

c   The sum of row 3 elements will tell us
the number of friends that girl 3 has,
which is 3 friends.

d   The girl with the least friends is girl 1,
with only 1 friend. The girl with the
most friends is girl 3, with 3 friends.
Solutions to Exercise 26C
1
a   The only matrices to be equal are C and F.

b   A and B have the same order as each other (1×2), C and F are of the same order (2×3), D and
E are of the same order (2×2).

c   Matrices that can be added and subtracted must be of the same order as each other.
Thus, the 3 pairs of matrices—A and B, C and F, and D and E—can all be added to and
subtracted from each other.

d   i A + B = [3 + 1 = 4          3 + 1 = 4]

ii D + E = ⎡                                ⎤
0+1=1           1+0=1
⎢                              ⎥
⎣-1 + 2 = 1       2+    -1 = 1 ⎦

iii C – F = ⎡
0–0=0            1–1=0          4 – 4 = 0⎤
⎢                                         ⎥
⎣3 – 3 = 0       2–2=0          1 – 1 = 0⎦

iv A – B = [1 – 3 = -2        3 – 1 = 2]

v E–D=⎡                                    ⎤
1–0=1          0 – 1 = -1
⎢                             ⎥
⎣2 – -1 = 3       -1 – 2 = -3 ⎦

vi 3B = [3 × 3 = 9     3 × 1 = 3]

⎡ 4×0=0              4×1=4          4 × 4 = 16 ⎤
vii 4F = ⎢
⎢
⎥
⎥
⎢
⎣ 4 × 3 = 12         4×2=8            4×1=4⎥
⎦

viii 3C + F = 4F (since C = F)
= same as viii
⎡0       4     16 ⎤
=
⎢                   ⎥
⎣12      8      4   ⎦
ix 4A – 2B = [4 × 1 – 2 × 3 = -2           4 × 3 – 2 × 1 = 10]

x E + F is not possible. The matrices are not of the same order, so this addition cannot be
performed and thus is undefined.
2
⎡1 + 4 = 5      2 + 3 = 5⎤
a
⎢                          ⎥
⎣4 + 1 = 5      3 + 2 = 5⎦

⎡1 – 4 = -3     2 – 3 = -1 ⎤
b
⎢                           ⎥
⎣4 – 1 = 3      3–2=1       ⎦

⎡1 + 8 = 9      2 + 6 = 8⎤
c
⎢                          ⎥
⎣4 + 2 = 6      3 + 4 = 7⎦

d   [1 + -1 = 0      – 1 + 1 = 0]

⎡0 + 1 = 1 ⎤
e
⎢          ⎥
⎣1 + 0 = 1 ⎦

⎡0 + 2 = 2 ⎤
f
⎢          ⎥
⎣3 + 0 = 3 ⎦
g   Matrices are not of the same order so this addition cannot be performed and thus is
undefined.

⎡0 – 2 = -2 ⎤
h
⎢           ⎥
⎣3 – 0 = 3 ⎦
I   [ 2 + 1 = 3 1 + -1 = 0 2 + -2 = 0]

j   Matrices are not of the same order so this addition cannot be performed and thus is
undefined.

3
⎡2.2 – 4.4 = -2.2       4.4 – 3.3 = 1.1  ⎤
a
⎢                                        ⎥
⎣8.8 – 1.1 = 7.7       6.6 – 2 – 2 = 4.4 ⎦

⎡1.2 – 1.4 = -0.2      0.2 – 14 = -13.8 ⎤
⎢                                         ⎥
b    ⎣ 4.5 – 3.5 = 1         3.3 – 7 = -3.7    ⎦

⎡ 5–0=5             10 – 2 = 8    5 + 2 = 7⎤
⎢                                          ⎥
c
⎢20 – 4 = 16        0–0=0         5–2 = 3⎥
⎢                                          ⎥
⎣ 0 – 1 = -1        5–0=5         0 + 4 = 4⎦

⎡0.8 + -0.2 = 0.6       1.6 + 0.4 = 2          0.8 + 0.2 = 1      3.2 + 0 = 3.2   ⎤
d
⎢                                                                                 ⎥
⎣ 0.8 + 0.2 = 1           0+0=0               -0.8 + 0.2 = -0.6   1.6 – 0.4 = 1.2 ⎦
4
⎡2.4 ⎤
⎢ ⎥
a   A=
⎢3.5 ⎥
⎢ ⎥
⎣1.6 ⎦
⎡2.8 ⎤
⎢ ⎥
B=
⎢3.4 ⎥
⎢ ⎥
⎣1.8 ⎦
⎡2.5 ⎤
⎢ ⎥
C=
⎢2.6 ⎥
⎢ ⎥
⎣1.7 ⎦
⎡3.4 ⎤
⎢ ⎥
D=
⎢4.1 ⎥
⎢ ⎥
⎣2.1 ⎦
b   The sum of the 4 matrices represents the total sale of CDs in a year for each of the three
stores.
⎡2.4 + 2.8 + 1.5 + 3.4 = 11.1 ⎤
⎢                             ⎥
⎢3.5 + 3.4 + 2.6 + 4.1 = 13.6 ⎥
⎢                             ⎥
⎣ 1.6 + 1.8 + 1.7 + 2.1 = 7.2 ⎦
5
⎡16      104    86 ⎤
a   A=
⎢                 ⎥
⎣75      34     94 ⎦

⎡24      124    100 ⎤
B=
⎢                     ⎥
⎣70      41     96    ⎦

⎡ 16 + 24 = 40       104 + 124 = 228       86 + 100 = 186 ⎤
b   C=A+B=
⎢                                                           ⎥
⎣75 + 70 = 145        34 + 41 = 75          94 + 96 = 190   ⎦
Matrix C represents the total numbers of males and females enrolled in all the programs for
the two years of data given.

⎡24 – 16 = 8        124 – 104 = 20      100 – 86 = 14 ⎤
c   D=B–A=
⎢                                                       ⎥
⎣70 – 75 = -5        41 – 34 = 7          96 – 94 = 2   ⎦
Matrix D represents the increase in the number of males and females in all the programs
from 2005 to 2006. The negative element in the matrix represents a decrease in numbers in
that category from 2005 to 2006.

⎡ 2 × 24 = 28       2 × 124 = 248        2 × 100 = 200 ⎤
d   E = 2B = ⎢
⎢
⎥
⎥
⎢
⎣ 2 × 70 = 140        2 × 41 = 82        2 × 96 = 192 ⎥
⎦
Solutions to Exercise 26D
1
a   Of the given matrix products, those that are defined are: AB, BA, CE, EC and FE, which are
denoted by i, ii, iv, v and vii.
Note that although FE is defined, EF is not defined.

b   i AB = [1 × 3 + 3 × 1 = 6]
ii CE = [1 × 2 + 0 × 1 + -1 × 0 = 2]
⎡ 0×3+1×1=1 ⎤
⎢                     ⎥
iii DB = ⎢
⎢
⎥
⎥
⎣ -1 × 3 + 2 × 1 = -1 ⎦
⎡ 0 × 2 + 1 × 1 + 4 × 0 = 1⎤
iv FE = ⎢⎢
⎥
⎥
⎢                          ⎥
⎣ 3 × 2 + 2 × 1 + 1 × 0 = 8⎦

c   i AB = [1 × 3 + 3 × 1 = 6]
⎡ 0 × 2 + 1 × 1 + 4 × 0 = 1⎤
ii FE = ⎢
⎢
⎥
⎥
⎢
⎣ 3 × 2 + 2 × 1 + 1 × 0 = 8⎥
⎦
iii AB – 3CE = [6 – 3 × 2 = 0]
⎡ 2 × 1 + 3 × 3 = 11 ⎤
iv 2FE + 3B = ⎢
⎢
⎥
⎥
⎢                    ⎥
⎣ 2 × 8 + 3 × 1 = 19 ⎦

2
a   [0 × 2 + 2 × 0 = 0]

b   [10 × 5 + -30 × 1 = 20]

c   [-2 × 1 + 0 × -2 + 1 × 0 = -2]

d   [2 × 100 + 0 × 200 + 1 × 0 = 200]

e   [0 × 1 + 2 × 2 + 1 × 3 + 3 × 2 = 13]

f   [0 × 10 + 0.2 × 20 + 0.1 × 30 + 0.3 × 20 = 13]

⎡ 0.5 × 2 = 1        0.5 × 4 = 2       0.5 × 6 = 3 ⎤
⎢                                                   ⎥
g    ⎢                                                   ⎥
⎢ -1.5 × 2 = -3     -1.5 × 4 = -6     -1.5 × 6 = -9 ⎥
⎢                                                   ⎥
⎢                                                   ⎥
⎣ 2.5 × 2 = 5       2.5 × 4 = 10      2.5 × 6 = 15 ⎦

⎡ 1×4+2×1=6                1×3+2×2=7⎤
h    ⎢                                            ⎥
⎢                                            ⎥
⎢
⎣ 4 × 4 + 3 × 1 = 19      4 × 3 + 3 × 2 = 18 ⎥
⎦

⎡ 1 × 0 + -1 × 1 = -1      1 × 3 + -1 × 2 = 1 ⎤
i    ⎢                                             ⎥
⎢                                             ⎥
⎢
⎣ 4×0+3×1=3                4 × 3 + 3 × 2 = 18 ⎥
⎦
⎡ 10 × 0 + -10 × 0.1 = -1        10 × 0.3 + -10 × 0.2 = 1 ⎤
j   ⎢                                                         ⎥
⎢                                                         ⎥
⎢
⎣ 40 × 0 + 30 × 0.1 = 3          40 × 0.3 + 30 × 0.2 = 18 ⎥
⎦

⎡ 1 × 2 + 3 × 0 + 1 × 1 = 3⎤
⎢                          ⎥
k   ⎢                          ⎥
⎢ 0 × 2 + 2 × 0 + 0 × 1 = 0⎥
⎢                          ⎥
⎢                          ⎥
⎣ 1 × 2 + 1 × 0 + 2 × 1 = 4⎦

⎡ 1 × 2 + 3 × 1 + 1 × -1 = 4       1 × 1 + 3 × -1 + 1 × 2 = 0 ⎤
⎢                                                              ⎥
l   ⎢                                                              ⎥
⎢ 0 × 2 + 2 × 1 + 0 × -1 = 2       0 × 1 + 2 × -1 + 0 × 2 = -2 ⎥
⎢                                                              ⎥
⎢                                                              ⎥
⎣ 1 × 2 + 1 × 1 + 2 × -1 = 1       1 × 1 + 1 × -1 + 2 × 2 = 4 ⎦

⎡ 1 × 0 + 3 × 1 + 1 × -2 = -5        1 × 2 + 3 × 4 + 1 × 1 = 15    1 × 1 + 3 × 2 + 1 × 2 = 9⎤
⎢                                                                                           ⎥
m   ⎢                                                                                           ⎥
⎢ 0 × 0 + 2 × -1 + 0 × -2 = -2       0×2+2×4+0×1=8                 0 × 1 + 2 × 2 + 0 × 2 = 4⎥
⎢                                                                                           ⎥
⎢                                                                                           ⎥
⎣ 1 × 0 + 1 × -1 + 2 × -2 = -5       1×2+1×4+2×1=8                 1 × 1 + 1 × 2 + 2 × 2 = 7⎦

3   RP =
⎡ 4 × 2 + 1 × 1 + 0 × 0 = 9⎤
⎢                          ⎥
⎢ 3 × 2 + 1 × 1 + 1 × 0 = 7⎥
⎢                          ⎥
⎢ 3 × 2 + 0 × 1 + 2 × 0 = 6⎥
⎢                          ⎥
⎢                          ⎥
⎢ 1 × 2 + 2 × 1 + 2 × 0 = 4⎥
⎢                          ⎥
⎢ 1 × 2 + 1 × 1 + 3 × 0 = 3⎥
⎢                          ⎥
⎢                          ⎥
⎣ 0 × 2 + 1 × 1 + 4 × 0 = 1⎦
4   TE =
⎡ 10 × 25 + 20 × 40 + 30 × 65 = 3000 ⎤
⎢                                    ⎥
⎢ 15 × 25 + 20 × 40 + 25 × 65 = 2800 ⎥
⎢                                    ⎥
⎢                                    ⎥
⎢ 20 × 25 + 20 × 40 + 20 × 65 = 2600 ⎥
⎢                                    ⎥
⎢                                    ⎥
⎣ 30 × 25 + 20 × 40 + 10 × 65 = 2200 ⎦
Solutions to Multiple–choice questions
1    W is the row matrix as it contains a single row. ⇒ C

2    U and Y are both 2×2 square matrices. ⇒ D

3    The order of matrix X is 2 rows by 3 columns = 2×3. ⇒ B

4    U and Y can be added as they are of the same order (2×2). ⇒ D

5    XY is undefined as X has 3 columns but Y has 2 rows, and these numbers must be the same
for XY to be defined. ⇒ D

⎡ -2 × 0 = 0      -2 × 1 = -2 ⎤
6    –2Y = ⎢
⎢
⎥
⎥   ⇒A
⎢
⎣ -2 × -1 = 2     -2 × 2 = -4 ⎥
⎦

7    X has 2 rows and Z has 1 column, so the order of XZ is 2×1. ⇒ B

8    a2,3 = 3 ⇒ D

⎡4 – -1 = 5        0 – 0 = 0⎤
9
⎢                           ⎥ ⇒E
⎣-2 – 1 = -3       2 – 1 = 1⎦

10   [1 × 3 + 2 × 2 + 3 × 1 = 10]    ⇒A

⎡ 1×2+2×1=4⎤
11   ⎢
⎢
⎥
⎥   ⇒D
⎢                    ⎥
⎣ 3 × 2 + 4 × 1 = 10 ⎦

12   While the first matrix has 3 columns, the second has 2 rows.
The matrix product is therefore undefined. ⇒ E

⎡ 2 × 0.5 + 0 × 0 + 2 × 0.5 = 2           2 × 0 + 0 × -1.5 + 2 × -1.5 = -3    2 × -1.5 + 0 × 0.5 + 2 × 0 = -3 ⎤
⎢                                                                                                            ⎥
13   ⎢                                                                                                            ⎥
⎢ 4 × 0.5 + -2 × 0 + 8 × 0.5 = 6          4 × 0 + -2 × -1.5 + 8 × -1.5 = -9   4 × -1.5 + -2 × 0 + 8 × 0 = -7 ⎥
⎢                                                                                                            ⎥
⎢                                                                                                            ⎥
⎣ 2 × 0.5 + 6 × 0 + 0 × 0.5 = 1           2 × 0 + 6 × -1.5 + 0 × -1.5 = -9    2 × -1.5 + 6 × 0.5 + 0 × 0 = 0 ⎦
⇒D

14   Representing connectivity as a matrix with the first column and row being point 1, the
second column and row being point 2, etc., we get:
⎡0     0       0    1
⎤
⎢0                   ⎥
⎢      0       1    0⎥
⎢                    ⎥⇒     C
⎢0     1       0    1⎥
⎢                    ⎥
⎣1     0       1    0⎦

15   Multiplying the matrices through, we get the equations 3x – 2y = 1 and x + 4y = 2. ⇒ B
Chapter 27 – Matrices and
applications II
Solutions to Exercise 27A
1
⎡1     0⎤
a   i 2×2 identity matrix =
⎢          ⎥
⎣0     1⎦

⎡1     0        0⎤
⎢                ⎥
ii 3×3 identity matrix =
⎢0     1        0⎥
⎢                ⎥
⎣0     0        1⎦

⎡1     0        0    0
⎤
⎢0                    ⎥
⎢      1        0    0⎥
iii 4×4 identity matrix =
⎢                     ⎥
⎢0     0        1    0⎥
⎢                     ⎥
⎣0     0        0    1⎦

⎡1 × 1 + 2 × 0 = 1      1 × 0 + 2 × 1 = 2⎤
b   AI = ⎢
⎢
⎥
⎥
⎢
⎣0 × 1 + 3 × 0 = 0      0 × 0 + 3 × 1 = 3⎥
⎦

⎡1 × 1 + 0 × 0 = 1      1 × 2 + 0 × 3 = 2⎤
IA = ⎢
⎢
⎥
⎥
⎢
⎣0 × 1 + 1 × 0 = 0      0 × 2 + 1 × 3 = 3⎥
⎦

Thus, we can see that AI = IA = A.

⎡1 × 1 + 2 × 0 + 0 × 0 = 1            1×0+2×1+0×0=2   1 × 0 + 2 × 0 + 0 × 1 = 0⎤
⎢                                                                              ⎥
c   CI = ⎢ 3 × 1 + 1 × 0 + 0 × 0 = 3
⎢                                     3×0+1×1+0×0=1
⎥
3 × 0 + 1 × 0 + 0 × 1 = 0⎥
⎢                                                                              ⎥
⎢                                                                              ⎥
⎣0 × 1 + 1 × 0 + 2 × 0 = 0            0×0+1×1+2×0=1   0 × 0 + 1 × 0 + 2 × 1 = 2⎦

⎡1 × 1 + 0 × 3 + 0 × 0 = 1            1×2+0×1+0×1=2   1 × 0 + 0 × 0 + 0 × 2 = 0⎤
⎢                                                                              ⎥
IC = ⎢ 0 × 1 + 1 × 3 + 0 × 0 = 3
⎢                                     0×2+1×1+0×1=1
⎥
0 × 0 + 1 × 0 + 0 × 2 = 0⎥
⎢                                                                              ⎥
⎢                                                                              ⎥
⎣0 × 1 + 0 × 3 + 1 × 0 = 0            0×2+0×1+1×1=1   0 × 0 + 0 × 0 + 1 × 2 = 2⎦

Thus, we can see that CI = IC = C.
2   Multiplying all the pairs of matrices   4   Using the matrix functions of the
results in the following 2×2 identity       graphics calculator, we are able to
matrix:                                     obtain the following inverse matrices:
⎡1    0⎤                                         ⎡ 10     – 2⎤
⎢      ⎥                                         ⎢ 11       3⎥
⎣0    1⎦                               a   A–1 = ⎢
⎢
⎥
⎥
⎢         1 ⎥
⎢0          ⎥
3                                                     ⎣         3 ⎦
a   det(A) = 1×3 – 0×2 = 3
⎡ 20           1⎤
b   det(B) = 0×4 – 1×3 = –3                         ⎢ 9           18 ⎥
b   B = ⎢
–1
⎢
⎥
⎥
c   det(C) = 1×4 – 2×2 = 0                          ⎢ 50           1⎥
⎢–              ⎥
⎣ 9            9⎦
d   det(D) = –1×4 – 2×2 = –8
⎡4    – 1⎤
e   det(E) = –1×4 – –2×2 = 0                           ⎢       2⎥
c   C –1 = ⎢
⎢
⎥
⎥
⎢      1 ⎥
f   det(F) = 1×3 – 0×0 = 3                             ⎢2       ⎥
⎣      2 ⎦
g   det(G) = 0×4 – –3×1 = 3
d   D–1 does not exist, since det(D) = 0.
h   det(H) = 5×2 – 2.5×4 = 0
⎡1     –1         – 1⎤
i   det(I) = 1×1 – 0×0 = 1                             ⎢2      2           2⎥
e   E –1 = ⎢
⎢0
⎥
⎥
j   det(J) = 0×4 – –2×–2 = –4                          ⎢      0          0
⎥
⎢                     ⎥
⎣0     0          0   ⎦
Solutions to Exercise 27B
1   The given sets of simultaneous            4   The matrices with no solutions are
equations can be written as per below:        those where the determinant of the first
⎡3    2⎤  ⎡x ⎤ ⎡2 ⎤                          matrix in the equation is 0.
a
⎢       ⎥×⎢ ⎥=⎢ ⎥
⎣2    5 ⎦ ⎣y ⎦ ⎣4 ⎦                      b   det = 4×1 – 2×2 = 0 = no unique
solution.
⎡3    5⎤  ⎡x ⎤ ⎡6 ⎤
b
⎢       ⎥×⎢ ⎥=⎢ ⎥                        e   det = 2.5×2 – 5×1 = 0 = no unique
⎣5    4 ⎦ ⎣y ⎦ ⎣3 ⎦
solution.

⎡1    2  ⎤ × ⎡x ⎤ = ⎡1 ⎤                 f   det = 1×2 – 2×1 = 0 = no unique
c
⎢        ⎥ ⎢⎥ ⎢⎥
⎣2    -3 ⎦ ⎣y ⎦ ⎣2 ⎦                         solution.

i   det as determined by the graphics
⎡1    -3 ⎤ ⎡x ⎤ ⎡7 ⎤
d
⎢        ⎥×⎢ ⎥= ⎢ ⎥                          calculator = 0 = no unique solution.
⎣-2    1 ⎦ ⎣y ⎦ ⎣4 ⎦                         This can also be seen by the fact that the
first two equations given by the matrices
⎡-3   -2 ⎤ ⎡x ⎤ ⎡2 ⎤                         (x+2y+3z = 1 and 2x+4y+6z = 0) are
e
⎢        ⎥×⎢ ⎥=⎢ ⎥                           parallel (inconsistent).
⎣1     2 ⎦ ⎣y ⎦ ⎣-1 ⎦
5
⎡3    4      -2 ⎤   ⎡x ⎤   ⎡5 ⎤          a   det = 1×3 – 1×2 = 1
⎢              ⎥× ⎢ ⎥= ⎢ ⎥                   ⎡x ⎤ = ⎡3       -1 ⎤ ⎡2 ⎤
f
⎢2    3      5 ⎥ ⎢y ⎥ ⎢2 ⎥                   ⎢⎥ ⎢               ⎥×⎢ ⎥
⎢              ⎥ ⎢⎥ ⎢⎥                       ⎣y ⎦ ⎣-2         1 ⎦ ⎣1 ⎦
⎣1    2      3 ⎦ ⎣z ⎦ ⎣3 ⎦                   x = 5, y = –3

⎡5    0      -2 ⎤  ⎡x ⎤ ⎡3 ⎤             b   det = 4×4 – 8×2 = 0. ⇒ no solution
⎢               ⎥× ⎢ ⎥= ⎢ ⎥
g
⎢1    -1      1 ⎥ ⎢y ⎥ ⎢2 ⎥
c   det = 1×1 – –2×–1 = –1
⎢               ⎥ ⎢⎥ ⎢ ⎥
⎣1    1       1 ⎦ ⎣z ⎦ ⎣1 ⎦                  ⎡x ⎤ = ⎡-1      -1 ⎤ ⎡2 ⎤
⎢⎥ ⎢               ⎥×⎢ ⎥
⎣y ⎦ ⎣-2        -1 ⎦ ⎣1 ⎦
⎡1    1       -2     1
⎤ ⎡ x ⎤ ⎡3 ⎤          x = –3, y = –5
⎢2                     ⎥ ⎢ ⎥ ⎢ ⎥
⎢     -1      1     -1 ⎥ ⎢ y ⎥ ⎢2 ⎥
⎥× ⎢ ⎥= ⎢ ⎥
d   det = 1×0 – 2×–1 = 2
h
⎢
⎢1    2       1      1 ⎥ ⎢ z ⎥ ⎢1 ⎥               ⎡0          1⎤
⎢                      ⎥ ⎢ ⎥ ⎢ ⎥             ⎡x ⎤ ⎢           2⎥  ⎡2 ⎤
⎣2    -3      2     -2 ⎦ ⎣w ⎦ ⎣0 ⎦           ⎢ ⎥=⎢              ⎥×⎢⎥
⎣y ⎦ ⎢-1         1 ⎥ ⎣4 ⎦
2   No unique solutions may exist if the
⎣           2⎦
x = 2, y = 0
equations are inconsistent (graphs
parallel with no intersection) or
e   det = 5×2 – 10×1 = 0. ⇒ no solution
dependent (graphs parallel intersecting
infinitely since they have the same y-
f   det = 3×5 – 5×3 = 0. ⇒ no solution
intercept).

3   The equation will have no unique          g   det = –1×–1 – 0×1 = 1
solution if det(A) = 0.                       ⎡x ⎤ = ⎡-1      -1 ⎤ ⎡2 ⎤
⎢⎥ ⎢               ⎥×⎢ ⎥
⎣y ⎦ ⎣0          1 ⎦ ⎣1 ⎦
x = –3, y = –1
h   det = 1                                       e   det = 1
6                                                   7
⎡    –2        –1      7  ⎤                    ⎡ -1                 0           2 ⎤
⎢                         ⎥                    ⎢                                ⎥
⎡x ⎤    ⎢
3         6      6
⎥    ⎡3 ⎤       ⎡x ⎤ ⎢                                  ⎡8 ⎤
6⎥ ⎢ ⎥
⎢ ⎥=    ⎢                              ⎢ ⎥        ⎢ ⎥ =⎢4                   1        – ⎥×
1 ⎥×              ⎢y ⎥ ⎢ 7                  7         7 ⎥ ⎢4 ⎥
⎢y ⎥                                   ⎢2 ⎥
1         –1
⎢                       6 ⎥               ⎢⎥ ⎢                                  ⎥          ⎢ ⎥
⎢⎥      ⎢    3          6
⎥    ⎢ ⎥        ⎣z ⎦ ⎢ 1                              ⎥          ⎣3 ⎦
⎣z ⎦    ⎢                         ⎥    ⎣2 ⎦            ⎢
⎣7
2        – 5⎥
7⎦
⎢    1          1      – 1⎥
7
⎣    3          3        3⎦               x = –2, y = 18 , z = 1
x = 0, y = 1, z = 1                                                     7           7

i   det = 0. ⇒ no solution                        f   det = – 1
2

det = 1                                                   ⎡ 3      1   -1 ⎤
j
2                                         ⎡x ⎤    ⎢ 2      2      ⎥ ⎡0 ⎤
⎢               ⎥
⎡ -1 1       -1 ⎤                          ⎢ ⎥ = ⎢ -1 0 1 ⎥ × ⎢ ⎥
⎢               ⎥ ⎡3 ⎤                     ⎢y⎥ ⎢                  ⎥ ⎢2 ⎥
⎡x ⎤   ⎢             1 ⎥ ⎢ ⎥                      ⎢⎥ ⎢                   ⎥ ⎢ ⎥
⎢ ⎥= ⎢ 0 2      1
⎥
2 ⎥ × ⎢2 ⎥
⎣z ⎦ ⎢ – 1 1 0 ⎥ ⎣-3 ⎦
⎢               ⎥
⎢y ⎥ ⎢                                                   ⎣ 2 2           ⎦
⎢⎥ ⎢                   ⎥ ⎢ ⎥
⎣z ⎦ ⎢ 0 1 – 1 ⎥ ⎣-1 ⎦
⎢               ⎥
x = 4, y = –3, z = 1
⎣        2     2⎦
g   det = 1
x = 0, y = 1 , z = 3
2         2
⎡x ⎤ ⎡-1                  1        2 ⎤ ⎡2 ⎤
⎢⎥=⎢                                 ⎥ ⎢ ⎥
6                                                     ⎢y ⎥ ⎢ 1                  0       -1 ⎥ × ⎢1 ⎥
a   det = 1                                           ⎢⎥ ⎢                                 ⎥ ⎢ ⎥
⎣z ⎦ ⎣-1                  1        1 ⎦ ⎣2 ⎦
⎡x ⎤ = ⎡-3         5⎤  ⎡9 ⎤
×
⎢⎥ ⎢                 ⎥ ⎢ ⎥                        x = 3, y = 0, z = 1
⎣y ⎦ ⎣-2           3 ⎦ ⎣12 ⎦
x = 33, y = 18                                h   det = – 1
4

b   det = 1                                                         ⎡   1           3             3       -2 ⎤
2                                                       ⎢   2           2             2            ⎥
⎢                                          ⎥
⎡ 5             – 3⎤                      ⎡ ⎤
x
⎢                                          ⎥    ⎡-5 ⎤
x⎤ ⎢ 2
⎡ =⎢                     2⎥                      ⎢y ⎥                                                          ⎢2 ⎥
⎥ × ⎡4 ⎤                              ⎢   –1          –1        –1              1⎥
⎢⎥ ⎢                     ⎥ ⎢ ⎥                   ⎢ ⎥=          ⎢    2           2         2               ⎥×   ⎢ ⎥
⎣y ⎦ ⎢ – 1
⎢               1 ⎥ ⎣10 ⎦
⎥                       ⎢ ⎥           ⎢                                          ⎥    ⎢ ⎥
⎣ 2             2 ⎦                       ⎢z ⎥          ⎢               –1                        3⎥    ⎢0 ⎥
⎢ ⎥           ⎢   0
2
-1
2⎥    ⎢ ⎥
x = –5, y = 3
⎣w ⎦          ⎢                                          ⎥    ⎣0 ⎦
⎢                                          ⎥
c   det = 0. ⇒ no solution                                          ⎢   0           –1            0           1⎥
⎣                2                        2⎦
x = 3 , y = 1 , z = –1, w = –1
d   det = 1                                                 2           2
2
⎡ -3    1⎤
det = – 5
⎡x ⎤ = ⎢           ⎥ × ⎡4 ⎤                 7
⎢⎥ ⎢              3⎥ ⎢ ⎥
6
⎣y ⎦ ⎢ -5
⎣
⎥ ⎣9 ⎦
2⎦                                   ⎡ –3    5 ⎤
⎢ 2     3 ⎥ ⎡150 ⎤
x = –3, y = – 13                                  ⎡x ⎤ =⎢           ⎥×
⎢⎥ ⎢             ⎥ ⎢ ⎥
2
⎣y ⎦ ⎢ 2
⎢        5 ⎥ ⎣150 ⎦
– ⎥
⎣        3⎦
x = 25, y = 50
8   det = –1 × 10–7

⎡ x ⎤ ⎡ 0.0075    0.01    -0.015⎤   ⎡ 175 000⎤
⎢ ⎥ ⎢                           ⎥ ⎢          ⎥
⎢ ⎥ =⎢                          ⎥ ⎢          ⎥
⎢ y ⎥ ⎢ -0.0235   -0.04   0.055 ⎥ × ⎢ 149 000⎥
⎢ ⎥ ⎢                           ⎥ ⎢          ⎥
⎢ ⎥ ⎢                           ⎥ ⎢          ⎥
⎣ z ⎦ ⎣ 0.004     0.02    -0.02 ⎦ ⎣ 183 500⎦

x = 50, y = 20, z = 10
Solutions to Exercise 27C
1   As determined by the graphics             3
calculator, the following powers can be   a   A + 2B – C2
calculated:                                   ⎡2    1⎤  ⎡-2     2⎤  ⎡1        -2 ⎤⎡-1           5⎤
⎡5      5  ⎤                           ⎢       ⎥+⎢         ⎥–⎢           ⎥=⎢              ⎥
A2 =
⎢          ⎥                           ⎣1    3 ⎦ ⎣2      4 ⎦ ⎣-2       5 ⎦ ⎣5            2⎦
⎣5      10 ⎦
b   AB – 2C2
A3 =   ⎡15      20 ⎤                          ⎡-1    4⎤  ⎡2      -4 ⎤ ⎡-3         8  ⎤
⎢           ⎥                          ⎢        ⎥–⎢          ⎥=⎢              ⎥
⎣20      35 ⎦                          ⎣2     7 ⎦ ⎣-4     10 ⎦ ⎣6          -3 ⎦

4     ⎡50      75  ⎤                     c   (A + B + 2C)2
A =
⎢            ⎥                         ⎡1    4⎤  ⎡1      4⎤  ⎡17       8  ⎤
⎣75      125 ⎦                         ⎢       ⎥×⎢         ⎥=⎢            ⎥
⎣4    1 ⎦ ⎣4      1 ⎦ ⎣8        17 ⎦

A7 =   ⎡2250       3625 ⎤
⎢                ⎥                 d   4A + 3B2 – C3
⎣3625       5875 ⎦
⎡8     4 ⎤ + ⎡6    3 ⎤ – ⎡-2        5   ⎤ = ⎡16        2  ⎤
⎢        ⎥ ⎢         ⎥ ⎢                ⎥ ⎢               ⎥
2   As determined by the graphics                 ⎣4    12 ⎦ ⎣3     15 ⎦ ⎣5           -12 ⎦ ⎣ 2          39 ⎦
calculator, the following powers can be
calculated:
⎡5      7  ⎤                       e   (A – B)3 – C3
A4 =
⎢          ⎥                           ⎡27    0⎤  ⎡-2     5  ⎤ = ⎡29        -5 ⎤
⎣7      26 ⎦                           ⎢        ⎥ ⎢–
⎥ ⎢               ⎥
⎣0     1 ⎦ ⎣5      12 ⎦ ⎣-5          13 ⎦

A5 =   ⎡2       19 ⎤
⎢           ⎥
⎣19      59 ⎦

A6 =   ⎡17      40  ⎤
⎢            ⎥
⎣40      137 ⎦

A7 =   ⎡23      97  ⎤
⎢            ⎥
⎣97      314 ⎦
Solutions to Exercise 27D
1   The transition matrices can be completed by subtracting the value given in the column of the
missing value from 1.
⎡0.75           0.05     ⎤
a
⎢                        ⎥
⎣0.25    1 – 0.05 = 0.95 ⎦

⎡ 0.90                    0.15 ⎤
b
⎢                                ⎥
⎣1 – 0.90 = 0.10          0.85 ⎦

⎡0.80    1 – 0.65 = 0.35 ⎤
c
⎢                                ⎥
⎣0.20           0.65             ⎦

⎡ 0.50                           0.33    ⎤
d
⎢                                        ⎥
⎣1 – 0.50 = 0.50         1 – 0.33 = 0.67 ⎦

2   Let the first rows and columns represent A and X, the second rows and columns B and Y, the
third rows and columns C and Z.
⎡0.40    0.55 ⎤
a
⎢               ⎥
⎣0.60    0.45 ⎦

⎡0.70    0.25 ⎤
b
⎢               ⎥
⎣0.30    0.75 ⎦

⎡0.60    0.15           0.22 ⎤
⎢                            ⎥
c
⎢0.10    0.70           0.23 ⎥
⎢                            ⎥
⎣0.30    0.15           0.55 ⎦

⎡0.45    0.35           0.15 ⎤
⎢                            ⎥
d
⎢0.25    0.45           0.20 ⎥
⎢                            ⎥
⎣0.30    0.20           0.60 ⎦

3
a   Let J be the first row and column and P be the second row and column.
⎡0.80    0.25 ⎤
⎢               ⎥
⎣0.20    0.75 ⎦

b   Let the first row be J and the second row be P.
⎡400 ⎤
⎢ ⎥
⎣400 ⎦
c   S1 = T1 × S0
⎡0.80    0.25 ⎤ ⎡400 ⎤ = ⎡420 ⎤
×
⎢             ⎥ ⎢ ⎥ ⎢ ⎥
⎣0.20    0.75 ⎦ ⎣400 ⎦ ⎣380 ⎦

This represents 420 to Jill’s and 380 to Pete’s.
d   S 5 = T5 × S 0
⎡0.5779        0.5276 ⎤ ⎡400 ⎤ = ⎡442.2 ⎤
×
⎢                     ⎥ ⎢ ⎥ ⎢           ⎥
⎣0.4221        0.4724 ⎦ ⎣400 ⎦ ⎣357.8 ⎦
This represents 442 to Jill’s and 358 to Pete’s.

e   Using very large values of n we find that an infinite value of n will return the following
matrix:
⎡444.4 ⎤
⎢      ⎥
⎣356.6 ⎦
This represents 444 to Jill’s and 356 to Pete’s.

4
a   Let H be the first row and column and U be the second row and column.
⎡0.90      0.40 ⎤
⎢                 ⎥
⎣0.10      0.60 ⎦

b   Let the first row be J and the second row be P.
⎡1500 ⎤
⎢ ⎥
⎣ 500 ⎦
c   S1 = T1 × S0
⎡0.90      0.40 ⎤ ⎡1500 ⎤ = ⎡1550 ⎤
×
⎢               ⎥ ⎢ ⎥ ⎢ ⎥
⎣0.10      0.60 ⎦ ⎣ 500 ⎦ ⎣ 450 ⎦
This represents 1550 happy and 450 sad.

d   S1 = T4 × S0
⎡0.8125        0.75 ⎤ ⎡1500 ⎤ ⎡1593.75 ⎤
⎢                   ⎥×⎢ ⎥= ⎢           ⎥
⎣0.1875        0.25 ⎦ ⎣ 500 ⎦ ⎣ 406.25 ⎦
This represents 1594 happy and 406 sad.

e   Using very large values of n we find that an infinite value of n will return the following
matrix:
⎡1600 ⎤
⎢ ⎥
⎣ 400 ⎦
5
a   Let H be the first row and column, N be the second row and column, and S be the third row
and column.
⎡0.80      0.40           0.35 ⎤
⎢                              ⎥
⎢0.15      0.30           0.40 ⎥
⎢                              ⎥
⎣0.05      0.30           0.25 ⎦

b   Let the first row be H, the second row be N, and the third row be S.
⎡1200 ⎤
⎢ ⎥
⎢ 600 ⎥
⎢ ⎥
⎣ 200 ⎦
c   S1 = T1 × S0
⎡0.80       0.40    0.35 ⎤   ⎡1200 ⎤ ⎡1270 ⎤
⎢                        ⎥ × ⎢ ⎥= ⎢ ⎥
⎢0.15       0.30    0.40 ⎥ ⎢ 600 ⎥ ⎢ 440 ⎥
⎢                        ⎥ ⎢ ⎥ ⎢ ⎥
⎣0.05       0.30    0.25 ⎦ ⎣ 200 ⎦ ⎣ 290 ⎦
This represents 1270 happy, 440 neither and 290 sad.

d   S1 = T5 × S0
⎡0.6606        0.6473   0.6459 ⎤  ⎡1200 ⎤ ⎡1310.33 ⎤
⎢                              ⎥× ⎢ ⎥ = ⎢          ⎥
⎢0.2123        0.2186   0.2193 ⎥ ⎢ 600 ⎥ ⎢ 429.82 ⎥
⎢                              ⎥ ⎢ ⎥ ⎢             ⎥
⎣0.1271        0.1341   0.1348 ⎦ ⎣ 200 ⎦ ⎣ 259.85 ⎦
This represents 1310 happy, 430 neither and 260 sad.

e   Using very large values of n we find that an infinite value of n will return the following
matrix:
⎡1311.7 ⎤
⎢       ⎥
⎢ 429.1 ⎥
⎢       ⎥
⎣ 259.1 ⎦
6
⎡0.90    0.20 ⎤  ⎡100 ⎤ ⎡130 ⎤
a   i S1 =
⎢              ⎥×⎢ ⎥=⎢ ⎥
⎣0.10     0.80 ⎦ ⎣200 ⎦ ⎣170 ⎦

ii S2 = ⎡
0.90     0.20 ⎤ ⎡130 ⎤ ⎡151 ⎤
⎢              ⎥×⎢ ⎥=⎢ ⎥
⎣0.10     0.80 ⎦ ⎣170 ⎦ ⎣149 ⎦

iii S3 = ⎡
0.90     0.20 ⎤ ⎡151 ⎤ ⎡165.7 ⎤
⎢              ⎥×⎢ ⎥=⎢         ⎥
⎣0.10     0.80 ⎦ ⎣149 ⎦ ⎣134.3 ⎦

T5 =   ⎡0.7227     0.5546 ⎤
b
⎢                    ⎥
⎣0.2773     0.4454 ⎦

⎡0.83    0.34 ⎤   ⎡100 ⎤ = ⎡151 ⎤
×
c   i S2 =
⎢             ⎥ ⎢ ⎥ ⎢ ⎥
⎣0.17    0.66 ⎦ ⎣200 ⎦ ⎣149 ⎦

ii S3 = ⎡
0.781    0.438 ⎤ ⎡100 ⎤ ⎡165.7 ⎤
⎢               ⎥×⎢ ⎥= ⎢          ⎥
⎣0.219    0.562 ⎦ ⎣200 ⎦ ⎣134.3 ⎦

iii S7 = ⎡
0.6941     0.6118 ⎤ ⎡100 ⎤ ⎡191.76 ⎤
⎢                  ⎥×⎢ ⎥= ⎢          ⎥
⎣0.3059     0.3882 ⎦ ⎣200 ⎦ ⎣108.24 ⎦

d   Using very large values of n we find that an infinite value of n will return the following
matrix:
⎡200 ⎤
⎢ ⎥
⎣100 ⎦
7
⎡0.7     0.4      0.1 ⎤    ⎡100 ⎤ ⎡180 ⎤
i S1 = ⎢0.2                      ⎥× ⎢ ⎥= ⎢ ⎥
a
⎢        0.1       0.3 ⎥ ⎢200 ⎥ ⎢130 ⎥
⎢                      ⎥ ⎢ ⎥ ⎢ ⎥
⎣0.1     0.5       0.6 ⎦ ⎣300 ⎦ ⎣290 ⎦
⎡0.7     0.4       0.1 ⎤ ⎡180 ⎤ ⎡207 ⎤
ii S2 = ⎢0.2                     ⎥ ⎢ ⎥ ⎢ ⎥
0.3 ⎥ × ⎢130 ⎥ = ⎢136 ⎥
⎢        0.1
⎢                      ⎥ ⎢ ⎥ ⎢ ⎥
⎣0.1      0.5      0.6 ⎦ ⎣290 ⎦ ⎣257 ⎦
⎡0.7      0.4      0.1 ⎤ ⎡207 ⎤ ⎡ 225 ⎤
iii S3 = ⎢0.2                   ⎥ ⎢ ⎥ ⎢             ⎥
0.3 ⎥ × ⎢136 ⎥ = ⎢132.1 ⎥
⎢         0.1
⎢                      ⎥ ⎢ ⎥ ⎢             ⎥
⎣0.1      0.5      0.6 ⎦ ⎣257 ⎦ ⎣242.9 ⎦

⎡0.58      0.37      0.25 ⎤⎡100 ⎤ ⎡207 ⎤
i S2 = ⎢0.19                      ⎥× ⎢ ⎥= ⎢ ⎥
b
⎢          0.24    0.23 ⎥ ⎢200 ⎥ ⎢136 ⎥
⎢                       ⎥ ⎢ ⎥ ⎢ ⎥
⎣0.23      0.39    0.52 ⎦ ⎣300 ⎦ ⎣257 ⎦
⎡0.505      0.394    0.319 ⎤ ⎡100 ⎤ ⎡ 225 ⎤
ii S3 = ⎢0.204                       ⎥ ⎢ ⎥ ⎢             ⎥
0.229 ⎥ × ⎢200 ⎥ = ⎢132.1 ⎥
⎢           0.215
⎢                          ⎥ ⎢ ⎥ ⎢             ⎥
⎣0.291      0.391    0.452 ⎦ ⎣300 ⎦ ⎣242.9 ⎦
⎡0.4210       0.4099    0.4027 ⎤ ⎡100 ⎤ ⎡244.9 ⎤
iii S7 = ⎢0.2145                        ⎥ ⎢ ⎥ ⎢             ⎥
0.2169 ⎥ × ⎢200 ⎥ = ⎢129.7 ⎥
⎢             0.2159
⎢                              ⎥ ⎢ ⎥ ⎢             ⎥
⎣0.3645       0.3742    0.3805 ⎦ ⎣300 ⎦ ⎣225.4 ⎦

c   Using very large values of n we find that an infinite value of n will return the following
matrix:
⎡247.1 ⎤
⎢      ⎥
⎢129.4 ⎥
⎢      ⎥
⎣223.5 ⎦
Solutions to Multiple-choice questions
1   V cannot be raised to a power as it is      10   det(II) = 2×4 – 4×2 = 0. Thus, system
not a square matrix. ⇒ B                         II does not have a unique solution.
Systems I and III do have unique
2   det(U) = 2×1 – 1×0 = 2 ⇒ D                       solutions. ⇒ D

3   det(Y) = 1×4 – 2×2 = 0.                     11   In matrix form, the equations can be
Thus, the inverse of Y is undefined.             written as:
⇒E                                               ⎡2      -3 ⎤  ⎡x ⎤ ⎡6 ⎤
⎢         ⎥ × ⎢ ⎥ = ⎢ ⎥ ⇒D
4   det(U) = 2                                       ⎣2      1 ⎦ ⎣y ⎦ ⎣3 ⎦

⎡ 1 = 0.5     0 = 0⎤
⎢ 2           2    ⎥                             ⎡0.6     0.5 ⎤ ⎡100 ⎤ ⎡160 ⎤
U –1 = ⎢
⎢
⎥ ⇒A
⎥
12   S1 =
⎢            ⎥×⎢ ⎥=⎢ ⎥
⎢ -1          2 = 1⎥
⎣0.4     0.5 ⎦ ⎣200 ⎦ ⎣140 ⎦
⎢ = -0.5           ⎥                      ⇒C
⎣2            2    ⎦

⎡2   0⎤  ⎡2        0⎤  ⎡4    0⎤                  ⎡0.56     0.55 ⎤
5   U2 =
⎢      ⎥ ⎢×
⎥ ⎢=
⎥
13   T2 =
⎢                ⎥ ⇒B
⎣1   1 ⎦ ⎣1        1 ⎦ ⎣3    1⎦                  ⎣0.44     0.45 ⎦
Thus, 3U 2 ⇒ C
⎡0.556     0.555 ⎤ ⎡100 ⎤ ⎡166.6 ⎤
14   S3 =
⎢                ⎥×⎢ ⎥= ⎢        ⎥
6   W could be a transition matrix since the                ⎣0.444     0.445 ⎦ ⎣200 ⎦ ⎣133.4 ⎦
values in each column sum to give 1.             ⇒B
⇒C
15   Using very large values of n we find
7   V could be a state matrix with two               that an infinite value of n will return
states since it is a single column matrix        the following matrix:
with two rows. ⇒ B
⎡166.7 ⎤ ⇒ C
⎢      ⎥
8   The relevant transition matrix                    ⎣133.3 ⎦
⎡0.75   0.05 ⎤
=
⎢              ⎥ ⇒B
⎣0.25   0.95 ⎦

9   The relevant transition matrix
⎡0.75   0.05     0.30 ⎤
⎢                     ⎥ ⇒A
=
⎢0.10   0.60     0.20 ⎥
⎢                     ⎥
⎣0.15   0.35     0.50 ⎦
Chapter 28 – Revision: Matrices and
applications
Solutions to Multiple-choice questions
1    Z is a column matrix. ⇒ E                           14   From the diagram, the adjacency
matrix can be seen to be:
U is a 3×3 square matrix. ⇒ A
2
⎡0      0      1          1
⎤
⎢0                         ⎥
3    X has 2 rows and 4 columns, so its                        ⎢       0      1          1⎥
order is 2×4. ⇒ B                                         ⎢                          ⎥ ⇒C
⎢1      1      0          1⎥
⎢                          ⎥
4    None of the matrices are of the same                      ⎣1      1      1          0⎦
order, which means none can be added
to each other. ⇒ E                                  15   Multiplying out the matrices gives us
2x = 1 as the first equation and x + 3y
5    Of the given matrix products, WZ is                      = 4 as the second. ⇒ B
defined, since W has 2 columns and Z
has 2 rows. ⇒ E                                     16   W cannot be raised to a power as it is
not a square matrix. ⇒ C
6    –4 × V =
[-4 × 4 = -16 – 4 × 1 = -4       – 4 × 0 = 0]
17   det(X) = 0.75×0.5 – 0.25×0.5
⇒C                                                             = 0.25 ⇒ C
7    U has 3 rows and Y has 2 columns, so                18   det(Y) = 1×4 – 1×2 = 2.
the order of UY will be (3 × 2). ⇒ D
Y–1 =   ⎡2            -1⎤ ⇒B
⎢               ⎥
8    a3,2 = –4 ⇒ A                                                    ⎣-0.5       0.5 ⎦

⎡6      0⎤  ⎡-2        0⎤  ⎡8        0⎤            19   det(U) = 2×10 – 4×5 = 0.
9
⎢         ⎥ ⎢–
⎥ ⎢=
⎥                 U–1 is undefined. ⇒ E
⎣-3     3 ⎦ ⎣2         2 ⎦ ⎣-5       1⎦
⇒C
(U–Y)2 =    ⎡1             3⎤
×   ⎡1   3⎤
⇒A
20
⎢               ⎥       ⎢     ⎥
10   [1 × 1 + 0 × 2 + -1 × 3 = -2]                                        ⎣3             6⎦       ⎣3   6⎦

= ⎡
10            21 ⎤
⇒B
⎡1      0⎤
× ⎡ ⎤ = ⎡ ⎤ ⇒B
1     1
⎢                 ⎥
11
⎢         ⎥ ⎢⎥ ⎢⎥                                                   ⎣21            45 ⎦
⎣-1     2 ⎦ ⎣2 ⎦ ⎣3 ⎦
21   X could be a transition matrix as the
⎡2      4⎤
× ⎡
-1    2⎤
= ⎡
-2         20 ⎤
values in each column sum to give 1.
12
⎢         ⎥ ⎢         ⎥ ⎢                 ⎥             ⇒D
⎣0      1 ⎦ ⎣0      4 ⎦ ⎣0           4    ⎦
⇒A
22   With A as the first row and column and
B as the second row and column, the
⎡1 0      -1 ⎤ ⎡1   0    1⎤ ⎡ -1      -1 -1 ⎤
13   ⎢            ⎥×⎢          ⎥=⎢                   ⎥        relevant transition matrix is given by:
⎢ 4 -2     2 ⎥ ⎢0   -1   2⎥ ⎢ 8       4        0⎥
⎡0.45        0.25 ⎤
⎢            ⎥ ⎢          ⎥ ⎢                   ⎥        ⎢                ⎥ ⇒B
⎣2 1       0 ⎦ ⎣2        0⎦ ⎣ 2                4⎦
⇒C
1                 -1
⎣0.55        0.75 ⎦
23   With X as the first row and column, Y     26   The missing element = 1 – 0.95
as the second row and column, and Z                               = 0.05 ⇒ A
as the third row and column, the
relevant transition matrix is given by:
⎡0.1      0.8 ⎤ ⎡100 ⎤ ⎡ 90 ⎤
⎡0.80     0.10   0.55 ⎤
27   S1 =
⎢             ⎥×⎢ ⎥=⎢ ⎥
⎢                     ⎥ ⇒C                           ⎣0.9      0.2 ⎦ ⎣100 ⎦ ⎣100 ⎦
⎢0.10     0.85   0.05 ⎥                       ⇒A
⎢                     ⎥
⎣0.10     0.05   0.40 ⎦
⎡0.5977       0.3576 ⎤
28   T4 =
⎢                    ⎥ ⇒B
24   det(III) = 2×–6 – 4×–3 = 0.                           ⎣0.4023       0.6424 ⎦
Hence, matrix III does not have a
unique solution while the other two do.   29   S5 = ⎡0.3816 0.5497 ⎤ × ⎡100 ⎤ = ⎡ 93.1 ⎤
⇒D                                                    ⎢                 ⎥ ⎢ ⎥ ⎢         ⎥
⎣0.6184    0.4503 ⎦ ⎣100 ⎦ ⎣106.9 ⎦
⇒B
25   Writing the equations in matrix form
gives us:
30   Using very large values of n we find
⎡1      -5 ⎤  ⎡x ⎤ ⎡4 ⎤
⎢         ⎥ × ⎢ ⎥ = ⎢ ⎥ ⇒B                    that an infinite value of n will return
⎣-2     1 ⎦ ⎣y ⎦ ⎣3 ⎦                         the following matrix:
⎡ 94.1 ⎤ ⇒ C
⎢      ⎥
⎣105.9 ⎦

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