VIEWS: 159 PAGES: 224 POSTED ON: 6/29/2011
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 constructed by reading percentages of 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 recorded. The majority, 54.5%, had brown eyes. Of the remaining children, 27.3% had blue eyes and 18.2% had 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 us more information. circumference is positively skewed. 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 possessions respectively. The spread of 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 broader, we can say: 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 broader, we can say: ‘On average, Brand A batteries have longer and less variable lifetimes.’ 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 2 assumptions made are: 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 gradient of 0.4. 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% with each additional centimetre from 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 between radial length and femur 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 relationship between radial and femur 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 quadrant of a circle. 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 quadrant of a circle. 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 steadily over the time period. 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 read directly off the equations. 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 gradients are equal: 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 Chapter 22 – Revision: Business- 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 ⎦