# Warm-Up 9

Document Sample

```					                              Warm-Up 9
1. ________ The perimeter of a rectangle is 24 inches. The length is 8½ inches. What is the
inches

2. ________ Mrs. Goodteacher is ordering supplies for her 28 students. The chart shows what
\$
she needs and how these items can be purchased, only in the quantities listed,
through her available supplier. A gross is 12 dozen, and a pound is 16 ounces. What
is the minimum total cost for her supplies?
Item    Quantity Needed                   Purchase Info
A      Each student needs 2 pieces       1 gross for \$2.38
B     Every 2 students share 1 piece    1 dozen for \$2.27
C     Each student needs 1 ounce        1 pound for \$2
D      Every 4 students share 1 piece    \$10 each

3. ________ Four people are sitting around a table. A name tag is made for each person. In how
ways
many ways can these name tags be passed out so that every person gets the wrong
name tag?

4. ________ Every other time that Cheri Linn saw a friend today, she gave away half of
ﬂowers
her ﬂowers. At the end of the day, she had ﬁve ﬂowers left. How many
ﬂowers did she start the day with if she saw eight friends?

5. ________ A recent autobiography used 2808 digits to number the pages of the book. How
pgs
many numbered pages are in the book?

6. ________ A positive two-digit integer less than 50 with two different digits, both of which
are odd, is chosen at random. What is the probability that the number is prime?

7. ________ Points A, B, E and F are the corners of an 8½-inch by 11-inch sheet of paper.
inches
Corners A and B are folded as shown below so that they meet at point D. The paper
is then unfolded. In the same manner, corners E and F meet at point G. What is the
B             E    B               B               E

11”                 C          D        G      D

A             F    A               A               F

8. ________ What is the least number of additional squares that must be drawn to
squares
make this ﬁgure have a vertical line of symmetry and a horizontal line
of symmetry? The ﬁgure may not be rotated.

9. ________ How many positive integer solutions does the inequality 2x + 9 > 4x + 1 have?
solns

10. _______ A sequence of 20 terms is formed by using the rule f (x ) = 4x − 1 for
x = 1, 2, 3, …, 20. What is the sum of the last eight terms?

MATHCOUNTS 2005-2006                                                                             41
Warm-Up 9

1.   3½            (F, G, M)        5. 972                (P, T)      8. 3                     (E, G, P)
5
2. 80.92                 (C)        6.                    (G, T)      9. 3                          (G)
8
3. 9            (E, G, S, T)        7. 2½                 (F, M)      10. 520                  (F, P, T)

4. 80                 (G, T)

Solution - Problem #4                                                             Figure 1
This is a great problem to illustrate how well working           Friends   Flowers Given     Flowers
backwards can help with difﬁcult problems. We know Cheri                         to Friend      Cheri Kept
gives away half of her ﬂowers every other time a friend stops          8                             5
by. We also know that she saw eight friends, and she has ﬁve           7
ﬂowers left at the end of the day. What we were not told               6
was whether she gave away ﬂowers to the ﬁrst, third, ﬁfth              5
and seventh friends she saw or the second, fourth, sixth and
4
eighth friends she saw. Either way, though, notice that she
3
would be giving away ﬂowers four times. Let‛s set up a table
2
with the information we have (Figure 1). We need to ﬁgure
1
out what number goes in the last box of the table. If Cheri
has ﬁve ﬂowers at the end, she must have given ﬁve ﬂowers            Start           -
to a friend and kept the ﬁve ﬂowers she currently has. That
means that she had 10 ﬂowers before giving ﬁve away. Let‛s                        Figure 2
assume she did not give ﬂowers to her eighth friend, but gave
Friends   Flowers Given     Flowers
ﬁve ﬂowers to her seventh friend and kept ﬁve ﬂowers. This                       to Friend      Cheri Kept
means she had 10 ﬂowers when her sixth friend came by, and             8             -               5
she didn‛t give any to the sixth friend. If she had 10 ﬂowers,         7            5                5
though, she must have previously given away 10 ﬂowers, too,
6             -              10
and had 20 ﬂowers before that. All of this is shown in Figure
5            10              10
2. If this logic is continued, we can ﬁll in the rest of the table
4                            20
and see that she started with 80 ﬂowers.
3
2
1
Start           -

Representation - Problem #9
Using a graphing calculator, we can graph both expressions and see where 2x + 9 is greater than
4x + 1. We would enter Y1 = 2x + 9 and Y2 = 4x + 1 into the “Y=” ﬁeld. On the GRAPH screen, we
can use CALC to calculate the intersection point at (4, 17). From the graph we see that
2x + 9 is greater than 4x + 1 to the left of the intersection point, which would include x = 1, 2 or
3. There are then three positive integer solutions to the inequality. We also could check out the
TABLE feature and see that again 2x + 9 is greater than 4x + 1 at x = 1, 2 or 3; at x = 4 the two
expressions are equal; and at every value of x greater than 4, the expression 2x + 9 is less than
4x + 1.

Rather than using a graphing calculator, we also can solve the inequality 2x + 9 > 4x + 1
algebraically. If we subtract 2x and 1 from both sides, we have 8 > 2x. Dividing both sides by 2
yields 4 > x or x < 4. There are only three positive integer values for x that are less than 4.

42                                                                                  MATHCOUNTS 2005-2006
Warm-Up 10
1. ________ The box-and-whisker plot below shows the speed of the 25 fastest roller coasters in
coasters
the world. What is the largest possible number of roller coasters that could have a
speed greater than 80 mph?
Speeds (mph) of the Fastest Roller Coasters in the World

74                                                                     120
75.5     80                   94

2. ________ Each curve in the logo is a semicircle with a
sq cm
radius of 8 cm. What is the area of the logo?

3. ________ A 10% potassium chloride solution is needed for use in a lab. How much water must
liters
be added to dilute 300 liters of a 50% potassium chloride solution for the lab‛s use?

4. ________ The sum of the digits of a positive two-digit integer is 45 less than the integer.
What is the tens digit of the integer?

5. ________ How many non-congruent triangles can be drawn if each one must have one side of
triangles
length four inches, one side of length ﬁve inches and one 90º angle?

6. ________ A square has two diagonals, and a convex pentagon has ﬁve diagonals. How many
diagonals
diagonals does a convex decagon have?

7. ________ Connie has 5 the amount of money as Joe. If Joe
\$
3                                                                   20

gives Connie \$3, then Connie would have three times
the money Joe has. How many dollars did Connie have
originally?

8. ________ Vanessa takes a test worth 85 points. If the minimum percent required to earn an
points
A is 90%, and no rounding can be done, what is the least number of whole points she
can earn on the test to receive an A?

9. ________ A college class has exactly enough students to form eight equal
students
rows. On Monday a student is absent, and the professor is
able to seat the students into ﬁve equal rows. On Tuesday two
students are absent, and the professor can seat the students
into nine equal rows. What is the least possible number of students in the class?

10. _______ Bag A contains four yellow marbles, two green marbles and three blue marbles.
Bag B contains four blue marbles and four green marbles. If you choose one marble
from each bag, what is the probability that you end up with one blue marble and one

MATHCOUNTS 2005-2006                                                                               43
Warm-Up 10

1.   12                  (C)       5. 2               (E, G, M)        8. 77                        (F)

2. 160π               (F, P)       6. 35         (F, M, P, S, T)       9. 56              (E, G, P, T)
2
3. 1200            (F, S, T)       7. 15               (F, G, T)       10.                (F, M, S, T)
9
4. 5               (F, G, T)

Solution - Problem #5
We are limited to right triangles with sides of four inches and ﬁve inches.              Figure 1
Because this is a right triangle, it must have a hypotenuse. From what we know of
right triangles, the hypotenuse is opposite the 90˚ angle and is the longest side                      4
of the right triangle. This means that our four-inch side can‛t be the hypotenuse
(Figure 1). We can see that it wouldn‛t be long enough to connect to any segment
placed on the dotted line that is perpendicular to the ﬁve-inch side. If our ﬁve-                  5
inch side is the hypotenuse, then the third side must be smaller than ﬁve inches.
We also know that the side lengths of right triangles work with the Pythagorean                   Figure 2
Theorem (Hypotenuse2 = Leg12 + Leg22). Using the ﬁve-inch side as the hypotenuse
(Figure 2), we can set up and solve 52 = 42 + x2. This is equivalent to 25 = 16 + x2 or
9 = x2 or x = 3. (This is a 3-4-5 right triangle.) Our only other option is for neither                    5
4
of our known sides to be the hypotenuse (Figure 3). That means the missing side
must be longer than ﬁve inches. Again using the Pythagorean Theorem, we can ﬁnd
the hypotenuse for this case: y2 = 42 + 52, which is y2 = 16 + 25 or y2 = 41 and ﬁnally                X
y = 41 inches. These two triangles are the only two options, and they are
deﬁnitely not congruent to each other, so there are two non-congruent triangles             Figure 3
that can be drawn with the given restrictions.

4                 Y

5
Representation - Problem #3
These are tricky! Perhaps a visual representation of what is going on will make this easier. We
currently have 300 liters of a solution that is 50% potassium chloride. We don‛t want to add any
more potassium chloride; instead, we want to add water to dilute the solution to only 10% potassium
chloride. The picture to the right shows the current situation.
We‛ve made the sodium chloride the black section, and the gray
section is the other half of the current 300 liters of solution. Each
of these sections is 150 liters of the solution. Now we need to add
water so that the black region is no longer half of what is in the
beaker but is instead 10% of what is in the beaker. To be 10% of
the solution in the beaker, it should be the one black section out of
10 equal sections. We need to add eight more gray sections, which is
8(150) = 1200 liters of water. (We‛ll need a bigger container!)
150
300                      150

44                                                                                MATHCOUNTS 2005-2006
Workout 5
1. ________ Hong scored 17 out of 20 on his quiz and 38 out of 50 on his test. What is the
%
positive difference of the percent of questions Hong answered correctly on the quiz
and the percent of questions he answered correctly on the test?

2. ________ Pi‛s value is often estimated to be 3.14. The 200 digits to the right of the decimal
point in pi are summarized in the frequency table below. For example, the digit 2
appears 24 times in those ﬁrst 200 digits. What is the mean of these 200 digits?
Digit           0      1      2       3      4       5       6      7        8        9
Frequency      19     20      24     20      22     20      16      12       24    23

3. ________ The vertices of triangle ABC are A(1, -2), B (4, 0) and C(2, 2). What is the area of
sq units
triangle ABC?

4. ________ Simon and Amanda had six children, each of whom had six children who each also
females
had six children. If exactly half of each generation is female and none of the
offsprings‛ spouses are included, how many females are represented in this family
tree?

5. ________ Of the students answering a survey, 45% of the 8th graders receive an allowance and
%
of those, 80% save part of their allowance. What percent of the responding

6. ________ The average of three numbers a , b and c is 15. The average of b and c is 18. The
average of a and b is 12.5. What is the average of a and c ? Express your answer as
a decimal to the nearest tenth.

7. ________ In the number puzzle to the right, each of the eight non-shaded          1    2
unit squares contains one digit. What is the answer to 1-Across?
Across                       Down                                3        4
1. A prime number            1. A multiple of 9
3. A perfect square          2. A perfect cube                   5
5. A perfect fourth power    4. A perfect square

8. ________ Logan wants to slightly change her pie graph
cm
SODA
with a radius of 8 cm (left) to a similar graph       SODA
67%
67%

(right). Instead of the Pop region being a                                 POP
sector of the pie graph, she wants it to be                                29%

a circular region within the circular graph.    OTHER
2%
POP
29%
COKE
2%
COKE
2%

The Pop region must still represent 29% of                                     OTHER
2%

the entire graph area. The radius of the
entire graph is to remain 8 cm. What should the radius of the Pop region be in her
new graph? Express your answer as a decimal to the nearest tenth.

9. ________ Every student in a 2nd grade class sends a valentine to each of the other students in
students
the class, for a total of 306 valentines. How many students are in the class?

10. _______ A ball with a 3”-radius ﬁts inside a cylinder with a 3”-radius so that it touches both
sq in
bases of the cylinder. The same ball ﬁts into a cube and is tangent to all of the
faces of the cube. What is the positive difference between the total surface areas
of the cylinder and the cube? Express your answer to the nearest whole number.
MATHCOUNTS 2005-2006                                                                               45
Workout 5

1.       9                        (C)       5. 36                    (C)       8. 4.3                  (F, M)

2. 4.475                          (C)       6. 14.5            (F, G, M)       9. 18        (F, G, M, P, S)

3. 5                           (F, M)       7. 61              (E, G, T)       10. 46                  (M, F)

4. 130                 (F, M, P, S, T)

Solution - Problem #8
The ﬁrst graph (left) is a typical pie graph. In both graphs, the area of the graph taken up by
the Pop section must be 29% of the total graphing area. However, in the ﬁrst graph, this Pop region
is a sector of the circle. When Logan switches to the second                                SODA
graph, she still needs the Pop section to take up 29% of the             SODA
67%
67%
area of the circle, but wants the Pop region to be a circular                                POP

region within the entire circle graph rather than a sector of                                29%

the circle graph. The radius of this new graph is still 8 cm,             POP    COKE                COKE
OTHER    29%                         2%
2%             2%

meaning that the area within that boundary is 64π sq cm.                                         OTHER
2%

The Pop section of the new graph is still supposed to take up
29% of this area. We can calculate the area of this Pop region to be 0.29 × 64π ≈ 58.31 sq cm. We
can now ﬁgure out the radius of this circular Pop region using πr2 = 58.31. Dividing both sides by
π and then taking the square root, we see that the radius of the Pop circular region in the second
graph is 4.3 cm, to the nearest tenth.

Representation - Problem #3
Let‛s ﬁrst graph the three vertices of this triangle and see what the triangle                         C
looks like. Unfortunately, it doesn‛t appear to be a triangle with an obvious
base and/or height measurement. None of the sides are vertical or horizontal.                                      B
However, notice that it is relatively easy to draw a rectangle such that the
triangle is inscribed in the rectangle. The rectangle is 3 by 4, or 12 square units.
It is comprised of the original triangle (shaded) and right triangles 1, 2 and 3.                 A
Since these last three triangles are right triangles, it is fairly easy to calculate
their areas. Triangle 1 is 2 (1)(4) = 2 sq units; triangle 2 is 2 (2)(2) = 2 sq units;
1                                   1
C
and triangle 3 is 2 (3)(2) = 3 sq units. These three triangles account for
1
1       2
2 + 2 + 3 = 7 sq units of the total 12 sq units, meaning that the area of the                                      B
shaded triangle is left to be 12 − 7 = 5 sq units.
3
A
x    y
1   −2                     Because this is a convex polygon, we can use another method to
4    0              determine the area of the triangle. Follow this algorithm, and then see how
it works with other convex polygons you create. Step 1: Write the ordered
−8           2    2       0      pairs of the vertices in two columns, repeating the ﬁrst listed ordered pair
0        1   −2       8      at the end. (We can start with any vertex, but we must then write the
2                    −4      other vertices in order as they appear clockwise or counterclockwise around
the ﬁgure.) Step 2: Write the products of the pairs of numbers found
−6                        4      along the dotted diagonal lines in separate columns out to the sides. Step 3:
Add these products together in each of the two columns. Step 4: Subtract
4 − (−6) = 10
these two sums and take the absolute value. Step 5: Divide this absolute
10 ÷ 2 = 5           value by 2. This is the area of the convex polygon.

46                                                                                        MATHCOUNTS 2005-2006
Warm-Up 11
1. ________ A ﬂight from San Francisco to New York takes ﬁve hours, but the return ﬂight
takes six hours. When it is 4 p.m. in New York, it is only 1 p.m. in San Francisco. The
ﬂight leaves San Francisco at noon and arrives in New York at x o‛clock, and the
return ﬂight the next day leaves New York at noon and arrives in San Francisco at
y o‛clock. What is the positive difference of x and y ?

2. ________ A 6” by 6” grid made from 1” pieces of wire glued together at their
pieces
ends is shown here. How many pieces of wire would be required to
create a similar 10” by 10” grid?

3. ________ What is the least positive integer that has each of the ﬁrst eight positive integers
as factors?

4. ________ Jackson and Lisa divide some money in a ratio of 3:2, respectively. If Lisa gives
\$
Jackson \$6, then Jackson will have twice as much money as Lisa will have. How much
money did Lisa have after the original division of money?

5. ________ The length of a rectangle is 2x + 5 units, the width is 4x + 15 units and the
sq units
perimeter of the rectangle is 100 units. What is the area of the rectangle?
°
6. ________ The consecutive angles of a particular trapezoid form an arithmetic sequence. If
the largest angle measures 120°, what is the measure of the smallest angle?

7. ________ Sandy originally beat Terry by 82 votes in a school election. In a
voters
re-vote, all of Terry‛s supporters voted for her again, but some
of Sandy‛s supporters switched to Terry. If Terry beat Sandy by
four votes in the re-vote, how many voters switched?

8. ________ Which of the following graphs represents the solution to x 2 > −4x − 3?

A.                                    B.
4    2     0     2     4            4    2     0     2     4
C.                                   D.
4    2     0     2     4            4   2     0      2     4
E.
4    2    0      2     4

9. ________ How many positive integers less than or equal to 250 are relatively prime with 250?
integers

10. _______ The positive integers 2 through 2006 have median 1004. If the odd integers are all
increased by 1 and the even integers are all decreased by 1, what is the new median?

MATHCOUNTS 2005-2006                                                                                 47
Warm-Up 11

1.   5             (C, M, T)        5. 525                (F, M)        8. D                    (E, G)

2. 220              (P, S, T)       6. 60            (F, G, M, P)       9. 100          (F, M, P, S, T)

3. 840             (G, S, T)        7. 43              (C, M, T)        10. 1004              (P, S, T)

4. 36           (F, G, M, T)

Solution - Problem #8
We are asked to ﬁnd the solution to the inequality x2 > −4x − 3. To solve this algebraically, we
can follow the regular steps for solving a quadratic equation. By adding 4x and 3 to both sides, we
have x2 + 4x + 3 > 0. After factoring the quadratic we have (x + 3)(x + 1) > 0. Now this becomes
a little different from a regular equation. We know that for x = −3 or x = −1, we would have an
expression on the left that is equal to 0 instead of greater than 0, so the values of −3 and −1 are
not part of our solution set. When we see where these two values
fall on a number line, we see they split the number line into three
sections: less than −3 (non-shaded), between −3 and −1 (shaded),              4    2      0    2     4
and greater than −1 (non-shaded). Notice that any number in the
region less than −3 will make (x + 3) a negative value, (x + 1) a negative value, and therefore
(x + 3)(x + 1) a positive value... which is what we want. Any number in the region between −3 and
−1 will make (x + 3) a positive value, (x + 1) a negative value, and therefore (x + 3)(x + 1) a negative
value... which is not what we want. Finally, any number in the region greater than −1 will make (x + 3)
a positive value, (x + 1) a positive value, and therefore (x + 3)(x + 1) a positive value... which is what
we want. Now we know our solution set contains the values less than, but not equal to, −3 and the
values greater than, but not equal to, −1, which is graph D.

The solution above is very thorough, but what if we didn‛t even know how to start? Multiple
choice problems like this are common on standardized tests, and even if we don‛t know how to solve
the problem, there is value in understanding how to use the answer options to ﬁgure out the answer.
In looking at our answer options, it seems that determining whether 0 is a solution would help to
narrow our choices. Substituting 0 into the inequality gives us 02 > (−4)(0) − 3, which is true since
0 > −3 is true. This means that only graphs A and D are possible answers since they are the only
ones that include 0 in the solution set. Now, to narrow it down between graphs A and D, let‛s ﬁnd
a number to test that is included in one and not the other. Let‛s use −2, which is included in A, but
not in D. By substituting, we have (−2)2 > (−4)(−2) − 3, which is false since 4 > 5 is false. This means
−2 should not be in the solution set. The only graph that includes 0 but not −2 is graph D.

Representation - Problem #7                                                  Sandy                Terry
The top row represents the outcome of the ﬁrst vote. We
don‛t know how many votes the large squares represent, but we                      82
know they are the same, and we see Sandy‛s extra 82 votes. In the
re-vote, we know Terry kept all of her supporters, so she keeps her
big square of votes. We then need to get Terry four more votes                                            4
than Sandy. If we let Sandy keep her big square of votes, we just
need to redistribute those extra 82 votes. Let‛s give Terry her                      39           39
four votes that she will win by, which leaves 82 − 4 = 78 votes to
evenly distribute between the two candidates. This leaves
39 votes for each. Now Sandy has her big square plus 39 votes,
and Terry has her big square plus 39 votes, plus four more votes.
Terry now has 43 votes from people who switched from Sandy.

48                                                                                   MATHCOUNTS 2005-2006
Warm-Up 12
1. ________ A particular positive three-digit integer is divisible by 5. The integer also is
divisible by 11. The sum of the three digits of the integer is 13. What is the
integer?

2. ________ A European train compartment has six seats. Four                      Maintenance Report
of the seats are broken. Wilhelm needs to ﬁll out
Door
a form like the one here to indicate that there are
broken seats. If he randomly checks off four of
the seats in the diagram, what is the probability
that he marked the correct seats? Express your

3. ________ At the ﬁsh market, a whole ﬁsh costs \$4.20. If the body costs \$1 more than the
\$
tail, and the body costs \$2 more than the head, how much does the head cost?

4. ________ What is the greatest integer value of n such that 635,040 is divisible by 2n ?

5. ________ A ream of paper consists of 500 sheets. If a standard box of 8½” by 11” paper holds
cu in
exactly ten reams with no extra space, and a sheet of paper is .004 inches thick,
what is the box‛s volume?

6. ________ Four positive four-digit integers with a sum of 23,750 each contain the digit 1 in a
different position. If the 1s are all removed, the list of integers becomes 982, 829,
982, 298. What was the last four-digit number in the original list?

7. ________ Scott and Deric are playing the Shade Game. Two standard
dice are rolled, and the sum of the numbers shown is the
number of blank squares that must now be shaded in. It
is Deric‛s turn, and his grid is shown here. What is the
probability that Deric will win on this turn by rolling the
exact sum needed to shade the rest of the grid? Express

8. ________ Given the function f (a , b , c , d ) = d ÷ a − a b c − b c , what is the value of
f (1, 1, 1, −1)?

9. ________ A positive integer is 24 less than half its square. What is the integer?

10. _______ A polygonal region is bounded by the
lines x = 0, x = 4, y = −1 and y = 4. The              A.                            B.
region is then rotated about the y -axis to
form a solid. Which of these nets is the
approximate net of the solid formed? (The
nets are not drawn to scale.)                          C.                            D.

MATHCOUNTS 2005-2006                                                                                     49
Warm-Up 12

1. 715          (E, G, P, T)       5. 1870               (F, M)        8. −3                    (C)
1
2.                       (F)       6. 2918               (E, G)        9. 8                     (G)
15
5
3. 0.40                  (F)       7.                       (F)        10. A              (G, M, S)
36
4. 5               (E, G, P)

Solution - Problem #1
This problem is like a puzzle with many clues that we have to piece together. We are looking
for a three-digit integer we will call ABC, with digits A, B and C. Our ﬁrst clue is that the number
is a multiple of 5. This tells us C = 5 or C = 0. We also are told that the integer is divisible by 11.
This means A − B + C is divisible by 11. So A − B + C = 0 or A − B + C = 11. (It would be impossible
for this value to be 22 or any greater multiple of 11 since we have only three digits.) We also are
told that the sum of the digits of our integer is 13. This tells us A + B + C = 13. If A − B + C = 0 and
A + B + C = 13, then 2A + 2C = 13, which is impossible since 2A + 2C must be even. Therefore, we
have A − B + C = 11 and A + B + C = 13, so 2A + 2C = 24 and A + C = 12. Remember that we knew C = 0
or C = 5 since ABC is a multiple of 5. If A + C = 12, then C ≠ 0, so C = 5 and then A = 7. Our number
is 7B5, and since 7 − B + 5 = 11, we can ﬁgure B = 1. Our three-digit integer is 715.

Representation - Problem #3
We are told that the body of the ﬁsh is \$1 more than the tail and \$2 more than the head. This
means we have the price of the head, we add a dollar to get the price of the tail, and then we add
another dollar to get the price of the body. We also know the tail, body and head cost \$4.20. This
scenario is similar to the problem, “If three numbers form an arithmetic sequence and have a sum
of 420, what is the value of the smallest number?” What we know about arithmetic sequences is
that the middle term is equal to the average of all of the terms. There are three terms in this
sequence (head, tail, body), and the tail is the middle value. This means the tail is \$4.20 ÷ 3 = \$1.40.
Adding a dollar gets us to the value of the body, \$2.40, and subtracting a dollar gets us to the value

Perhaps setting up a picture of the scenario would be a helpful approach. We‛re told only the
price of the body in relation to the head and the tail. The head sounds like the least amount, and
it is the value that we need to ﬁnd for our answer. So
let‛s let H be the cost of the head of the ﬁsh. We are
told that the body costs \$2 more than the head, so                         H+2
let H + 2 be the cost of the body. Finally, we know the
body costs \$1 more than the tail. So if we take a dollar
off of the price of the body, which is H + 2, we see
that the cost of the tail is (H + 2) − 1 = H + 1. We‛re
told the cost of all three parts is \$4.20, so we can set
H+1
up the equation (H) + (H + 2) + (H + 1) = 4.20, which       H
simpliﬁes to 3H + 3 = 4.20. Subtracting 3 from both
sides and then dividing both sides by 3 gives us
H = \$0.40.

50                                                                                MATHCOUNTS 2005-2006
Workout 6                                    10
1. ________ What is the volume of the solid shown?
cu units
22

11
18
68
2. ________ In 2003, Chris Cagle recorded the song “What a Beautiful Life,” which includes the
days
lyric, “Day 18,253, well, honey, that‛s 50 years.” If we start counting the number
of days in a 50-year period starting Jan. 1, 2003, by how many days is 18,253 days
shorter or longer than the 50 years?

3. ________ A wall in Victoria‛s room is 83 inches wide. Victoria wants to put in wood paneling
panels
that comes in either 2½-inch or 1¾-inch widths and is the exact height of the wall.
She wants the panels to cover the width of the wall exactly without cutting any of
the panels. What is the fewest number of panels she will need?

4. ________ The length of a rectangle is decreased by three feet, and the width is increased
sq ft
by one foot, forming a square region having an area of 25 square feet. What is the
area of the original rectangular region?

5. ________ Using A, B, C, D and E, list the corresponding speeds in order from fastest to
slowest. Use 1 mile = 5280 ft, 1 km = 0.62 miles and 1 m = 3.28 ft.

A. 10,000 ft/min     B. 60 miles/hr   C. 100 ft/sec   D. 100 km/hr     E. 30 m/sec

6. ________ An anteater can eat 30,000 ants in a day. What is the average number of ants per
ants
minute that he eats during the 24-hour period? Express your answer to the nearest
whole number.

7. ________ The length of segment AB is 6 cm, and points A, B and C are
cm
collinear. What is the sum of the arc lengths bounding the
shaded region in the ﬁgure? The boundary is formed by three
6 cm

8. ________ Lin estimates she will use her phone 220 minutes per month for September through
\$
April, 360 minutes in May and 300 minutes during each of the three remaining
months of the
year. How          Company             Basic Plan            Rate per Add‛l Min.
much will the                                                (or fraction thereof)
May bill be        Cell Light          \$20/month,            \$0.30
using the most                         up to 200 min.
economical         Walk About Phones   \$50/month,            \$0.40
plan for the                           up to 350 min.
year?
Phones-R-Us        \$70/month, unlimited   N/A

9. ________ Sixty marbles are placed into boxes A, B, C, D and E. Together boxes A and B
marbles
contain 24 marbles. Together boxes B and C contain 15 marbles. Together boxes C
and D contain 18 marbles. Together boxes D and E contain 30 marbles. How many
marbles are in box A?

10. _______ An isosceles triangle has an area of 200 square units, and the altitude to the base
units
measures 10 units. What is the perimeter of the triangle? Express your answer in

MATHCOUNTS 2005-2006                                                                                51
Workout 6

1.   15,444          (F, M)       5. ACEDB                 (T)       8. 68                 (S, T)

2. 10                   (T)       6. 21                    (C)       9.   15               (F, G)

3. 35          (E, G, M, T)       7. 6π                 (F, P)       10. 40 + 20 5         (F, M)

4. 32                (F, M)

Solution - Problem #7
We can see the shaded area, and we are asked to ﬁnd the sum of the arc
lengths that bound the shaded area. We know that the arcs in question are
semicircles. These arc lengths are then going to be half of the circumferences
of their circles. The largest arc that bounds the top of the shaded region is a
semicircle with a diameter of 6 cm. The circumference of the circle is 6π cm,      A    C         B
so the arc length is 3π cm. Now we need the arc lengths of the two semicircles            6 cm
forming the lower boundary of the shaded region. We don‛t know, though, what
the diameters of those circles are. We know only that the sum of the two diameters is 6 cm. Let‛s
let the diameter AC be x cm, and the diameter CB will be (6 - x) cm. The circumferences of the
two circles are xπ cm and (6 - x)π cm, and the arc lengths are then half of each of these. Notice
the sum of the circumferences is xπ + (6 - x)π = xπ + 6π - xπ = 6π cm. The sum of the two arc
lengths will then be half of this amount, which is 3π. So the sum of the arc lengths forming the
boundary of the shaded region is 3π + 3π = 6π.

Representation - Problem #3
For this problem we need to determine what combinations of 2½-inch and 1¾-inch panels result
in a total of 83 inches. Since we want the least number of panels, we would prefer to make the 83
inches out of only the 2½-inch panels, or large panels (LP). However, when we divide 83 by 2½, we
see that it does not have an integer result, meaning that we would need to cut a panel to make it
ﬁt exactly. Rather than using the fractions, let‛s use the decimal representations of 2.5 and 1.75.
Since we now know we are going to have to use a 1.75-inch panel, or small panel (SP), notice that we
would need two or four of them so that the sum of the 1.75-lengths ends in .50 or .00 and can be
combined with a group of 2.5-inch panels. Making a chart, we can keep track of our attempts at
making this work. We know using zero small
panels will not work. Now we can check if        # of SP Length of SP Length of LP # of LP
using two of the small panels results in a
0             0             83           33.2
“left-over“ amount that is divisible by 2.5.
(We try 2 SPs, which is 3.5 inches, which            2            3.5           79.5          31.8
then leaves 83 − 3.5 = 79.5 inches, and then         4             7             76           30.4
we see if this is a whole number of large
panels by calculating 79.5 ÷ 2.5. We get             6           10.5           72.5           29
31.8 LPs, which is not satisfactory.) If two
SPs don‛t work, we can go to four SPs. The chart here shows we will use the least number of panels
(35) if we use a combination of six small panels and 29 large panels.

A graphing calculator also can be used for this problem. We know we need x small panels and
y large panels, we want x + y to be as small as possible, and 1.75x + 2.5y = 83. If we work with this
equation by subtracting 1.75x from both sides and then dividing by 2.5, we have y = −1.75x + 2.5 . If
2.5
83

we enter this into the “Y=” screen of our calculator, and then go to the TABLE function, we can see
there are ﬁve ordered pairs where x and y are integers. We need the ordered pair with the lowest
sum, which is when y is as large as possible. This happens at (6, 29), so there are 35 panels.

52                                                                              MATHCOUNTS 2005-2006

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