2002 STATE COMPETITION
SPRINT ROUND QUESTIONS
1.
Segments AB and CD are as shown.
5.
There are 4 marks dividing each length of 1.
1
or 0.2.
Therefore, each interval is worth
5
Segment AB starts at the third mark between
0 and 1 or 0.6. Segment AB ends at the
second mark between 2 and 3 or 2.4.
2.4 - 0.6 = 1.8
Segment CD starts at the second mark
between 1 and 2, or 1.4. Segment CD ends
at 4.
4 - 1.4 = 2.6
2.6 - 1.8 = 0.8 Answer
2.
3.
4.
one mile. It must then take her
1
15
=
14 × 7 = 14 ×
2
2
7 × 15 = 105 minutes to run 14 miles.
105 Answer
1
A rectangle whose dimensions are 1 yards
2
1
by 1 feet can be re-expressed in inches as
2
54 inches by 18 inches.
54 × 18 = 972 Answer
6.
A number n is randomly selected from the
set {1, 2, ..., 10}. We must find the
probability that 2n - 4 > n.
n-4>0
n>4
This means 5 ≤ n ≤ 10.
There are 6 values in the set that can be
chosen to fulfill this requirement.
6 3
=
Answer
10 5
The sum of the first n positive integers is 28.
The sum of the first n positive integers can
be represented by the formula:
(n + 1)× n
2
(n + 1)× n = 28
2
(n + 1) × n = 56
The product of two consecutive positive
integers is 56. By inspection, these are 7
and 8. Therefore, n = 7 Answer
Note: we could carry out the multiplication
and factor:
n2 + n - 56 = 0
(n + 8)(n - 7) = 0
n = -8, 7
But n must be positive so n = 7
7.
48 non-overlapping square tiles, each 1 inch
by 1 inch, fit within a rectangle. To find the
minimum number of inches in the perimeter,
inspect the factors whose product is 48.
1 × 48
2 × 24
3 × 16
4 × 12
6×8
The perimeter is minimized when the sum of
the length and width are minimized. The
sum of the length and width are minimized
the closer you get to the square root of 48.
Thus, the smallest perimeter occurs when
the sides of the rectangle are 6 and 8.
2 × (6 + 8) = 2 × 14 = 28 Answer
8.
The temperature at 5 AM was -28°F. The
temperature at 2 PM was 17°F. The
temperature rose
17 - (-28) = 17 + 28 = 45
degrees in 7 + 2 = 9 hours.
The average increase was
45
= 5. Answer
9
Angela’s school day consists of 4 45-minute
periods (4 × 45 = 180 min), one 90-minute
period, one 25-minute period, and 5 4minute breaks (5 × 4 = 20 min). The total is:
180 + 90 + 25 + 20 = 315 min. = 5 hr. and
15 min.
8:50 + 5 hours = 1:50
1:50 + 15 min = 2:05 Answer
9.
Two of the central angles are 80° and 130°.
Therefore, the third central angle is:
360 - (80 + 130) =
360 - 210 = 150
This piece has the largest area. The pizza
has a radius of 6 inches so the area is 36π.
The piece with a central angle of 150° is:
Edwina runs at a speed of 8 miles per hour
8
2
=
miles per minute.
or a speed of
60 15
1
Thus, it takes her 7 minutes to complete
2
2002 State Mathcounts Solution Set
Page 1 of 9
150 15 5
=
=
360 36 12
of the area of the pizza.
5
× 36π = 15π Answer
12
10. The units digit of 1! is 1.
The units digit of 3! is 6.
5! = 5 × 4 × 3 × 2 × 1 =
(5 × 2) × (4 × 2 × 1) =
10 × (4 × 2 × 1).
Thus, the units digit of 5! must be 0.
Similarly, the units digit of 7!, 9! and 11! are
all 0.
1 + 6 + 0 + 0 + 0 + 0 = 7 Answer
11. f (x) = 5x2 + 3x + 4
f(-2) = 5 × (-2)2 + (3 × -2) + 4 =
5 × 4 + (-6) + 4 = 20 - 6 + 4 =
24 - 6 = 18 Answer
12. Every fourth pot is marked with a red
balloon. Every sixth pot is marked with a
blue balloon. Every tenth pot is marked
with a yellow balloon. Pots with all three
occur on the least common multiple of 4, 6,
and 10.
Divisor
4
6
10
2
2
3
5
2
1
3
5
3
1
1
5
5
1
1
1
(The value in the left, or divisor, column is
divided into the values in the right column,
one row up. If the divisor does not divide
the value one row up with zero remainder,
the value is brought down to the next row
unchanged.
The LCM is 2 × 2 × 3 × 5 = 60
So all multiples of 60 have all three colors.
600
= 10 Answer
60
13. If two standard 6-sided dice are tossed, what
is the probability that the sum of the
numbers shown on the dice is a prime
number? What are all the primes ≤ 12?
They are 2, 3, 5, 7, and 11.
For 2, there is only the combination:
(1,1).
For 3, there are two combinations:
(2,1) and (1,2).
For 5, there are 4 combinations:
(1,4), (2,3), (3,2) and (4,1).
2002 State Mathcounts Solution Set
For 7, there are 6 combinations:
(1,6), (2, 5), (3,4), (4,3), (5,2), and (6,1).
For 11, there are 2 combinations:
(5, 6) and (6,5).
This makes a total of 1 + 2 + 4 + 6 + 2 = 15.
There are a total of 6 × 6 = 36 combinations.
15 5
=
Answer
36 12
14. Every student takes history or science. 85
take both. 106 students take history and 109
take science. We can use a Venn diagram to
see the solution.
106 students take history but 85 take both
history and science. Therefore, 21 take
history but not science. 109 take science,
but 85 take both. Therefore, 24 take science
only. 21 + 85 + 24 = 130 Answer
15. a + b + c = 88 (Eq. 1)
a - 5 = N (Eq. 2)
b + 5 = N (Eq. 3)
5c = N (Eq. 4)
b + 5 = a - 5 (Eq. 2, Eq. 3)
b = a - 10 (Eq. 5)
5c = a - 5 (Eq. 4, Eq. 2)
a−5
c=
(Eq. 6)
5
a−5
a + a − 10 +
= 88 (Eq. 1, Eq. 5, Eq. 6)
5
5a + 5a - 50 + a - 5 = 440
11a = 495
a = 45
N = a - 5 = 45 - 5 = 40 Answer
16. 6% of the eight graders and 3% of the
seventh graders participate in
MATHCOUNTS. There are 1.5 times as
many 8th graders as 7th graders.
Let x = the number of 7th graders.
Then 1.5x = the number of 8th graders.
6% of the 8th graders = .06 × 1.5x = .09x.
3% of the 7th graders = 0.3x.
The total number of students are 2.5x.
.09 x + .03 x 0.12 x 12 x
12
=
=
=
= .048 =
2.5 x
2 .5 x
250 x 250
4.8 % Answer
Page 2 of 9
17. Each of the circles P and Q is tangent to
rectangle ABCD at three points as shown.
Each circle has a radius of 6 cm. RS = 2
The length of AD is the same as the length
of a line drawn parallel to AD through P.
This means AD is the same as the diameter
of P, or 2 × 6 = 12. Since the length of RS
is 2, the length of PR must be 4 as must the
length of SQ. If you draw a line through
PRSQ starting at E and F you have a line
whose length is the same as AB or CD. The
length of EF is EP + PR + RS + SQ + QF.
EP and QF are both radii or 6. Therefore
AB or CD = 6 + 4 + 2 + 4 + 6 = 22.
22 × 12 = 264 Answer
18. How many nine-digit numbers can be made
using each of the digits 1 through 9 exactly
once with the digits alternating between odd
and even? If they must alternate between
odd and even, then we must start and end
with odd digits because there are 5 different
odd digits (1, 3, 5, 7, 9) but only 4 different
even digits (2, 4, 6, 8). So, there are 5
different choices for the first odd digit, 4
different choices for the first even digit, 4
different choices for the second odd digit, 3
different choices for the second even digit,
etc.
5×4×4×3×3×2×2×1×1=
20 × 12 × 6 × 2 = 240 × 12 =
2880 Answer
19. Marvelous Mondays are defined as the fifth
Monday in the month. To be a Marvelous
Monday, the date must be at least the 29th of
the month. September 4 is a Monday
making the last Monday of the month
September 25. The first Monday in October
would be October 2. The last Monday in
October would be October 30, which is a
Marvelous Monday. October 30 Answer
20. The whole numbers are written
consecutively in the pattern shown.
Row A:
Row B:
Row C:
Row D:
Row E:
5 6 7
17 18 19
4
8
16
20
3
9
15
21
2
10 14
22
1
11 12 13
23 24
2002 State Mathcounts Solution Set
Note that odd multiples of 6 appear in Row
A and even multiples of 6 appear in Row E.
500
= 83 R2
6
83 is an odd multiple of 6. Therefore, 83 ×
6 = 498 is in Row A. Since the multiple of 6
is always followed by one more value in
Row A and then one in Row B (look at 7,8
or 19,20, for example). The 500th number
must be in Row B. Answer
21. What is the smallest three-digit number that
is a multiple of 4 and whose digits sum to
23? Determine first a sum of three digits
that equals 23 but whose hundreds unit is as
small as possible. If that smallest digit were
1, could we get a sum to 23? No, since 23 1 = 22 and using two 9’s would get us a sum
of 18. How about 2? Then 23 - 2 = 21 and
again this is a problem. The largest value
that the 2 other digits can be is 18. Thus, the
hundreds unit would be 5. But 599 is not
divisible by 4. So try 17. Then, the
hundreds unit is 6 and we get either 689 or
698. Neither are divisible by 4. (A number
is divisible by 4 if its two lowest order digits
are divisible by 4. E.g., 04, 08, 12, 16, 20,
etc.) So we must try 16. This means the
hundreds digit is 7. We can have either 779
or 788 or 797. 779 is not divisible by 4 (nor
is 797) but 788 is. 788 Answer
22. a * b = b a b
(3 * 2) = 2
32 = 6
(2 * 6) = 6
2 6 = 6 × 2 3 = 48 Answer
23. How many numbers in the range 1-100 are
multiples of 2? All even numbers, or 50 of
them. How many numbers in the range 1100 are multiples of 3? 33 of them, but
every other one is an even number. The
1
33rd number is 99. Therefore,
× 32 = 16
2
of them are repeats of the numbers divisible
by 2. 33 - 16 = 17 more. How many
numbers in the range 1-100 are multiples of
5? There are 20 of them. How many are
repeats? 10 of them end in 0 and so are also
multiples of 2. That leaves 10 more, i.e.,
those that end in 5. But, of those, the values
15, 45, and 75 are also multiples of 3. That
leaves 10 - 3 = 7 that we haven’t counted
Page 3 of 9
before.
50 + 17 + 7 = 74
74 37
=
Answer
100 50
24. A jet that flies 540 mph flies 9 miles per
minute. A jet that flies 480 mph flies 8
miles per minute.
Let x = the expected number of minutes it
takes to fly between both airports.
9(x + 8) is the number of miles between
both airports (when flying at a rate of 9
miles per minute).
8(x + 53) is also the number of miles
between both airports (when flying at a rate
of 8 miles per minute).
9(x + 8) =8(x + 53)
9x + 72 = 8x + 424
x = 352
It takes 352 + 8 = 360 minutes to fly at a
rate of 9 miles per minute.
360 × 9 = 3240 Answer
25. The area of a triangle bounded by the lines
y = 0, x = 12 and y = mx is 1872.
The figure looks as follows:
21 + 3 5
= has a remainder of 5.
7
7
22 + 3 7
= has a remainder of 0.
7
7
2 3 + 3 11
= has a remainder of 4.
7
7
2 4 + 3 19
=
has a remainder of 5.
7
7
2 5 + 3 35
=
has a remainder of 0.
7
7
2 6 + 3 67
=
has a remainder of 4.
7
7
And the pattern emerges. When the
exponent is divided by 3 and the remainder
is 1, the remainder of the expression is 5.
When the exponent is divided by 3 and the
remainder is 2, the remainder of the
expression is 0. When the exponent is
divided by 3 and the remainder is 0, the
remainder of the expression is 4.
87
= 29 R0
3
Therefore, the remainder is 4. Answer
27. Triangle ABC is a right triangle with angle
B a right angle. Points P and Q are on BC
so that segments AP and AQ trisect angle
BAC as in the figure below.
When y is 0:
0 = mx
m=0
When the x-coordinate is 12:
y = 12m
So the three coordinates are (12,0), (0, 0)
and (12, 12m). The length of the vertical
side of the triangle is 12m. The length of
the horizontal side of the triangle is 12.
1
× 12 × 12m = 1872
2
6 × 12m = 1872
12m = 312
m = 26 Answer
26. What is the remainder when 287 + 3 is
divided by 7? Determine whether there is a
pattern when dividing by 7.
2002 State Mathcounts Solution Set
The measure of angle APC is 114°.
Therefore, the measure of angle APB is 180
- 114 = 66°. This makes the three angles in
triangle ABP, 66, 90 and 180 - (90 + 66) =
180 - 156 = 24°. Remember that segments
AP and AQ trisect angle BAC. This means
1
of angle A. Therefore,
that angle BAP is
3
angle A is 24 × 3 = 72°. Angle B is 90°.
Angle C = 180 - (72 + 90) =
180 - 162 = 18 Answer
28. A line segment drawn from a vertex of a
unit square to a point on the square forms
1
of the
two regions, one whose area is
3
other region. See the diagram below.
Page 4 of 9
This means that the area of smaller region is
1
of the area of the square. If we were to
4
draw a line through the square parallel to a
vertical edge and bisecting the horizontal
edges of the cube, the two resulting regions
would have equal area. See the diagram.
Thus, the diagonal is the line segment in
question and it is the hypotenuse of a right
1
triangle whose other sides are 1 and .
2
Let x = the length of the hypotenuse.
1
x 2 = 12 +  
 2
x=
2
=1+
30. The mean of six positive integers is 5, and
the median is 6. The sum of all 6 integers
must be 6 × 5 = 30. Since there are an even
number of items, the average of the third
and fourth item must be the median, i.e., 6.
If we want the largest mode, then we have to
have the rest of the values very small. We
know that the middle 2 numbers must add to
12. This leaves 18 to be divided between
the largest 2 numbers and the smallest 2. If
we assume that the first 2 numbers are 1,
then the two largest values would each be 8.
To make the largest mode, the middle two
values, which must sum to 12, would be 4
and 8, giving us the sequence:
1, 1, 4, 8, 8, 8. 8 Answer
1.
1
5
=
4
4
This makes the radii of the three semicircles
3, 4, and 5, respectively. The area of a
1
semicircle is πr2
2
The area of the semicircle with radius 3 is:
1
9
π × 32 = π
2
2
The area of the semicircle with radius 4 is:
1
π × 42 = 8π
2
The area of the semicircle with radius 5 is:
1
25
π × 52 =
π
2
2
9
25
34
π + 8π +
π = 8π +
π=
2
2
2
8π + 17π = 25π Answer
5
5
=
Answer
4
2
29. There are 6 jugs that contain 13, 15, 16, 21,
21 and 22 gallons, respectively. One jug has
blue punch. The other five jugs have red
1
of the red punch is sold to
punch. If
3
2
Randy and of the red punch is sold to
3
Lisa, this implies that Lisa’s amount of
punch is twice Randy’s amount of punch.
The sum of all the punch is:
13 + 15 + 16 + 21 + 21 + 22 = 108 gallons.
If the blue punch is 22 gallons, then there
would be 86 gallons of red punch to be
divided into thirds. But 86 is not divisible
by 3. If the blue punch is 21 gallons, then
there would be 87 gallons to be divided into
thirds. This works out to 29 + 58. Do we
have combinations to create 29 and 58?
16 + 13 = 29
15 + 21 + 22 = 58
Thus, there were 21 gallons of blue punch.
21 Answer
2002 State Mathcounts Solution Set
TARGET ROUND
Given that two sides of the right triangle are
6 and 8, the hypotenuse must be 10.
2.
The product of 4 consecutive positive
integers is 1 less than 4612.
Let x = 461
Then we can express 1 less than 4612 as
x2 - 1.
(x - 1) × (x + 1) = x2 - 1 or:
460 × 462 = 4612 - 1
Now we need to factor 460 and 462 into two
numbers such that all 4 resulting values can
be placed in such an order to make them
consecutive positive integers.
Page 5 of 9
460 = 2 × 5 × 2 × 23
462 = 2 × 3 × 7 × 11
460 = 20 × 23
462 = 21 × 22
Therefore:
20 × 21 × 22 × 23 = 4612 - 1
The least of the 4 numbers is 20. Answer
5.
We must determine for how many integers n
between 1 and 100 is the greatest common
divisor of 15 and n equal to 3. This would
be all the integers that are multiples of 3 but
not multiples of 5. There are 33 multiples of
3 between 1 and 100. Of those, 6 are
multiples of 5 (15, 30, 45, 60, 75, and 90).
Therefore, there are 33 - 6 = 27 integers
between 1 and 100 whose greatest common
divisor of 15 and themselves is 3.
27 Answer
6.
A 6 inch by 9 inch rectangle is cut into three
portions of equal area as shown.
Alternate solution: Since calculators are
permitted in this round, we could find the
value of
4
4612 − 1 ≈ 2147
. . Therefore
2147
. 4 ≈ 4612 − 1 , so it would make sense
to see if the product of 20, 21, 22 and 23
(the four consecutive integers closest to
21.47) is equal to 4612- 1, which it is. The
least of the four numbers is 20.
3.
4.
The world’s population passed 6 billion
people on July 18, 1999. An average or 4.2
births and 1.7 deaths occurs every second.
Thus, the world’s population increases by
4.2 - 1.7 = 2.5 people every second. This is
an increase of 2.5 × 60 × 60 × 24 = 216,000
people per day or approximately 78,840,000
people per year (if you don’t deal with leap
year). It would take:
1,000,000,000
≈ 12.68391679 years to get to
78,840,000
7 billion people from July 18, 1999. A half
year would place us in 2000. Twelve more
years would place us in early 2012. Does
the leap year factor in? No. We’re only
talking about 2 or 3 days more which is less
than .01 year. 2012 Answer
S is a set of 3 positive integers.
Property I says that S has mean of 10. Then,
the sum of the 3 numbers is 30. So far,
property I holds. Property II says that S has
median 8. Can that happen? Sure. 7 + 8 +
15 works fine. Property III says that S has
sum 45. No! If S has a mean of 10 the sum
must be 30. So Property III can not hold if
Property I does.
Property IV says that S contains the number
35. That couldn’t be true if Property I holds.
Suppose Property I doesn’t hold. Then S has
a sum of 45. Could S have a median of 8?
Sure. {2, 8, 35} works. What about if S
contains the number 35? {2, 8, 35} still
works fine. Thus we can have at most 3
properties holding. 3 Answer
2002 State Mathcounts Solution Set
The area of the rectangle is 6 × 9 = 54. The
1
area of each portion is of the total area or
3
18. We know that one side of the light gray
triangle is 6. Let x = the other side of the
light gray triangle that is not the hypotenuse.
Then:
1
× 6 × x = 18
2
3x = 18
x=6
Therefore, the light gray triangle is half of a
6 inch square and its hypotenuse is
6 2 ≈ 8.485281374
Now, look at the black triangle. One side is
9. Let y = the other side of the black
triangle that is not the hypotenuse. Then:
1
× 9 × y = 18
2
9y = 36
y=4
Let z = the length of the hypotenuse of the
black triangle.
9 2 + 4 2 = z2
81 + 16 = z2
z2 = 97
z = 97 ≈ 9.848857802
Now we can determine the perimeters of
each of the triangles. The perimeter of the
light gray triangle is 6 + 6 + 8.485281374 =
20.48528137. The perimeter of the dark
gray triangle is (9 - 6) + (6 - 4) +
8.485281374 + 9.848857802 =
23.334139176. The perimeter of the black
Page 6 of 9
triangle is 9 + 4 + 9.848857802 =
22.8488578. The dark gray triangle has the
greatest perimeter and to the nearest tenth
this is 23.3. Answer
7.
8.
The ratio of boys to girls in the upper classes
is 3 to 4 and there are 560 upperclassmen.
There are 110 freshman boys and 90
freshman girls.
First determine the number of upper class
boys. There are 3 boys for every 4 girls in
the upper classes, or 3 boys out of every 7
students.
3x + 4x = 560
7x = 560
x = 80
80 × 3 = 240
There are 240 upperclass boys and 110
freshmen boys for a total of 350 boys. There
are a total of 560 + 110 + 90 = 760 students
in the school.
350
≈ 0.460526316
760
0.460526316 × 100 = 46.0526316
46 Answer
The only lines of symmetry that will be
possible for the completed pattern will be
horizontal and vertical lines. The first 2
squares shown in the problem (the "+" and
"x" squares) have both lines of symmetry,
but it’s their horizontal line of symmetry that
allows us to create any pattern of three
squares with any combination of these two.
The three squares in the pattern can be 3 "+"
squares (figure 2), 2 "+" squares and an "x"
square (figures 4, 11, 12), 2 "x" squares and
a "+" square (figures 3, 9, 10), or 3 "x"
squares (figure 1). This is a total of 8
different patterns. The remaining two
squares, I’ll call them the "diagonal" squares,
do not have a horizontal line of symmetry
and will only give the pattern a vertical line
of symmetry if one of the "diagonal" squares
is placed first and the other is placed last. In
the middle can be the "+" square (figures 5,
6) or the "x" square (figures 7, 8). There are
four different such arrangements, for a total
of 12 arrangements.
1
2
3
4
2002 State Mathcounts Solution Set
5
6
7
8
9
10
11
12
Pictures 1, 2, 3, 4, 9, 10, 11, and 12 have a
line of symmetry through their horizontal
middle. Pictures 1, 2, 3, 4, 5, 6, 7, and 8
have a line of symmetry through their
vertical middle.
12 Answer
TEAM ROUND
1.
How many three-digit whole numbers have
no 7’s and no 9’s as digits? First, examine
the numbers from 100 to 199. In 100 - 109,
8 do not (everything but 107 and 109). This
is also true of 110 - 119, 120 - 129, 130 139, 140 - 149, 150 - 159, 160 - 169 and 180
- 189. That’s a total of 8 × 8 = 64 numbers.
Clearly 170 - 179 and 190 - 199 all contain
7’s or 9’s. In the same way, there are 64
numbers containing no 7’s or 9’s in the
ranges 200 - 299, 300 - 399, 400 - 499, 500 599, 600 - 699, and 800 - 899 for a total of
64 × 7 = 448. Of course, all numbers in the
700 - 799 and 900 - 999 ranges have either
7’s and/or 9’s. 448 Answer
Another solution would be to see that to
create such a number, there are 7 choices for
the first digit (1-6, 8), and 8 choices for each
of the second and third digits (0-6, 8), so
there are 7 × 8 × 8 = 448 such numbers.
2.
Let x = the number of meters that Kevin
must walk for the race to end in a tie.
x x − 50
=
3
1.75
1.75x = 3x - 150
1.25x =150
125x = 15000
x = 120 Answer
Page 7 of 9
3.
a
b

c
d
1
2

4
3
4
=1⋅2-4⋅3
= 2 - 12 = -10
Since the car travels a circular path the inner
wheels travel the circumference of a circle.
The inside wheels are rotating half as fast as
the outside wheels so the circumference of
the inner circle must be half as much as the
circumference of the circle traveled by the
outer wheels.
Let x = the radius of the circle traveled by
the inner wheels. Then x + 6 is the radius of
the circle traveled by the outer wheels.
2πx × 2 = 2π(x + 6)
4πx = 2πx + 12π
2πx = 12π
x=6
The number of feet in the path traced by the
inside front wheel is:
2 × π × 6 = 12π Answer
2

-1
3
5
= 4 ⋅ 2 - (-1) ⋅ 3
= 8 - (-3) = 8 + 3 = 11
1

0
-2
2
= 5 ⋅ 1 - 0 ⋅ (-2)
=5-0=5
1

-1
3
-10
= 2 ⋅ 1 - (-1) ⋅ 3
= 2 - (-3) = 2 + 3 = 5
6.
11

5
outside wheels. The front two wheels are
six feet apart. We must determine the
number of feet in the path traced by the
inside front wheel in one trip around the
circle.
=a⋅b-c⋅d
5
The easiest way is to list the possible paths.
= -10 ⋅ 11 - 5 ⋅ 5
= -110 - 25 =
-135 Answer
4.
5.
A square region has perimeter 60 inches.
When a unit tile is removed the remaining
tiles create a rectangular region with
minimum perimeter.
First determine the length of the square.
Let x = the length of the square.
4x = 60
x = 15
Therefore, there are 15 × 15 = 225 unit tiles
in the square. Subtracting 1 gives us 224
unit tiles to make the rectangle. A minimum
perimeter occurs when the length and width
of the rectangle are as close to each other as
possible i.e., as close to the square root of
the area, or 15.
224 = 2 × 112 = 2 × 2 × 56 =
2 × 2 × 2 × 28 =
2 × 2 × 2 × 2 × 14 =
2×2×2×2×2×7=
16 × 14
The perimeter is 2 × (16 + 14) =
2 × 30 = 60. Answer
The inside wheels of a car traveling on a
circular path are rotating half as fast as the
2002 State Mathcounts Solution Set
From A, the mouse can go to B and from
there to E and F. Of those two possibilities
only one leads to F. From A, the mouse can
go to C and then F. That is one possibility
out of one possibility. From A, the mouse
can go to D and then to F, making one
possibility out of 3 possibilities.
The probability that the mouse goes from A
to B to F is:
1 1 1
× =
3 2 6
The probability that the mouse goes from A
to C to F is:
1
1
×1 =
3
3
The probability that the mouse goes from A
to D to F is:
1 1 1
× =
3 3 9
1 1 1
+ + =
6 3 9
Page 8 of 9
3 6 2 11
+ +
=
Answer
18 18 18 18
7.
Marianna has some nickels and quarters and
their combined value is $9.15. Their
combined weight is one pound. 80 quarters
weigh one pound and ninety nickels weigh
one pound. We must find the number of
nickels.
Since 80 quarters weight one pound, one
1
quarter weighs
pound. Since 90 nickels
80
1
weigh one pound, one nickel weighs
90
pound.
Let x = the number of quarters.
Let y = the number of nickels.
25x + 5y = 915 (Eq. 1 -- from the cost)
1
1
x+
y = 1 (from the weight)
80
90
90x + 80y = 7200 (Eq. 2)
400x + 80y = 14640 (Eq. 3: Eq. 1 × 16)
310x = 7440 (Eq. 3 - Eq. 2)
x = 24
25 × 24 + 5y = 915
600 + 5y = 915
5y = 315
y = 63 Answer
8.
What is the base 4 representation of the base
2 number:
110110002
The solution is easy in this case. Instead of
transposing to base 10 and then to base 4,
you can simply, working from right to left,
take groups of 2 numbers and make the
translation:
00 => 0, 01 => 2, 10 => 1, 11 => 3
110110002 = 11 01 10 002 =
31204 Answer
9.
A farmer has fenced a field as shown.
The field now looks like this:
Instead of a single trapezoid with sides 40,
40, 80 and 40 2 , there are 4 trapezoids
with sides 20, 20, 40 and 20 2 .
What is new in creating the 4 congruent
trapezoids are two lines of length 20, a line
of length 40 and a line of length 20 2 .
20 + 20 + 40 + 20 2 = 80 + 20 2 =
108.2842712 ≈ 108 Answer
10. All the faces on Alison’s cube have a 5. The
faces on Brian’s cube have the numbers 1 6. Three of the faces on Noah’s cube have a
2 and the other three have a 6. If Allison
rolls her cube she always gets a 5. If Brian
1
rolls his cube, he has a probability of
6
getting any one of the values from 1 to 6. If
1
Noah rolls his cube he has a
probability
2
1
of getting a 2 and a
probability of getting
2
a 6. For Allison to roll greater than Brian,
Brian must roll any value between 1 and 4.
4 2
This has a probability of = . For
6 3
Allison to roll greater than Noah, Noah must
1
roll a 2 which has a probability of .
2
2 1 1
× =
Answer
3 2 3
The slanted line in the trapezoid (i.e., the
one that isn’t 40 or 80 m long) is the
hypotenuse of a 40 by 40 right triangle.
Therefore, its size is 40 2 . The farmer
divides the field into 4 congruent trapezoids.
2002 State Mathcounts Solution Set
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