2013 Mathcounts School Competition
Sprint Round Solutions
1. Solution:
2 + 3  4  5 = 2 + 12  5 = 9.
2: Solution:
When four people split the bill, each person pays: $71.20  4 = $17.80
3. Solution:
The sum of two smallest angles is 90. So the mean of them is 90  2 = 45.
4. Solution:
The fraction is
8  10 2
 .
10  12 3
5. Solution:
Let x be the number of cups of sugar needed.
12 30
30 5
1

x
 2 .
We use proportions:

1
x
12 2
2
6. Solution:
Let x be the number of minutes it will take 32 birds to eat 32 worms.
4 worms
32 worms

We use proportions:

4 birds  4 minutes 32 birds  x minutes
x  4.
7. Solution:
It takes two days and nights for Gretchen to row 10 miles. One more day she will get to
the finish line. The answer is 2 + 1 = 3.
8. Solution:
Let the number be x.
x 1
2
x  13 .

7
9. Solution:
There are 100  45% = 45 people who prefer chocolate and 100  25% = 25 people who
prefer strawberry. The difference is 45 - 25 = 20.
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2013 Mathcounts School Competition
Sprint Round Solutions
10. Solution:
We know that the ending point should have an odd number of line segments (odd node).
The point is I.
11. Solution:
0.6
6
3
0.6% 


.
100 1000 500
12. Solution:
n( B  C  O)  n( B)  n(C)  n(O)  n( B  C)  n(C  O)  n(O  B)  n( B  C  O)
= 14  13  17  5  7  6  4  30 .
13. Solution:
Method 1:
She has
1 way to jump the step one (1).
2 ways to jump the step 2 (1 + 1, 2).
3 ways to jump the step 3 (1 + 1 + 1, 1+ 2, 2 + 1).
5 ways to jump the step 4: 2 + 3.
8 ways to jump the step 5: 3 + 5 = 8.
Method 2:
There is a recurrence relation as follows:
an  an 1  an  2 with a1  1 and a2  2 .
So a5  a4  a3  a3  a2  a2  a1  a2  a1  a2  a2  a1  3a2  2a1  3  2  2 1  8 .
14. Solution:
Let p be the number of pencils and q be the number of pens.
p + 2q = 110
(1)
2p + q = 100
(2)
(1) + (2): 3 p + 3q = 210
(3)
(3)  2: 6(p + q) = 420.
So Edward spends $4.20.
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2013 Mathcounts School Competition
Sprint Round Solutions
15. Solution:
2  18  20 = 720 days.
16. Solution:
There are 303 + 312 + 285 = 900 students who responded.
The number of students who responded is 10%  900 = 90.
The number of students who responded not in favor of the Field Day is
(1  30%)  90 = 63.
17. Solution:
Let F be the number of field goals and T be the number of touchdowns the team got last
season.
7F + 7T = 36  7
(1)
3F + 7T = 212
(2)
(1) – (2): 4T = 40

T = 10.
18. Solution:
Let n be the number of deliveries.
9n  3286 + 3  298

9n  4180
So the smallest number of deliveries is 465.
n  464.4
19. Solution:
5x(1 + 7y) + x(y – 5) = 5x + 35x y + xy – 5x = 36xy.
20. Solution:
Let A be the event that John picks a peanut butter sandwich from bag A.
Let B be the event that John picks a peanut butter sandwich from bag B.
6
6 1
4
4 1
  and P( B) 

 .
We have P( A) 
6  4  2 12 2
4  2  4 12 3
Since two events are independent, the desired probability is
1 1 1
P  P( A)  P( B)    .
2 3 6
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2013 Mathcounts School Competition
Sprint Round Solutions
21. Solution:
The shaded area is the same as the area of the quarter circle O minus the area of the half
circle P.
  42
4

  22
2
 4  2  2 .
22. Solution:
Let x be the price before tax and y be the original price of the running shoes.
110
79.20 
x.
x  72 .

100
(1  40%) y  72
y  72  0.6  $120 .

23. Solution:
2
 2 miles.
60
The police car will intercepts the robber in t hours.
2
t
(75 – 60) t = 2

hours.
15
2
In this time, the robber travelled 60   8 miles.
15
Therefore, the robber is 8 + 2 = 10 miles away from the bank.
In 2 minutes, the robber travelled 60 
24. Solution:
 1 = 180  60  50 = 70.
25. Solution:
Let P be Paul's age, M be Manny's age, A be Alex's age, and T be Tim's age.
P = 3M
(1)
A = 2M
(2)
1
T  ( P  A)
(3)
4
T  P  A  75
(4)
1
1
5
Substituting (1) and (2) into (3): T  ( P  A)  (3M  2M )  M
(5)
4
4
4
4
2013 Mathcounts School Competition
Substituting (1), (2), and (5) into (4):
Sprint Round Solutions
5
25
M  3M  2M  75 
M  75  M  12 .
4
4
26. Solution:
2n  2(3m  4)  12
(1)
(2)
n  6  5m
We write (1) as 2n  6m  8  12

2n  6m  4 
(3)
n  3m  2
m  2 .
(3) – (2):  3m  6  2  5m 
Substituting m  2 into (2): n  6  5(2) 
n  10  6  4 .
Thus 3m  n  3(2)  (4)  10 .
27. Solution:
Let d be the distance for each commute and v be the average speed for the entire round
trip.
2d
v
 40 .
d
d

30 60
28. Solution:
Method 1:
Case I: We draw a red marble first.
The probability to draw 2 red marbles is
20 19
 .
32 31
Case II: We draw a blue marble first.
12 11
 .
32 31
20 19  12 11
512
16

 .
The desired solution is P 
32  31
32  31 31
The probability to draw 2 blue marbles is
Method 2:
5
2013 Mathcounts School Competition
Sprint Round Solutions
 20 
 
2
The probability to draw 2 red marbles is P    .
 32 
 
2 
12 
 
2
The probability to draw 2 blue marbles is P    .
 32 
 
2 
 20  12  20  19 12  11
    

2
2
2
2  512  16 .
The desired solution is P      
32  31
32  31 31
 32 
 
2
2 
29. Solution:
Let AQ = PQ = n and QB = m.
n2  m2  252
n2  252  m2

Since we want the smallest value for n, we let m = 24 in (1):
n2  252  242  (25  24)(25  24)  49
n  7.

Hence AB = m + n = 24 + 7 = 31.
30. Solution:
54 3
9 3.
6
By the area formula for an equilateral triangle:
The shaded area is
3 2
3 2
a
a
a  6.
a 2  36 
9 3 
4
4
We see that the radius of the circle is a. So the circumference of the circle
is 2r  2  6  12 .
A
6
(1)
2013 Mathcounts School Competition
Target Round Solutions
1. Solution:
There are two choices for the tens place and three choices for the units place. There are 2
 3 = 6 integers.
2. Solution:
We label D as shown in the figure. Since triangle XYZ is isosceles, XD = DZ.
Applying the Pythagorean Theorem to triangle XYD: XD 2  YD 2  XY 2
XD  XY  YD  85  68  2601
2
2
2
2
2


XD  2601  51.
So AB  XD  DZ  2 XD  2  51  102 .
3. Solution:
Let x be the minutes the computer has been power on. By proportions:
3  60  40
x

97%
(1  97%)
x

3  60  40
 0.03  6.8  7 .
0.97
4. Solution:
The probability that the spinner stops in the black section is 1/4. The probability that the
spinner stops in the white section is 3/4. The probability that the first four spins stop in
the black section and the fifth spin stops in the white section is:
P
1 1 1 1 3
3
    
.
4 4 4 4 4 1024
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2013 Mathcounts School Competition
Target Round Solutions
5. Solution:
Let x be the price before tax.
x(1  6.25%)  51.51

x
51.51
 $48.48 .
(1  6.25%)
6. Solution:
We label the figure as shown with AD as the height of
equilateral triangle ABC. Then DB = 1.
Applying the Pythagorean Theorem to triangle ABD:
AD 2  AB 2  BD 2  2  1  3 .
So AD  3 .
We see that DO = 1 which is the half of the length CB.
So the radius of the circle is AO  AD  DO  3  1 .
7. Solution:
The hose has the cylinder shape. The radius of the cylinder is ½
inches or 1/12 feet.
The volume of the cylinder shaped hose is
1
then: V  r 2 h    ( )2  100  2.18  2 .
12
8. Solution:
Let the number of years be x.
The number of permutations of six letters is 266.
So x = 266 /(100  106 )  3.089  3.
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2013 Mathcounts School Competition
Team Round Solutions
1. Solution:
Let d be the number of dimes and q be the number of quarters
10d + 25q = 715.
We use as many as quarters as possible.
715 = 25  28.6 = 25  28 + 15 = 25  27 + 10  4.
So the least number of coins is 27 + 4 = 31.
2. Solution:
Cost for bats: (12 + 15)  80 = 2160
Cost for balls: (45 + 38)  6 = 498
Cost for gloves: (15 + 17)  60 = 1920.
Total cost: 2160 + 498 + 1920 = $4578.
3. Solution:
20 chickens need 20/ 5 = 4 pounds and the rest (10 – 4 = 6) can feed 6  2 = 12 pigs.
So the answer is 12.
4. Solution:
100 yards = 100  3 feet = 100  3 12 inches
One revolution is 2    8.
The number of revolutions will be 100  3 12/ (2    8)  71.66  72.
5. Solution:
a
a3
5

. Then we have

7a  1
7a  1  7 8
8a  24  35a  30 
35a  8a  24  30

a
2
 .
So a  2 and
7a  1 15
Let the fraction be
6. Solution:
6 ▼ 4 = 6  4 4 =  10.
20 ▼ (10) = 20  4 (10) = 60.
9
8(a  3)  5(7a  6)

27a  54 .
2013 Mathcounts School Competition
Team Round Solutions
7. Solution:
The diameter of the second circle is d1 and d1 = ST = 2.
The diameter of the third circle is d2 and d2 = AB =
The diameter of the shaded circle is d3 and
d3 = FG  FB 2  BG 2  (
So the shaded area is
AT 2  TB2  12  12  2 .
2 2
2
)  ( ) 2  1.
2
2
1

d32 
4
4
8. Solution:
With the car, the cost will be $4  120/30 = $16.
With the truck, the cost will be $4  120/20 = $24.
The positive difference is $24 - $16 = $8.
9. Solution:
543  (54)3 = 2  543 = 314,928
10. Solution:
192 = 180 + 12.
The smallest number in S should be greater than 12.
We see that 180 = 22  32  5.
The factors are 15, 18, 20, 30, 36, 45, 60, 90, 180.
The median of them is 36.
Copyright © 2013 by Sam Chen and Yongcheng Chen
Special thanks to
Professor Harold Reiter, chair of 2005 Mathcounts Question Writing Committee, and
Mrs. Tonya McLawhorn, coach of 2007 North Carolina Mathcounts National Team
for reviewing our solutions.
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