American Math Competition 10 Practice Test 10
American Mathematics Competitions
Practice 10
AMC 10
(American Mathematics Contest 10)
INSTRUCTIONS
1. This is a twenty-five question multiple choice test. Each question is followed by
answers marked A, B, C, D and E. Only one of these is correct.
2. You will have 75 minutes to complete the test.
3. No aids are permitted other than scratch paper, graph paper, rulers, and erasers. No
problems on the test will require the use of a calculator.
4. Figures are not necessarily drawn to scale.
5. SCORING: You will receive 6 points for each correct answer, 1.5 points for each
problem left unanswered, and 0 points for each incorrect answer.
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American Math Competition 10 Practice Test 10
1. One can holds 15 ounces of soda. What is the minimum number of cans needed to
provide three gallons (1 gallon = 128 ounces) of soda?
(A) 27 (B) 26 (C) 25 (D) 20 (E) 18
2. The sums of three whole numbers taken in pairs are 17, 18, and 25. What is the
middle number?
(A) 11 (B) 12 (C) 13 (D) 14 (E) 15
1
3. Simplify as a common fraction: .
1
1
2
1
3
(A) 1/8 (B) 2/8 (C) 3/8 (D) 5/8 (E) 7/8
4. Thirty percent less than 70 is two-fifth more than what number?
(A) 26 (B) 30 (C) 32 (D) 35 (E) 48
4. D.
5. Kathy has two younger twin brothers. The product of their three ages is 1024. What is
the smallest possible sum of their three ages?
(A) 32 (B) 24 (C) 22 (D) 18 (E) 16
6. In a class of 42 students, 18 students are in the Math Club, 5 students are in both the
Math Club and the Science Club, and 14 are in neither. How many students are in the
Science Club?
(A) 5 (B) 15 (C) 20 (D) 35 (E) 30
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American Math Competition 10 Practice Test 10
7. The number of centimeters in the length, width and height of a rectangular carton are
consecutive integers. Find the smallest 4-digit number that could represent the number of
cubic centimeters in the volume)
(A) 1001 (B) 1320 (C) 1331 (D) 1025 (E) 1216
8. A majority of the 40 students in Ms. Li’s class bought pencils at the school bookstore.
Each of these students bought the same number of pencils, and this number was greater
than 1. The cost of a pencil in cents was greater than the number of pencils each student
bought, and the total cost of all the pencils was $20.15. How many students in the class
bought the pencils?
(A) 35 (B) 11 (C) 33 (D) 13 (E) 31
9. Which of the following is equal to Simplify 6 11 + 6 11 .
(A) 22 (B) 12 (C) 12 22 (D) 6 22 (E) 2 3
10. As shown below, convex pentagon ABCDE has sides AB = 3, BC = 4, CD = 6,
DE = 2, and EA = 7. The pentagon is originally positioned in the plane with vertex A at
the origin and vertex B on the positive x-axis. The pentagon is then rolled clockwise to
the right along the x-axis. Which side will touch the point x = 2015 and be completely on
the x-axis?
(A) AB (B) BC (C) CD (D) DE (E) EA
11. ABCD is a parallelogram. All the line segments inside
the figure either parallel to AD, AD, or BE. How many
parallelograms are there that contain the shaded triangle?
(A) 6 (B) 7 (C) 8 (D) 9 (E) 12
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American Math Competition 10 Practice Test 10
12. A 5 × 5 × 5 wooden cube is painted on five of its faces and is then cut into 125 unit
cubes. One unit cube is randomly selected and rolled. What is the probability that the face
showing is painted? Express your answer as a fraction.
1 1 1 2 28
(A) (B) (C) (D) (E)
6 3 5 7 125
13. The lengths of the parallel bases of a trapezoid are 14 cm and 7 cm. One of the legs
has length 8 cm. How many integer values are possible for the length of the other leg?
(A) 6 (B) 7 (C) 8 (D) 9 (E) 13
14. It takes Amy 4 hours to paint a house, it takes Bill 6 hours, and it takes Chandra 8
hours. Amy starts to paint the house for one hour, then Bill continues the job for another
hour, and Chandra follows Bill and works for one hour. If the pattern continues until the
job is completed, how many hours does Chandra paint the house?
4 1 5 2
(A) (B) (C) (D) 2 (E)
3 3 3 3
15. Alex can jog 120 meters in 1 minute, Bob can jog 80 meters in 1 minute, and Charlie
can jog 70 meters in 1 minute. The circular path has a circumference of 1000 meters.
They start running together at 10:00 a.m. at point A. At what time will they first all be
together again at point A?
(A) 10: 40 a.m. (B) 10:50 a.m. (C) 11:20 a.m. (D) 11:40 a.m. (E) 11:55 a.m.
16. We are choosing a committee of 6 animals from 3 cats, 4 dogs, and 5 pigs. If Alex
Cat, Bob Dog and Charles Pig do not like each other and they will not work in the same
group, how many compatible committees are there?
(A) 84 (B) 220 (C) 304 (D) 378 (E) 462
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American Math Competition 10 Practice Test 10
17. What is the remainder when 31 + 32 + · · · + 32015 is divided by 8?
(A) 7 (B) 6 (C) 5 (D) 4 (E) 3
18. The capacity of a car's radiator is nine liters. The mixture of antifreeze and water is
40% antifreeze. The temperature is predicted to drop rapidly requiring the mixture to be
70% antifreeze. How much of the mixture in the radiator must be drawn off and replaced
with pure antifreeze?
A. 3.5 liters B. 4.5 liters C. 5.0 liters D. 6.0 liters E. none of these
19. The sum of a four-digit positive integer and its digits is exactly 2015. Find the sum of
all possible values of the four-digit positive integer.
(A) 2015 (B) 2012 (C) 4004 (D) 4008 (E) 4030
20. Pipe A will fill a tank in 6 hours. Pipe B will fill the same tank in 4 hours. Pipe C will
fill the tank in the same number of hours that it will take Pipes A and B working together
to fill the tank. What fraction of the tank will be filled if all three of the pipes work
together for one hour?
4 5 5 2
(A) (B) (C) (D) 1 (E)
3 3 6 3
21. In ABC points D and E lie on BC and AC, respectively. If AD and BE intersect at T
so that AT /DT = 5 and BT /ET = 7, what is CD /BD?
3 5 4 4 1
(A) (B) (C) (D) (E)
17 7 17 7 7
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American Math Competition 10 Practice Test 10
22. The area of right triangle ABC is 360. BAC = 90. AD is the median on BC. DE
AB. AD an CE meet at F. Find the area of triangle AEF.
(A) 60 (B) 70 (C) 80 (D) 90 (E) 100
23. P is a point inside the equilateral triangle ABC. PA = 2, PB 2 3 , and PC = 4. Find
the area of triangle ABC.
(A) 7 3 (B) 4 3 (C) 7 38 (D) 8 38
(E) 7 2
24. A special deck of cards contains cards numbered 1 through 5 for each of five suits.
Each of the 25 cards has a club, diamond, heart or spade on one side and the number 1, 2,
3, 4 or 5 on the other side. After a dealer mixed up the cards, three were selected at
random. What is the probability that of these three
randomly selected cards, displayed here, one of the
cards showing the number 3 has a spade printed on the
other side? Express your answer as a common fraction.
4 4 25 5 5
(A) (B) (C) (D) (E)
11 9 54 11 9
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American Math Competition 10 Practice Test 10
25. Circles A, B, and C are externally tangent to each other and internally tangent to circle
D. Circles B and C are congruent with radius 8. Circle A
passes through the center of D. What is the radius of circle
A?
(A) 6 (B) 7 (C) 8 (D) 9 (E) 10
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American Math Competition 10 Practice Test 10
ANSWER KEYS
1. B.
2. B.
3. D.
4. D.
5. A.
6. B.
7. B.
8. E.
9. A.
10. C.
11. E.
12. A.
13. E.
14. A.
15. D.
16. E.
17. A.
18. B.
19. C.
20. C.
21. A.
22. A.
23. A.
24. B.
25. D.
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American Math Competition 10 Practice Test 10
SOLUTIONS:
1. B.
Because 3 128 / 13 = 25.6, there must be 26 cans.
2. B.
Let three numbers be a, b, and c.
a + b =17 (1)
b + c = 25 (2)
c + a = 18 (3)
(1) + ( 2) + (3): 2(a +b + c) = 60 a +b + c = 30 (4)
(4) – (3): b = 12.
3. D.
1 1 1 1 5
.
1 1 3 8 8
1 1 1
2 5 5 5
1
3 3
4. D.
30 2 7
Thirty percent less than 70 is 70 70 x x x 49 x 35
100 5 5
5. A.
The age of each person is a factor of 1024 = 210. Since we want the smallest sum, the
three numbers should be as close as possible.
210 = 23 23 24.
The twins could be 8, and 8. Kathy could 16.
The smallest sum of their ages is 8 + 8 + 16 = 32.
6. B.
There are 42 – 14 = 28 students participated in the two clubs. Let S be the number of
students in the Science Club.
By the Two Events Union Formula n( A B) n( A) n( B) n( A B) , we have 28 = 18 +
S – 5. So S = 15.
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American Math Competition 10 Practice Test 10
7. B.
Let the smallest value of the length, width and height be a – 1. Since the numbers are
consecutive integers, then the other two dimensions are a and a + 1.
V = (a – 1) a (a + 1) = a3 – a
We are seeking for a 4-digit number (a3) that is just over 1000.
103 = 1000 and 113 = 1331.
So a = 11. V = a3 – a =113 – 11 = 1320.
8. E.
Let C be the cost of a pencil in cents, N be the number of pencils each student bought,
and S be the number of students who bought pencils. Then C ·N · S = 2015 = 5 · 13 · 31,
and C > N > 1. Because a majority of the students bought pencils, 40 S > 40/2 = 20.
Therefore S = 31, N = 5, and C = 13.
9. A.
Method 1:
Let 6 11 + 6 11 = x (1)
Squaring both sides of (1): ( 6 11 + 6 11 )2 = x2
( 6 11 + 6 11 )2 = 6 11 2 (6 11)(6 11) 6 11
= 12 2 (6 11)(6 11) 12 2 62 11 12 2 25 12 10 22 .
x2 = 22 x= 22 .
Method 2:
We see that a = 6, b = 11, and a2 – b = 62 – 11 = 25 = 52. Therefore the given nested
radical can be denested.
By the formula, we have
6 62 11 6 62 11 11 1
6 11 .
2 2 2 2
6 62 11 6 62 11 11 1
Similarly, 6 11 .
2 2 2 2
11 1 11 1 11 2 11
Thus 6 11 + 6 11 = + = 2 2 22 .
2 2 2 2 2 2 2
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American Math Competition 10 Practice Test 10
10. C.
One complete rotation goes 3 + 4 + 6 + 2 + 7 = 22 unit length.
2015 = 22 × 91 + 13.
We only need to roll the pentagon 13 units. 3 + 4 + 6 = 13. Therefore CD will touch x =
2015 and be completely on the x-axis.
11. E.
2 2 2 1
We have 8
1 1 1 1
Parallelograms with sides parallel to AB and BC.
1 2 2 1
We have 4 more parallelograms with sides parallel to BE and BC.
1 1 1 1
Total number of parallelograms that contain the shaded triangle is 12.
12. A.
Method 1:
There are 5 · 25 = 125 painted faces all of which are equally likely. There are 125 · 6 =
750 faces altogether. Therefore the probability is 125/750 = 1/6.
Method 2:
Number of cubes painted 1 face: 12 4+ 9 = 57.
Number of cubes painted 2 faces: 12 2 + 4 = 28.
Number of cubes painted 3 faces: 4.
Number of cubes painted 0 side: 27 (We do not need this information).
57 1 28 2 4 3 1
The probability is P .
125 6 125 6 125 6 6
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American Math Competition 10 Practice Test 10
13. E.
a a 8 8
a 8.
7 14 7
a 8
b x.
b x
By the triangle inequality theorem,
AE AB EB AE AB
16 14 x x 16 14 2 2 x 30
1 x 15 2 x 14 .
The number of integer values for the length of the other
leg is then 14 – 2 + 1 = 13.
14. A.
1 1 1 13
In the first 3 hours, each person works one hour and they finish ( ) 1 of
4 6 8 24
the job.
11
The job left is .
24
Let x be the number of hours Chandra works. In the next round, we have
1 1 1 11 1 11 1 1 11 6 4 1 1
1 1 x x x
4 6 8 24 8 24 4 6 24 24 3
1 4
So Chandra paints the house 1 hours.
3 3
15. D.
Method 1:
We find the time needed for each person to complete the path.
1000 25
Alex needs minutes.
120 3
1000 25
Bob needs minutes.
80 2
1000 100
Charlie needs minutes.
70 7
We now find the least common multiple of them.
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American Math Competition 10 Practice Test 10
a c ac LCM (ad , bc)
LCM ( , )
b d GCF (ad , bc) bd
25 25 25 25 25 25
LCM ( , ) 25
3 2 GCF (25 2,25 3) 25
100 25 100 25 100 25 100
LCM (25, ) LCM ( , ) 100
7 1 7 GCF (25 7,100 1) 25
So at 11:40 a.m. they will first all be together again at point A.
Method 2:
Time needed for Alex to catch Bob is 1000 (120 – 80) = 25 minutes
Time needed for Alex to catch Charlie is 1000 (120 –70) = 20 minutes
Time needed for Bob to catch Charlie is 1000 (80 –70) = 100 minutes
LCM (25, 20, 10) = 100.
So at 11:40 a.m. they will first all be together again at point A.
16. E.
A compatible group will either exclude all these three animals or include exactly one of
9 3 9
them. This can be done in 84 378 462 committees.
6 1 5
17. A.
The remainder is 1 when 3n is divided by 8, where n is even. So the sum of every 8 terms
will have a remainder of 0 when divided by 8.
2014/2 = 1007 = 121 × 8 + 7. So the remainder is 7 when 32 + 34 + · · · + 32014 is divided
by 8.
The remainder is 3 when 3m is divided by 8, where m is odd. So the sum of every 8 terms
will have a remainder of 0 when divided by 8.
(2015 – 1)/2 + 1 = 1008 = 126 × 8 + 0. So the remainder is 0 when 31 + 33 + · · · + 32015
is divided by 8.
So the required remainder is 7 + 0 = 7.
18. B.
Let x be the amount of antifreeze to be drained off.
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American Math Competition 10 Practice Test 10
Name C V S
A1 0.4 9 3.6
0.4 x 0.4 x
A2 0.4 9x 0.4(9 x)
B 1.0 x 1.0 x
Mixture 0.7 9 0.7(9)
0.4(9 x) + 1.0 x = 0.7(9) x = 4.5 liters.
19. C.
The four-digit positive integer can be written as 1000a +100b + 10c + d, where a, b, c,
and d are digits.
We have 1000a +100b + 10c + d + a + b + c + d = 2015
1001a +101b + 11c + 2d = 2015.
Case 1: a = 2.
1001a +101b + 11c + 2d = 2015 101b + 11c + 2d = 2015 – 2002 = 13.
So b = 0, c = 1, and d = 1.
The four-digit positive integer is 2011.
Case 2: a = 1.
1001a +101b + 11c + 2d = 2015 101b + 11c + 2d = 2015 – 1001 = 1014.
So b = 9, c = 9, and 2d = 1014 – 909 – 99 = (105 – 99) = 6. So d = 3.
The four-digit positive integer is 1993.
The answer is 2011 + 1993 = 4004.
20. C.
1 1
Pipe A works at a rate of (tank/hour) and pipe B works at a rate of . Working
6 4
1 1
together, they fill the tank at a rate of . Let x be the time needed to fill the tank if
6 4
pipes A and B worked
together.
1 1
x ( ) 1 x = 2 hours.
6 4
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American Math Competition 10 Practice Test 10
Since pipe C will fill the tank in the same number of hours that it will take pipes A and B
1
working together to fill the tank, Ppipe C works at a rate of . When all three pipes work
2
1 1 1
together, they fill the tank at a rate of . Working together for one hour, they fill
6 4 2
1 10 5
1 ( ) of the tank.
1 1 1 12 6
6 4 2
21. A.
Method 1:
Draw DF // BE.
AD DF 6a DF 6b
ADF ATE. DF
AT TE 5a b 5
BE BC 8b BC
CBE CDF.
DF CD DF CD
8b BC 20 BC
6b CD 3 CD
5
20 BD CD 20 BD
1
3 CD 3 CD
20 BD 17 CD 3
1 .
3 CD 3 BD 17
Method 2:
Draw EF // AD.
BE EF 8b EF 8a
BEF BTD. EF
BT TD 7b a 7
AD CD 6a CD
ACD EFC.
EF CF 8a CF
7
CF 4 4 CD DF DF
1
CD 21 21 CD CD
4 DF 17 21
1 CD DF
21 CD 21 17
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American Math Competition 10 Practice Test 10
21
DF
CD 17 3
.
BD 7 DF 17
Method 3:
Draw DF // AC.
AT TE 5a b b
AET DFT. TF
DT TF a TF 5
BE BC 8b BC
BEC BFD.
BF BD BT TF BD
8b BC 8b BD CD
b BD 34 BD
7b b
5 5
20 CD CD 20 3
1 1
17 BD BD 17 17
22. A.
Method 1:
Draw FG AB.
FG // DE // AC.
1 1 1 1 1 1
DE AC FG 1 AC FG
AC
2
3 1 AC
FG .
AC FG 3
1 1 1 AC 1 1
The area of triangle AEF is AE FG AB ( AB AC )
2 2 2 3 6 2
1 1
SABC 360 60
6 6
Method 2:
Draw FG AB.
FG // DE // AC.
FG FG AG
(1)
1 DE AE
AC
2
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American Math Competition 10 Practice Test 10
FG EG
(2)
AC AE
3FG AE AC
(1) + (2): 1 FG .
AC AE 3
The area of triangle AEF is
1 1 1 AC 1 1 1 1
AE FG AB ( AB AC ) SABC 360 60 .
2 2 2 3 6 2 6 6
23. A.
We rotate BAP anti clock wise 60 such that BA and BC overlap. PB will be in the
position of BM, and PA will be in the position of MC. So BM = BP, MC = PA, PBM =
60.
Thus BPM is an equilateral triangle.
Therefore PM = PB 2 3 .
In MCP, PC = 4. MC = PA = 2, PM = 2 3 . We see that
PC 2 PM 2 MC 2 , and PC = 2MC.
So MCP is a right triangle with CMP = 90 and CPM =
30.
We also know that PBM is an equilateral triangle with BPM = 60. Then we know
that BPC is a right triangle with BPC = 90.
Thus BC 2 BP 2 PC 2 (2 3)2 42 28 BC 2 7 .
BC 2 (2 7 ) 2
The area of triangle ABC is 3 3 7 3.
4 4
24. B.
Case 1: The middle card, showing the spade, does not have a 3 printed on the other side.
There are four possible ways for the back of the middle card: 1, 2, 4, or 5.
There are 4 possible ways for the back of the left side card: 3/club, 3/diamond, 3/heart or
3/spade. There are 3 possible ways for the back of the right side card since one is already
taken by the left side card.
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American Math Competition 10 Practice Test 10
There are 4 × 4 3 = 48 different ways for this scenario to occur.
Among them, there are 1 × 4 3 = 12 ways such that the left side card has the spade on
the back, and 3 × 4 1 = 12 ways such that the right side card has the spade on the back.
That is, we have 24 ways that one of the cards showing the number 3 has a spade printed
on the other side.
Case 2: The middle card, showing the spade, has a 3 printed on the other side.
There is only one way for the back of the middle card: 3/spade. There are 3 possible ways
for the back of the left side card: 3/club, 3/diamond, or 3/heart. There are 2 possible ways
for the back of the right side card since two are taken by the left side card and the middle
card.
There are 3 × 1 2 = 6 different ways for this scenario to occur. But none of these
scenarios has a spade printed on the other side of one of the cards showing the number 3.
So of the 48 + 6 = 54 possible scenarios, the probability that one of the cards showing the
number 3 has a spade printed on the other side is 24/54 = 4/9.
25. D.
Let E, H, and F be the centers of circles A, B, and D, respectively, and let G be the point
of tangency of circles B and C.
Connect EH, EC, CH, and EG.
Since circles B and C are congruent, we know that EG is the perpendicular bisector of
HC.
Let x = EF and y = FG.
Since the center of circle D lies on circle A and the
circles have a common point of tangency, the radius of
circle D is 2x, which is the diameter of circle A.
Draw the line l tangent to circles D and B at K.
Connect KF and we know that KF goes through H.
Applying the Pythagorean Theorem to right triangles
EGH and FGH gives:
( x 8)2 ( x y)2 82 (1)
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American Math Competition 10 Practice Test 10
(2 x 8) 2 y 2 82 (2)
(1) can be simplified to y 2 2 xy 16 x 0 (3)
(2) can be simplified to 4 x 2 32 x y 2 0 (4)
(3) + (4): 4 x 2 48x 2 xy 0 2 x 2 24 x xy 0
x(2 x 24 y) 0 .
We know tha x ≠ 0. So we have 2 x 24 y 0 y 24 2 x (5)
Subsituting (5) into (4): 4 x 32 x (24 2 x) 0
2 2
4 x 2 32 x 242 4 x 2 2 2 24 x 0 64 x 242 0 x 9.
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