Paper Set

2010

2010 AMC10 Paper & Solutions Pick Paper A

25 questions with solutions, easier difficulty, good for foundational practice.

Sample Problems Download PDF

Scan to get free 2010 PDF + solutions

25 Questions

40 Minutes

Max Score 25

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Exam Overview

This exam emphasizes multiple math topics. Overall difficulty is relatively basic.

DDifficulty

EasyQ1-10
MediumQ11-20
HardQ21-25

TTopics

Algebra35%

Geometry30%

Number Theory15%

Combinatorics20%

AAwards

AIME Qualification

AIME Qualification (Top 2.5%)

113+

Honor Roll

Honor Roll (Top 5%)

95+

Achievement Roll

Grade 6 and below · 15+ points

89+

Sample Problems

2010 AMC10 Sample Problems (12 questions)

Sample reference problems by difficulty — click an option to check your answer

Q1EasyPolynomial

If f(x) = 2x² - 3x + 2, find f(4).

A) 22

B) 16

C) 19

D) 25

E) 28

Steps

1f(4) = 2×4² - 3×4 + 2

2= 2×16 - 12 + 2 = 22

Answer: ASubstitute, then exponentiate, multiply/divide, add/subtract

Q3EasyArithmetic Seq

Arithmetic sequence: a₁=5, d=2, find the 14-th term.

A) 23

B) 31

C) 27

D) 35

E) 39

Steps

1aₙ = a₁ + (n-1)d

2a_14 = 5 + 13×2 = 31

Answer: BGeneral term: aₙ=a₁+(n-1)d

Q5EasyExponent

If 3ˣ = 243, find x.

A) 3

B) 4

C) 5

D) 6

E) 7

Steps

1243 = 3^5

2so x = 5

Answer: CConvert to same base, compare exponents

Q7EasyVieta's

Sum of roots of x² - 5x + 6 = 0?

A) 1

B) 3

C) 7

D) 5

E) 9

Steps

1Vieta's:sum of roots = -(-5)/1 = 5

2product of roots = 6

Answer: DFor x²-px+q=0, sum of roots = p

Q10MediumDivisibility

How many integers from 1 to 40 are divisible by 4?

A) 6

B) 8

C) 12

D) 14

E) 10

Steps

1⌊40 / 4⌋ = 10

Answer: ECount divisible by k: ⌊n/k⌋

Q12MediumDiagonals

How many diagonals does a regular 11-gon have?

A) 44

B) 38

C) 41

D) 47

E) 50

Steps

1diagonals = n(n-3)/2

2= 11×8/2 = 44

Answer: An-gon diagonals: n(n-3)/2

Q15MediumCombination

Compute C(6, 3).

A) 14

B) 20

C) 17

D) 23

E) 26

Steps

1C(n,3) = n(n-1)(n-2)/6

2= 6×5×4/6 = 20

Answer: BCombination: C(n,k)=n!/(k!(n-k)!)

Q17MediumPythagorean

Right triangle legs 3 and 4, find the hypotenuse.

A) 1

B) 3

C) 5

D) 7

E) 9

Steps

1c² = 3² + 4² = 9 + 16 = 25

2c = √25 = 5

Answer: CPythagorean theorem a²+b²=c²

Q20HardModular Arith

Find the remainder of 5^5 divided by 7.

A) 1

B) 2

C) 4

D) 3

E) 5

Steps

1compute 5^5 mod 7

2simplify step by step (modular arithmetic)

3=3

Answer: DModular arithmetic simplifies large exponents

Q22HardCompleting Sq

Find the minimum of f(x) = x² - 8x + 20.

A) 0

B) 2

C) 6

D) 8

E) 4

Steps

1complete the square:f(x) = (x - 4)² + 4

2when x = 4 , min at 4

Answer: ECompleting the square for quadratic extrema

Q24HardInclusion-Excl

5 distinct balls into 3 distinct boxes, each ≥1 ball, how many ways?

A) 150

B) 118

C) 134

D) 166

E) 182

Steps

1total 3^5 = 243

2subtract empty boxes: -C(3,1)×2^5 = -96

3add back 2 empty boxes: +C(3,2)×1 = +3

4total 243 - 96 + 3 = 150

Answer: AInclusion-exclusion handles 'at least'

Q25HardLaw of Cosines

In △ABC, a=6, b=7, cos C=1/2, find c².

A) 33

B) 43

C) 38

D) 48

E) 53

Steps

1Law of cosines c² = a² + b² - 2ab·cos C

2= 36 + 49 - 42 = 43

Answer: BLaw of cosines is key for solving triangles

Q1EasyPolynomial

If f(x) = 2x² - 5x + 2, find f(4).

A) 14

B) 10

C) 12

D) 16

E) 18

Steps

1f(4) = 2×4² - 5×4 + 2

2= 2×16 - 20 + 2 = 14

Answer: ASubstitute, then exponentiate, multiply/divide, add/subtract

Q3EasyArithmetic Seq

Arithmetic sequence: a₁=7, d=2, find the 14-th term.

A) 25

B) 33

C) 29

D) 37

E) 41

Steps

1aₙ = a₁ + (n-1)d

2a_14 = 7 + 13×2 = 33

Answer: BGeneral term: aₙ=a₁+(n-1)d

Q5EasyExponent

If 3ˣ = 243, find x.

A) 3

B) 4

C) 5

D) 6

E) 7

Steps

1243 = 3^5

2so x = 5

Answer: CConvert to same base, compare exponents

Q7EasyVieta's

Sum of roots of x² - 7x + 6 = 0?

A) 3

B) 5

C) 9

D) 7

E) 11

Steps

1Vieta's:sum of roots = -(-7)/1 = 7

2product of roots = 6

Answer: DFor x²-px+q=0, sum of roots = p

Q10MediumDivisibility

How many integers from 1 to 42 are divisible by 4?

A) 6

B) 8

C) 12

D) 14

E) 10

Steps

1⌊42 / 4⌋ = 10

Answer: ECount divisible by k: ⌊n/k⌋

Q12MediumDiagonals

How many diagonals does a regular 13-gon have?

A) 65

B) 59

C) 62

D) 68

E) 71

Steps

1diagonals = n(n-3)/2

2= 13×10/2 = 65

Answer: An-gon diagonals: n(n-3)/2

Q15MediumCombination

Compute C(8, 3).

A) 44

B) 56

C) 50

D) 62

E) 68

Steps

1C(n,3) = n(n-1)(n-2)/6

2= 8×7×6/6 = 56

Answer: BCombination: C(n,k)=n!/(k!(n-k)!)

Q17MediumPythagorean

Right triangle legs 6 and 8, find the hypotenuse.

A) 6

B) 8

C) 10

D) 12

E) 14

Steps

1c² = 6² + 8² = 36 + 64 = 100

2c = √100 = 10

Answer: CPythagorean theorem a²+b²=c²

Q20HardModular Arith

Find the remainder of 5^5 divided by 7.

A) 1

B) 2

C) 4

D) 3

E) 5

Steps

1compute 5^5 mod 7

2simplify step by step (modular arithmetic)

3=3

Answer: DModular arithmetic simplifies large exponents

Q22HardCompleting Sq

Find the minimum of f(x) = x² - 8x + 22.

A) 2

B) 4

C) 8

D) 10

E) 6

Steps

1complete the square:f(x) = (x - 4)² + 6

2when x = 4 , min at 6

Answer: ECompleting the square for quadratic extrema

Q24HardInclusion-Excl

5 distinct balls into 3 distinct boxes, each ≥1 ball, how many ways?

A) 150

B) 118

C) 134

D) 166

E) 182

Steps

1total 3^5 = 243

2subtract empty boxes: -C(3,1)×2^5 = -96

3add back 2 empty boxes: +C(3,2)×1 = +3

4total 243 - 96 + 3 = 150

Answer: AInclusion-exclusion handles 'at least'

Q25HardLaw of Cosines

In △ABC, a=8, b=7, cos C=1/2, find c².

A) 45

B) 57

C) 51

D) 63

E) 69

Steps

1Law of cosines c² = a² + b² - 2ab·cos C

2= 64 + 49 - 56 = 57

Answer: BLaw of cosines is key for solving triangles

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2010 AMC10 Sample Problems (12 questions)

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