Paper Set

2016

2016 AMC10 Paper & Solutions Pick Paper A

25 questions with solutions, balanced algebra and geometry, inclusion-exclusion in combinatorics. Moderate difficulty.

Sample Problems Download PDF

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25 Questions

40 Minutes

Max Score 25

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Exam Overview

This exam emphasizes algebra, geometry, combinatorics. Overall difficulty is moderate.

DDifficulty

EasyQ1-10
MediumQ11-20
HardQ21-25

TTopics

Algebra35%

Geometry30%

Number Theory15%

Combinatorics20%

AAwards

AIME Qualification

AIME Qualification (Top 2.5%)

106+

Honor Roll

Honor Roll (Top 5%)

93+

Achievement Roll

Grade 6 and below · 15+ points

89+

Sample Problems

2016 AMC10 Sample Problems (12 questions)

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

Q1EasyPolynomial

If f(x) = 2x² - 4x + 2, find f(2).

A) 0

B) 2

C) 4

D) 6

E) 8

Steps

1f(2) = 2×2² - 4×2 + 2

2= 2×4 - 8 + 2 = 2

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

Q3EasyArithmetic Seq

Arithmetic sequence: a₁=3, d=3, find the 14-th term.

A) 32

B) 37

C) 42

D) 47

E) 52

Steps

1aₙ = a₁ + (n-1)d

2a_14 = 3 + 13×3 = 42

Answer: CGeneral term: aₙ=a₁+(n-1)d

Q5EasyExponent

If 3ˣ = 27, find x.

A) 1

B) 2

C) 4

D) 3

E) 5

Steps

127 = 3^3

2so x = 3

Answer: DConvert to same base, compare exponents

Q7EasyVieta's

Sum of roots of x² - 6x + 4 = 0?

A) 2

B) 4

C) 8

D) 10

E) 6

Steps

1Vieta's:sum of roots = -(-6)/1 = 6

2product of roots = 4

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

Q10MediumDivisibility

How many integers from 1 to 46 are divisible by 4?

A) 11

B) 7

C) 9

D) 13

E) 15

Steps

1⌊46 / 4⌋ = 11

Answer: ACount divisible by k: ⌊n/k⌋

Q12MediumDiagonals

How many diagonals does a regular 11-gon have?

A) 38

B) 44

C) 41

D) 47

E) 50

Steps

1diagonals = n(n-3)/2

2= 11×8/2 = 44

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

Q15MediumCombination

Compute C(7, 3).

A) 27

B) 31

C) 35

D) 39

E) 43

Steps

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

2= 7×6×5/6 = 35

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

Q17MediumPythagorean

Right triangle legs 5 and 12, find the hypotenuse.

A) 9

B) 11

C) 15

D) 13

E) 17

Steps

1c² = 5² + 12² = 25 + 144 = 169

2c = √169 = 13

Answer: DPythagorean theorem a²+b²=c²

Q20HardModular Arith

Find the remainder of 3^5 divided by 5.

A) 1

B) 2

C) 4

D) 5

E) 3

Steps

1compute 3^5 mod 5

2simplify step by step (modular arithmetic)

3=3

Answer: EModular arithmetic simplifies large exponents

Q22HardCompleting Sq

Find the minimum of f(x) = x² - 4x + 8.

A) 4

B) 0

C) 2

D) 6

E) 8

Steps

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

2when x = 2 , min at 4

Answer: ACompleting the square for quadratic extrema

Q24HardInclusion-Excl

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

A) 118

B) 150

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: BInclusion-exclusion handles 'at least'

Q25HardLaw of Cosines

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

A) 33

B) 38

C) 43

D) 48

E) 53

Steps

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

2= 36 + 49 - 42 = 43

Answer: CLaw of cosines is key for solving triangles

Q1EasyPolynomial

If f(x) = 2x² - 6x + 2, find f(2).

A) -6

B) -2

C) -4

D) 0

E) 2

Steps

1f(2) = 2×2² - 6×2 + 2

2= 2×4 - 12 + 2 = -2

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

Q3EasyArithmetic Seq

Arithmetic sequence: a₁=5, d=3, find the 14-th term.

A) 34

B) 39

C) 44

D) 49

E) 54

Steps

1aₙ = a₁ + (n-1)d

2a_14 = 5 + 13×3 = 44

Answer: CGeneral term: aₙ=a₁+(n-1)d

Q5EasyExponent

If 3ˣ = 27, find x.

A) 1

B) 2

C) 4

D) 3

E) 5

Steps

127 = 3^3

2so x = 3

Answer: DConvert to same base, compare exponents

Q7EasyVieta's

Sum of roots of x² - 8x + 4 = 0?

A) 4

B) 6

C) 10

D) 12

E) 8

Steps

1Vieta's:sum of roots = -(-8)/1 = 8

2product of roots = 4

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

Q10MediumDivisibility

How many integers from 1 to 48 are divisible by 4?

A) 12

B) 8

C) 10

D) 14

E) 16

Steps

1⌊48 / 4⌋ = 12

Answer: ACount divisible by k: ⌊n/k⌋

Q12MediumDiagonals

How many diagonals does a regular 13-gon have?

A) 59

B) 65

C) 62

D) 68

E) 71

Steps

1diagonals = n(n-3)/2

2= 13×10/2 = 65

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

Q15MediumCombination

Compute C(9, 3).

A) 66

B) 75

C) 84

D) 93

E) 102

Steps

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

2= 9×8×7/6 = 84

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

Q17MediumPythagorean

Right triangle legs 8 and 15, find the hypotenuse.

A) 13

B) 15

C) 19

D) 17

E) 21

Steps

1c² = 8² + 15² = 64 + 225 = 289

2c = √289 = 17

Answer: DPythagorean theorem a²+b²=c²

Q20HardModular Arith

Find the remainder of 3^5 divided by 5.

A) 1

B) 2

C) 4

D) 5

E) 3

Steps

1compute 3^5 mod 5

2simplify step by step (modular arithmetic)

3=3

Answer: EModular arithmetic simplifies large exponents

Q22HardCompleting Sq

Find the minimum of f(x) = x² - 4x + 10.

A) 6

B) 2

C) 4

D) 8

E) 10

Steps

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

2when x = 2 , min at 6

Answer: ACompleting the square for quadratic extrema

Q24HardInclusion-Excl

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

A) 118

B) 150

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: BInclusion-exclusion handles 'at least'

Q25HardLaw of Cosines

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

A) 45

B) 51

C) 57

D) 63

E) 69

Steps

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

2= 64 + 49 - 56 = 57

Answer: CLaw of cosines is key for solving triangles

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2016 AMC10 Sample Problems (12 questions)

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