2025

2025 AMC8 Paper & Solutions Hot

12 selected sample solutions, algebra and geometry each ~40%. Final questions involve recursive sequences and area dissection.

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12 Sample Questions

40 Minutes

Max Score 25

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Exam Overview

This exam emphasizes multiple math topics. Overall difficulty is moderate.

DDifficulty

EasyQ1-10
MediumQ11-20
HardQ21-25

TTopics

Algebra40%

Geometry40%

Number Theory10%

Combinatorics10%

AAwards

Distinguished Honor Roll

Distinguished Honor Roll (Top 1%)

23+

Honor Roll

Honor Roll (Top 5%)

19+

Achievement Roll

Grade 6 and below · 15+ points

15+

Sample Problems

2025 AMC8 Sample Problems (12 questions)

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

Q1EasySpeed Calc

Calculate the value of 997 + 998 + 999 + 1000 + 1001 + 1002 + 1003.

A) 6998

B) 7000

C) 6999

D) 7001

E) 7002

Steps

Center the terms around 1000: (1000-3)+(1000-2)+(1000-1)+1000+(1000+1)+(1000+2)+(1000+3)

= 7 × 1000 = 7000

Answer: BThe sum of a symmetric sequence = middle term × number of terms

Q3EasyKey concept: fractions

A rope is cut by 1/3 of its total length, then another 2 meters are cut off, leaving 6 meters. What was the original length of the rope in meters?

A) 10

B) 14

C) 12

D) 16

E) 18

Steps

Let the original length be x meters

x - x/3 - 2 = 6

2x/3 = 8 → x = 12

Answer: CSetting up an equation is the general method for fraction word problems

Q5EasyApply the relevant mathematical concept

In triangle ABC, ∠A = 50° and ∠B = 65°. Find the measure of ∠C.

A) 55°

B) 60°

C) 65°

D) 70°

E) 75°

Steps

The sum of the interior angles of a triangle = 180°

∠C = 180° - 50° - 65° = 65°

Answer: CThe interior angles of a triangle always add up to 180°

Q7EasyAverage

A class has 12 students with an average age of 13 years. After a 14-year-old student joins, what is the new average age?

A) 13

B) 13 1/12

C) 13 1/6

D) 13 1/13

E) 13.5

Steps

Original total age = 12 × 13 = 156

Total age after joining = 170, number of students = 13

New average = 170/13 = 13 and 1/13

Answer: DAverage = sum ÷ count

Q11MediumKey concept: equations

Solve the equation 3(x - 2) + 5 = 2(x + 4).

A) x = 5

B) x = 7

C) x = 9

D) x = 11

E) x = 13

Steps

Expand: 3x - 6 + 5 = 2x + 8

3x - 1 = 2x + 8

x = 9

Answer: CFirst remove the parentheses, then move terms and combine

Q13MediumApply the relevant mathematical concept

Find the remainder when 2^20 is divided by 7.

A) 1

B) 2

C) 3

D) 5

E) 4

Steps

The powers of 2 mod 7 are periodic: 2¹≡2, 2²≡4, 2³≡1 (period 3)

20 = 3×6 + 2, remainder 2

2²⁰ ≡ 2² = 4 (mod 7)

Answer: EModular exponentiation is periodic

Q15MediumArea

In rectangle ABCD, AB=8 and BC=6. E is the midpoint of AB, and F is the midpoint of CD. Find the area of quadrilateral AECF.

A) 18

B) 20

C) 24

D) 28

E) 30

Steps

AECF is a parallelogram

Base AE = 4, height = BC = 6

Area = 4 × 6 = 24

Answer: CConnecting the midpoints in a rectangle forms a parallelogram

Q18MediumProbability

A bag contains 3 red, 2 blue, and 1 green ball, for a total of 6 balls. Two balls are drawn. What is the probability of drawing exactly one red and one blue?

A) 1/5

B) 1/3

C) 1/2

D) 2/5

E) 3/5

Steps

Total number of ways C(6,2) = 15

One red and one blue: C(3,1)×C(2,1) = 6

Probability = 6/15 = 2/5

Answer: DProbability with combinations = favorable ways ÷ total ways

Q21HardArea Dissection

Square ABCD has side length 10. E and F lie on AB and CD respectively, with AE=3 and DF=4. Find the area of quadrilateral AEFD.

A) 30

B) 32

C) 38

D) 40

E) 35

Steps

AEFD is a trapezoid (AE and DF are unequal but both lie on the parallel sides)

AE=3 lies on AB, DF=4 lies on CD

Top base AE=3, bottom base DF=4, height=AD=10

Area = 1/2 × (3+4) × 10 = 35

Answer: EArea of a trapezoid = 1/2 × (top base + bottom base) × height

Q23HardRecursive Seq.

A sequence is defined by a₁=1, a₂=1, aₙ₊₂ = aₙ₊₁ + aₙ. Find the remainder when a₁₀ is divided by 3.

A) 0

B) 1

C) 2

D) 3

E) Cannot be determined

Steps

Compute each term mod 3: 1,1,2,0,2,2,1,0,1,1

a₁₀ mod 3 = 1

Answer: BThe Fibonacci sequence is periodic under modular arithmetic

Q24HardNumber Theory

Find the smallest positive integer n such that n! is divisible by 100.

A) 5

B) 7

C) 9

D) 10

E) 15

Steps

100 = 4 × 25 = 2² × 5²

The power of 5 in n!: ⌊n/5⌋ + ⌊n/25⌋ + ...

We need the power of 5 to be ≥ 2: when n=10, ⌊10/5⌋=2 ✓

When n=10, the power of 2 is far ≥ 2 ✓

Answer: DUse Legendre's formula for the power of a prime factor in a factorial

Q25HardCombinatorics

Five distinct balls are placed into 3 distinct boxes, with each box containing at least one ball. How many ways are there to do this?

A) 60

B) 90

C) 120

D) 180

E) 150

Steps

Use the inclusion-exclusion principle: total arrangements - arrangements with an empty box

Total arrangements = 3⁵ = 243

At least one empty box: C(3,1)×2⁵ - C(3,2)×1⁵ = 96 - 3 = 93

No empty box = 243 - 93 = 150

Answer: EThe inclusion-exclusion principle handles "at least" problems

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2025 AMC8 Paper & Solutions Hot

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2025 AMC8 Sample Problems (12 questions)

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