2001
2001 AMC8 Paper & Solutions Pick
25 questions with solutions, early AMC8 exam, classic problem types.

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12 Sample Questions
40 Minutes
Max Score 25
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Exam Overview
Exam Overview
This exam emphasizes multiple math topics. Overall difficulty is relatively basic.
Difficulty
- EasyQ1-10
- MediumQ11-20
- HardQ21-25
Topics
- Algebra35%
- Geometry30%
- Number Theory15%
- Combinatorics20%
Awards
Distinguished Honor Roll
Distinguished Honor Roll (Top 1%)
18+
Honor Roll
Honor Roll (Top 5%)
14+
Achievement Roll
Grade 6 and below · 15+ points
15+
Sample Problems
2001 AMC8 Sample Problems (12 questions)
Sample reference problems by difficulty — click an option to check your answer
Q1EasyArithmetic
Calculate the value of 500-247.
Steps
500-247=253
Answer: DSubtraction with borrowing
Q3EasyKey concept: fractions
1/3+1/6=?
Steps
2/6+1/6=3/6=1/2
Answer: CFind a common denominator, then add
Q5EasyArea
What is the area of a square with side length 7?
Steps
7²=49
Answer: CArea of a square
Q8EasyApply the relevant mathematical concept
What is the mean of the data 3, 5, 7, 5, 10?
Steps
(3+5+7+5+10)/5=30/5=6
Answer: DSum ÷ count
Q11MediumKey concept: equations
Two numbers have a sum of 30 and a difference of 6. What is their product?
Steps
a+b=30, a-b=6
a=18, b=12
product=216
Answer: DSum-and-difference problem
Q14MediumKey concept: triangles
Two angles of a triangle are 60° and 80°. What is the third angle?
Steps
180-60-80=40°
Answer: DThe angles of a triangle sum to 180°
Q16MediumProbability
A number is chosen at random from 1 to 8. What is the probability of getting an even number?
Steps
There are 4 even numbers: 2, 4, 6, 8
4/8=1/2
Answer: EBasic probability
Q18MediumNumber Theory
What is the greatest common divisor of 24 and 36?
Steps
24=2³×3, 36=2²×3²
GCD=2²×3=12
Answer: DCommon factors
Q21HardArea
A right triangle has a hypotenuse of 17 and one leg of 8. What is its area?
Steps
the other leg=√(289-64)=√225=15
area=1/2×8×15=60
Answer: A8-15-17 Pythagorean triple
Q22HardKey concept: permutations
Five people A, B, C, D, E are arranged in a row, with A required to be in front of B. How many arrangements are there?
Steps
total 5!=120
A in front of B=120/2=60
Answer: BSymmetry
Q24HardNumber Theory
How many zeros are at the end of 1×2×3×...×10?
Steps
Power of 5 in 10!: ⌊10/5⌋=2
2 zeros at the end
Answer: ELegendre's formula
Q25HardCombinatorics
From 6 boys and 5 girls, 4 people are chosen with at least 2 girls. How many ways are there?
Steps
total ways C(11,4)=330
0 girls (all boys) C(6,4)=15
1 girl C(5,1)×C(6,3)=5×20=100
at least 2 girls = 330 − 15 − 100 = 215
Answer: EComplement method: subtract the cases that do not satisfy the condition from the total
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