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2018

2018 AMC8 Paper & Solutions Pick

12 selected sample solutions, geometry covers Pythagorean theorem and area transforms, algebra focuses on inequalities and functions.

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12 Sample Questions
40 Minutes
Max Score 25
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Exam Overview

Exam Overview

This exam emphasizes geometry, area calculations. Overall difficulty is moderate.

DDifficulty

  • EasyQ1-10
  • MediumQ11-20
  • HardQ21-25

TTopics

  • Algebra35%
  • Geometry35%
  • Number Theory15%
  • Combinatorics15%

AAwards

Distinguished Honor Roll
Distinguished Honor Roll (Top 1%)
22+
Honor Roll
Honor Roll (Top 5%)
18+
Achievement Roll
Grade 6 and below · 15+ points
15+
Sample Problems

2018 AMC8 Sample Problems (12 questions)

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

Q1EasyArithmetic
Calculate the value of 125 × 8 ÷ 10.
A) 80
B) 90
C) 100
D) 110
E) 120

Steps

125 × 8 = 1000
1000 ÷ 10 = 100
Answer: CRounding trick: 125 × 8 = 1000
Q4EasyApply the relevant mathematical concept
How many integers satisfy x + 3 > 7 and x < 8?
A) 2
B) 3
C) 4
D) 5
E) 6

Steps

1x > 4 and x < 8
Integers: 5, 6, 7 → 3 of them
Answer: BSolve the system of inequalities and take the intersection
Q6EasyArea
A triangle has a base of 10 cm and a height of 8 cm. What is its area?
A) 30
B) 36
C) 48
D) 80
E) 40

Steps

Area = 1/2 × 10 × 8 = 40
Answer: EArea of a triangle = base × height ÷ 2
Q9EasyKey concept: factors
How many positive divisors does 36 have?
A) 6
B) 7
C) 9
D) 8
E) 10

Steps

36 = 2² × 3²
Number of divisors = (2+1)(2+1) = 9
Answer: CFormula for the number of divisors
Q11MediumPythagorean Thm.
A right triangle has legs of 5 and 12. What is the length of the hypotenuse?
A) 11
B) 12
C) 14
D) 13
E) 15

Steps

Hypotenuse = √(25+144) = √169 = 13
Answer: D5-12-13 is a classic Pythagorean triple
Q13MediumArea
Triangle ABC has area 24. D is the midpoint of BC, and E is a trisection point of AC (the one nearer to C). Find the area of triangle ADE.
A) 4
B) 8
C) 6
D) 10
E) 12

Steps

S(ADC) = S(ABC)/2 = 12
AE/AC = 2/3
S(ADE) = 12 × 2/3 = 8
Answer: BFor triangles sharing a vertex, the ratio of their areas equals the ratio of their bases
Q15MediumProbability
A coin and a die are tossed. What is the probability of getting heads and an even number?
A) 1/3
B) 3/8
C) 1/2
D) 2/3
E) 1/4

Steps

P(heads) = 1/2
P(even) = 3/6 = 1/2
Independent events: 1/2 × 1/2 = 1/4
Answer: EThe probability that independent events both occur = the product of their individual probabilities
Q17MediumKey concept: equations
Students in a class line up. With 6 per row there are 2 left over, and with 8 per row there are 4 short. What is the smallest possible number of students?
A) 14
B) 26
C) 32
D) 38
E) 20

Steps

n ≡ 2 (mod 6) and n ≡ 4 (mod 8)
That is, n ≡ 2 (mod 6) → 2, 8, 14, 20, 26...
n ≡ 4 (mod 8) → 4, 12, 20, 28...
Smallest common value = 20
Answer: ESolving a system of congruences
Q21HardInclusion-exclusion
How many numbers from 1 to 100 are not multiples of 2, 3, or 5?
A) 24
B) 28
C) 30
D) 32
E) 26

Steps

|A∪B∪C| = 50+33+20-16-10-6+3 = 74
None of them = 100 - 74 = 26
Answer: EInclusion-exclusion principle for three sets
Q22HardGeometry
A rhombus has diagonals of 16 and 12. Find the perimeter of the rhombus.
A) 36
B) 38
C) 40
D) 42
E) 44

Steps

The diagonals are perpendicular bisectors of each other
Half-diagonals are 8 and 6
Side length = √(64+36) = 10
Perimeter = 40
Answer: CThe diagonals of a rhombus are perpendicular bisectors of each other
Q24HardKey concept: permutations
Six people line up in a row, with A not in the first position and B not in the last position. How many arrangements are there?
A) 504
B) 360
C) 400
D) 432
E) 480

Steps

Total arrangements 6! = 720
A in the first position: 5! = 120
B in the last position: 5! = 120
A first and B last: 4! = 24
Inclusion-exclusion: 720 - 120 - 120 + 24 = 504
Answer: AUse inclusion-exclusion for "not in" problems in permutations
Q25HardNumber Theory
Find the number of primes p such that p² + 2 is also prime.
A) 2
B) 3
C) 4
D) 1
E) Infinitely many

Steps

p=2: 4+2=6 is not prime
p=3: 9+2=11 is prime ✓
p≥5: p² mod 3 = 1, so p²+2 mod 3 = 0; it is divisible by 3 and greater than 3, hence not prime
Only p=3
Answer: DAnalysis modulo 3 is an important technique in number theory

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