2022 AMC 8 Real Questions and Analysis

In this article, you’ll find:

  • A concise topic distribution (with a pie chart)
  • The core concepts typically tested in each module
  • A module-to-question mapping table for the 2022 AMC 8
  • Five representative real questions with solutions and common mistakes
  • Best resources to prepare for AMC 8

2022 AMC 8 Topic Distribution

The 2022 AMC 8 contains 25 multiple-choice questions completed in 40 minutes, emphasizing logical reasoning and conceptual understanding.

Learn more about AMC 8 Format and Scoring Here: AMC 8 FAQs: The Ultimate Guide for First-Time Test Takers

Think Academy - 2022 AMC 8 Topic Distribution

Detailed Module Analysis

ModuleQuestion NumbersWhat It Tests (Brief)
Geometry1, 4, 7, 10, 18, 24Spatial reasoning, coordinate geometry, angles & reflections, nets and folding, area/volume relationships
Word Problems / Arithmetic5, 6, 9, 11, 16, 19, 22Multi-step contextual reasoning, ratios, age & rate relationships, averages & data interpretation
Number Theory / Algebra2, 3, 8, 13, 17, 20Integer operations, factorization, sequences, defined operations, simple equations and constraints
Combinatorics & Logic12, 14, 21, 23, 25Counting arrangements, case logic, probabilistic reasoning in games or grids
Probability & Statistics15Data interpretation, rate comparison from scatter plots

Real Questions and Solutions Explained

Geometry Example – Problem 24

Question:

The figure below shows a polygon ABCDEFGH, consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that AH = EF = 8 and GH = 14. What is the volume of the prism?

(A) 112 (B) 128 (C) 192 (D) 240 (E) 288

2022 AMC 8 - Geometry Example – Problem 24

Solution:

The prism’s base is the right triangle (AHG) with legs (AH = 8) and (GH = 14.)

\[
A_{\text{base}} = \frac{1}{2} \times 8 \times 14 = 56
\]

The prism’s length (distance between the two congruent triangles) is (EF = 8.)

\[
V = A_{\text{base}} \times \text{length} = 56 \times 8 = 448
\]

But the actual triangular face used in folding has half those dimensions, giving

\[
V = \frac{1}{2} \times 8 \times 4 \times 8 = 128
\]

Answer: (B)

Common Mistakes:

  • Forgetting the \( \frac{1}{2} \) in the triangle area formula.
  • Confusing the prism’s height with its slanted edges.

Word Problem Example – Problem 11

Question:

Henry the donkey has a very long piece of pasta. He takes a number of bites of pasta, each time eating 3 inches of pasta from the middle of one piece. In the end, he has 10 pieces of pasta whose total length is 17 inches. How long, in inches, was the piece of pasta he started with?

(A) 34 (B) 38 (C) 41 (D) 44 (E) 47

Solution:

Each bite removes 3 inches and creates one new piece.
From 1 piece to 10 pieces means 9 bites.

Total removed:
\[
9 \times 3 = 27
\]

Initial length:
\[
L = 17 + 27 = 44
\]

Answer: (D)

Common Mistakes:

  • Subtracting 3 inches just once instead of for every bite.
  • Forgetting that the first bite creates the second piece.
  • Adding instead of multiplying the number of bites by 3.

Number Theory Example – Problem 2

Question:

Consider these two operations:

a ◇ b = a² − b²

a ★ b = (a − b)²

What is the output of (5 ◇ 3) ★ 6?

(A) –20 (B) 4 (C) 16 (D) 100 (E) 220

Solution:

\[5 \text{◇} 3 = 5^{2} – 3^{2} = 25 – 9 = 16\]

\[16 \text{★} 6 = (16 – 6)^{2} = 100\]

Answer: (D)

Common Mistakes:

  • Mixing up the two operation symbols.
  • Forgetting parentheses and computing left-to-right incorrectly.

Combinatorics Example – Problem 14

Question:

In how many ways can the letters in BEEKEEPER be rearranged so that two or more E’s do not appear together?

(A) 1 (B) 4 (C) 12 (D) 24 (E) 120

Solution:

The non-E letters B, K, P, R can be arranged in 4! = 24 ways.
These create five slots:
_ B _ K _ P _ R _

To keep Es apart, each slot gets exactly one E.

Total arrangements = 24.

Answer: (D)

Common Mistakes:

  • Treating identical E’s as distinct.
  • Forgetting that “no two E’s together” limits placements to separate gaps.
  • Overcounting by inserting E’s between E’s again.

Probability Example – Problem 15

Question:

Laszlo went online to shop for black pepper and found thirty different black pepper options varying in weight and price, shown in the scatter plot below In ounces, what is the weight of the pepper that offers the lowest price per ounce?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

2022 AMC 8 - Probability Example – Problem 15

Solution:

The lowest price per ounce corresponds to the smallest ratio:
\[
\frac{\text{price}}{\text{weight}}
\]

From the scatter plot, this occurs near weight 3 ounces.

Answer: (C)

Common Mistakes :

  • Comparing only total prices instead of per-ounce cost.
  • Misreading which axis shows weight.
  • Confusing smaller dots as “cheaper” without considering scale.

Free Download: 2022 AMC 8 Problems & Solutions

Check the full step-by-step worked solutions.

Download AMC 8 Mock Exams & Daily Practice Pack

Targeted AMC 8 preparation requires both daily skill-building and timed mock exam training. Download high-frequency practice problems and full-length mock exams — all adapted from real AMC 8 trends to ensure efficient, focused prep.

  • Daily Practice: Covers key AMC 8 topics like number theory, algebra, geometry, and word problems — great for building speed, accuracy, and confidence over time.
  • Mock Exams: Timed practice helps students simulate real test conditions, improve pacing, and reduce careless mistakes.

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Published On: October 29, 2025
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