2018 AMC 10B Real Questions and Analysis

In this article, you’ll find:

• Representative real questions from each module with detailed solutions.
• The complete 2018 AMC 10B Answer Key.
• The best resources to prepare effectively for the AMC 10.
• A concise topic distribution chart showing which areas appeared most in the 2018 AMC 10B.
• A module-to-question mapping table highlighting the core concepts tested in each module for the 2018 AMC 10B.

Real Question and Solutions Explained

Algebra Example – Problem 2

Question:

Sam drove 96 miles in 90 minutes. His average speed during the first 30 minutes was 60 mph, and his average speed during the second 30 minutes was 65 mph. What was his average speed, in mph, during the last 30 minutes?

(A) 64 (B) 65 (C) 66 (D) 67 (E) 68

Solution:

First 30 minutes = 0.5 hours → distance = \( 0.5 \times 60 = 30 \) miles.
Second 30 minutes = 0.5 hours → distance = \( 0.5 \times 65 = 32.5 \) miles.
Remaining distance = \( 96 – (30 + 32.5) = 33.5 \) miles.

During the last 0.5 hour:

\[
v = \frac{33.5}{0.5} = 67 \text{ mph.}
\]

Answer:(D)

Common Mistakes:

• Taking the average of the three speeds instead of using distance.
• Mixing minutes and hours.
• Subtracting time instead of distance.

Number Theory Example – Problem 5

Question:

How many subsets of \( \{2,3,4,5,6,7,8,9\} \) contain at least one prime number?

(A) 128 (B) 192 (C) 224 (D) 240 (E) 256

Solution:

Total subsets: \( 2^8 = 256 \).
Non-prime elements: \( \{4,6,8,9\} \) → subsets with no prime numbers = \( 2^4 = 16 \).

Thus, subsets containing at least one prime:

\[
256 – 16 = 240.
\]

Answer:(D)

Common Mistakes:

• Forgetting the empty set counts among “no-prime” subsets.
• Misclassifying 9 as prime.
• Subtracting \( 2^3 \) instead of \( 2^4 \).

Geometry Example – Problem 4

Question:

A three-dimensional rectangular box with dimensions \( X, Y, Z \) has faces whose surface areas are 24, 24, 48, 48, 72, and 72 square units.
What is \( X + Y + Z \)?

(A) 18 (B) 22 (C) 24 (D) 30 (E) 36

Solution:

Pair the equal faces:
\[
XY = 24, \quad XZ = 48, \quad YZ = 72.
\]

Then:
\[
X = \sqrt{\frac{(XY)(XZ)}{YZ}} = \sqrt{\frac{24 \times 48}{72}} = \sqrt{16} = 4,
\]
\[
Y = \sqrt{\frac{(XY)(YZ)}{XZ}} = \sqrt{\frac{24 \times 72}{48}} = \sqrt{36} = 6,
\]
\[
Z = \sqrt{\frac{(XZ)(YZ)}{XY}} = \sqrt{\frac{48 \times 72}{24}} = \sqrt{144} = 12.
\]

So:
\[
X + Y + Z = 4 + 6 + 12 = 22.
\]

Answer: (B)

Common Mistakes:

• Adding face areas instead of working with their products.
• Pairing the wrong faces.
• Forgetting to take square roots after dividing products.

Combinatorics Example – Problem 6

Question:

A box contains chips numbered 1, 2, 3, 4, 5. Chips are drawn uniformly at random without replacement until the sum of the values drawn exceeds 4. What is the probability that 3 draws are required?

(A) \( \frac{1}{15} \) (B) \( \frac{1}{10} \) (C) \( \frac{1}{6} \) (D) \( \frac{1}{5} \) (E) \( \frac{1}{4} \)

Solution:

To require 3 draws:

• After two draws, the sum \( \le 4 \);
• After the third, it becomes \( > 4 \).

Possible first-two-draw pairs with \( \text{sum} \le 4 \):
\( (1,2), (2,1), (1,3), (3,1) \).

For \( (1,2) \): sum = 3 → third chip can be from \( \{3,4,5\} \), 3 choices.
For \( (1,3) \): sum = 4 → third chip can be from \( \{2,4,5\} \), 3 choices.

\[
4 \text{ valid ordered pairs} \times 3 \text{ third choices} = 12 \text{ favorable sequences.}
\]

Total ordered triples (without replacement): \( 5 \times 4 \times 3 = 60. \)

Therefore:
\[
P = \frac{12}{60} = \frac{1}{5}.
\]

Answer:(D)

Common Mistakes:

• Using combinations instead of ordered sequences.
• Allowing repetition of chips.
• Misreading “exceeds 4” as “\( \ge 4 \)”.

2018 AMC 10B Answer Key

Question	Answer
1	A
2	D
3	B
4	B
5	D
6	D
7	D
8	C
9	D
10	E
11	C
12	C
13	C
14	D
15	A
16	E
17	B
18	D
19	E
20	B
21	C
22	C
23	B
24	C
25	C

Free Download: 2018 AMC 10B Problems & Solution

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Last 10 Years AMC 10 Real Questions and Analysis

Think Academy provides in-depth breakdowns of the past decade of AMC 10 exams. Click below to explore:

• Year-by-year topic trend insights and concept distributions

• Real AMC 10 exams from the last 10 years

• Official answer keys

• Representative questions, detailed solutions, and common mistakes

AMC 10A	AMC 10B
2024 AMC 10A	2024 AMC10B
2023 AMC 10A	2023 AMC10B
2022 AMC 10A	2022 AMC10B
2021 AMC 10A	2021 AMC10B
2020 AMC 10A	2020 AMC10B
2019 AMC 10A	2019 AMC10B
2018 AMC 10A	2018 AMC10B
2017 AMC 10A	2017 AMC10B
2016 AMC 10A	2016 AMC10B

2018 AMC 10B Topic Distribution

The 2018 AMC 10B featured 25 questions to be completed in 75 minutes, emphasizing advanced problem-solving and proof-based reasoning skills.

Learn more about AMC 10 Format and Scoring Here: AMC 10 FAQ and Resources: Your Ultimate Guide

Detailed Module Analysis

Module	Question Numbers	What It Tests (Brief)
Algebra	1, 2, 3, 8, 10, 14, 20, 22, 25	Linear, quadratic, inequalities
Number Theory	5, 11, 13, 16, 19, 21, 23	Primes, divisibility
Geometry	4, 7, 12, 15, 17, 24	Solids, polygons
Combinatorics / Probability	6, 9, 18	Counting, probability

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