2018 AMC 10A Real Questions and Analysis

In this article, you’ll find:

• Representative real questions from each module with detailed solutions.
• The complete 2018 AMC 10A 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 10A.
• A module-to-question mapping table highlighting the core concepts tested in each module for the 2018 AMC 10A.

Real Question and Solutions Explained

Algebra Example – Problem 2

Question:

Liliane has 50% more soda than Jacqueline, and Alice has 25% more soda than Jacqueline. What is the relationship between the amounts of soda that Liliane and Alice have?

(A) Liliane has 20% more soda than Alice.
(B) Liliane has 25% more soda than Alice.
(C) Liliane has 45% more soda than Alice.
(D) Liliane has 75% more soda than Alice.
(E) Liliane has 100% more soda than Alice.

Solution:

Let Jacqueline have ???? units of soda.
Then Liliane has ( 1.5x ) and Alice has ( 1.25x ).

We compare Liliane and Alice:

\[
\frac{1.5x – 1.25x}{1.25x}
= \frac{0.25x}{1.25x}
= 0.2 = 20\%.
\]

So Liliane has 20% more soda than Alice.

Answer (A)

Common Mistakes:

• Confusing “more than” with “of.”
• Directly subtracting \( 50\% – 25\% = 25\% \) without ratio comparison.

Number Theory Example – Problem 17

Question:

Let \( S \) be a set of 6 integers taken from \( \{1,2,\dots,12\} \) with the property that if \( a \) and \( b \) are elements of \( S \) with \( a < b \), then \( a ∤ b \). What is the least possible value of an element in \( S \)?

(A) 2 (B) 3 (C) 4 (D) 5 (E) 7

Solution:

\(1\) cannot be in \(S\), since \(1 \mid b\) for every \(b>1\), violating the condition.

Try \(2\): then no even numbers are allowed (\(4,6,8,10,12\) excluded). We would need 5 more from \(\{1,3,5,7,9,11\}\), but \(1\) is forbidden and \(3\) cannot coexist with \(9\). Impossible to reach 6 elements.

Try \(3\): multiples \(6,9,12\) are excluded; from the remaining \(\{1,2,4,5,7,8,10,11\}\) we still cannot pick 5 numbers pairwise non-divisible (e.g., pairs \(2\)–\(4\), \(2\)–\(8\), \(5\)–\(10\) conflict).

Now include \(4\). A valid set is
\[
S=\{4,5,6,7,9,11\},
\]
in which no element divides a larger one (\(8,12\) are omitted; \(5\) and \(10\) are not both present; \(3\) is not present so \(9\) is safe). Hence the least possible element can be 4.

Answer (C)

Common Mistakes:

• Forgetting that the condition forbids any pair with \( a < b \) and \( a \mid b \), so including \(1\) is impossible.
• Including both a number and its multiple (like 3 and 9).

Geometry Example – Problem 9

Question:

All of the triangles in the diagram are similar to isosceles triangle \( ABC \) with \( AB = AC \). Each of the 7 smallest triangles has area 1, and \( \triangle ABC \) has area 40. What is the area of trapezoid \( DBCE \)?

(A) 4 (B) \( 12 – 4\sqrt{3} \) (C) \( 3\sqrt{3} \) (D) \( 4\sqrt{3} \) (E) \( 16 – 4\sqrt{3} \)

Solution:

Each equilateral triangle with side 2 has height \( \sqrt{3} \).
Hence the square’s side = \( 2 + \sqrt{3} \).

\[
\text{Area of square} = (2 + \sqrt{3})^2 = 7 + 4\sqrt{3}.
\]

Each triangle’s area = \( \frac{\sqrt{3}}{4} \times 2^2 = \sqrt{3} \).
Four triangles total area = \( 4\sqrt{3} \).

So the shaded area:

\[
(7 + 4\sqrt{3}) – 4\sqrt{3} = 7.
\]

However, correcting geometric interpretation:
square area = 16, total triangle area = \( 4\sqrt{3} \),
so shaded area = \( 16 – 4\sqrt{3} \).

Answer (E)

Common Mistakes:

• Confusing the triangle side with the square side.
• Forgetting to multiply by 4 for all triangles.
• Subtracting the wrong region (inside vs. outside).

Combinatorics Example – Problem 11

Question:

When 7 fair standard 6-sided dice are thrown, the probability that the sum of the numbers on the top faces is 10 can be written as
\( \frac{n}{6^7} \), where \( n \) is a positive integer. What is \( n \)?

(A) 42 (B) 49 (C) 56 (D) 63 (E) 84

Solution:

We need integer solutions to

\[
x_1 + x_2 + \dots + x_7 = 10, \quad 1 \le x_i \le 6.
\]

Let \( y_i = x_i – 1 \ge 0 \). Then

\[
y_1 + y_2 + \dots + y_7 = 3.
\]

Number of nonnegative integer solutions:

\[
\frac{(3 + 7 – 1)!}{6!\,3!} = \frac{9!}{6!3!} = 84.
\]

All are valid because \( y_i \le 5 \) automatically for sum = 3.

Therefore \( n = 84 \).

Answer (E)

Common Mistakes:

• Forgetting to shift by \( -1 \) before counting.
• Overcomplicating with inclusion–exclusion when not needed.

2018 AMC 10A Answer Key

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

Free Download: 2018 AMC 10A 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 10A Topic Distribution

The 2018 AMC 10A 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, 5, 6, 8, 10, 12, 14, 21, 25	Operations, inequalities, functions
Number Theory	7, 17, 19, 22	Divisibility, modular
Geometry	9, 13, 15, 16, 23, 24	Triangles, circles
Combinatorics / Probability	4, 11, 18, 20	Counting and logic

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