2016 AMC 10A Real Questions and Analysis

In this article, you’ll find:

• Representative real questions from each module with detailed solutions.
• The complete 2016 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 2016 AMC 10A.
• A module-to-question mapping table highlighting the core concepts tested in each module for the 2016 AMC 10A.

Real Question and Solutions Explained

Algebra Example – Problem 3

Question:

For every dollar Ben spent on bagels, David spent 25 cents less. Ben paid \$12.50 more than David. How much did they spend in the bagel store together?

(A) $37.50 (B) $50.00 (C) $87.50 (D) $90.00 (E) $92.50

Solution:

Let the number of bagels David bought be \( n \).
David spent \( n \) dollars, while Ben spent \( n + 0.25n \) if “25% less” was meant —
but since the problem says 25 cents less, it means a fixed difference per unit, not a percentage.

So Ben paid \$0.25 more per bagel. We are told Ben paid \$12.50 more in total:

\[
0.25n = 12.5 \Rightarrow n = 50.
\]

David’s total = \( 50 \times 0.75 = 37.5 \),
Ben’s total = \( 50 \times 1.00 = 50.0 \).
Together, they spent \( 37.5 + 50.0 = 87.5 \).

Answer:(C)

Common Mistakes:

• Misinterpreting “25 cents less” as “25% less.”
• Forgetting the difference applies per bagel, not per dollar.
• Subtracting \( 0.25 \times \text{total} \) instead of a fixed amount per item.

Number Theory Example – Problem 14

Question:

How many ways are there to write 2016 as the sum of 2’s and 3’s, ignoring order?

(A) 236 (B) 336 (C) 337 (D) 403 (E) 672

Solution:

We want nonnegative integer solutions to

\[
2x + 3y = 2016.
\]

Solve for \( x \):

\[
x = \frac{2016 – 3y}{2}.
\]

For \( x \) to be an integer, \( 2016 – 3y \) must be even.
Since 3 is odd, \( 3y \) is even only when \( y \) is even.

Let \( y = 2k \). Then

\[
2x + 6k = 2016 \Rightarrow x = 1008 – 3k.
\]

The condition \( x \ge 0 \) gives \( k \le 336 \).

Thus \( k = 0, 1, 2, \dots, 336 \), giving 337 solutions.

Answer:(C)

Common Mistakes:

• Forgetting \( y \) must be even for \( x \) to remain integer.
• Counting both \( (x,y) \) and \( (y,x) \) though order is ignored.
• Stopping at \( k = 335 \) instead of including 336.

Geometry Example – Problem 11

Question:

Find the area of the shaded region.

(A) \( \frac{3}{5} \) (B) 5 (C) \( 5\frac{1}{4} \) (D) \( 6\frac{1}{2} \) (E) 8

Solution:

The rectangle’s total area:

\[
8 \times 5 = 40.
\]

Two diagonals drawn from offset corners form two shaded triangles.
Using coordinate geometry (placing rectangle from \( (0,0) \) to \( (8,5) \)):

Top diagonal through \( (1,5) \) and \( (8,1) \):
\[
y = 5 – \frac{4}{7}(x – 1).
\]

Bottom diagonal through \( (0,4) \) and \( (7,0) \):
\[
y = 4 – \frac{4}{7}x.
\]

Vertical difference = 1 at both ends, so shaded total area:

\[
\frac{1}{2} \times (1 + 1) \times \frac{7}{2} = 5.25.
\]

Answer: (C)

Common Mistakes:

• Treating diagonals as intersecting when they’re parallel.
• Computing only one triangle instead of both.
• Confusing base–height alignment in coordinate setup.

Combinatorics Example – Problem 12

Question:

Three distinct integers are selected at random between 1 and 2016, inclusive. What is the probability \( p \) that the product of the three integers is odd?

(A) \( p < \frac{1}{8} \) (B) \( p = \frac{1}{8} \) (C) \( \frac{1}{8} < p < \frac{1}{3} \) (D) \( p = \frac{1}{3} \) (E) \( p > \frac{1}{3} \)

Solution:

For the product to be odd, all 3 numbers must be odd.
There are 1008 odd and 1008 even numbers from 1 to 2016.

\[
p = \frac{\text{number of ways to choose 3 odd}}{\text{number of ways to choose 3 total}}
= \frac{\frac{1008!}{3!(1008-3)!}}{\frac{2016!}{3!(2016-3)!}}.
\]

Approximation:

\[
p \approx \left( \frac{1008}{2016} \right)^3
= \left( \frac{1}{2} \right)^3
= \frac{1}{8}.
\]

Since the odd count is slightly smaller due to integer rounding, actually:

\[
p < \frac{1}{8}.
\]

Answer:(A)

Common Mistakes:

• Forgetting the numbers must be distinct.
• Using \( \frac{1}{2} \) for one number being odd and cubing directly.
• Assuming \( p = \frac{1}{8} \) exactly instead of slightly less.

2016 AMC 10A Answer Key

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

Download AMC 10 Mock Exam & Practice Pack

Download AMC 10 mock exams with full solutions, key AR/HR knowledge reviews, and daily practice sets — perfect for focused and efficient prep.

• Mock Exams: Timed practice with full explanations to build test readiness and review mistakes.
• Knowledge Points: Covers must-know Algebra, Geometry, and Number Theory concepts for AMC 10.
• Daily Practice: Sharpen accuracy and speed with high-frequency AMC-style problems.

*Please enter a valid email address to access the materials and receive future prep updates.

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

2016 AMC 10A Topic Distribution

The 2016 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	1–6, 8–10, 23	Algebraic expressions, ratios, series
Number Theory	14, 22, 25	Divisibility, LCM, GCD
Geometry	11, 15, 16, 19, 21, 24	Triangles, circles, polygons
Combinatorics / Probability	7, 12, 13, 17, 18, 20	Counting and probability

About Think Academy

At Think Academy, a subsidiary of TAL Education Group, we specialize in advanced coaching and rigorous preparation for math competitions, including AMC 10 and AIME. Guided by experienced educators and supported by partnerships with renowned institutions, Think Academy empowers high-school students to reach their full competitive potential in mathematics. Notably, in 2024, our students achieved exceptional recognition, including selections for prestigious programs such as USA(J)MO (17 students), the Mathematical Olympiad Program (3 students), and the International Mathematical Olympiad teams for the U.S. and South Africa. Explore our AMC 10 training programs to help your child excel.

Check our math programs:

• Early Childhood — PreK
• Lower Elementary — Grades K–2
• Upper Elementary — Grades 3–5
• Middle School — Grade 6-10

Want more insights on math learning and parenting? Fill out the form on this page and subscribe to our newsletter for weekly tips and the latest resources.