2020 AMC 10B Real Questions and Analysis

In this article, you’ll find:

• Representative real questions from each module with detailed solutions.
• The complete 2020 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 2020 AMC 10B.
• A module-to-question mapping table highlighting the core concepts tested in each module for the 2020 AMC 10B.

Real Question and Solutions Explained

Algebra Example – Problem 3

Question:

The ratio of 𝑤 to 𝑥 is 4 : 3, the ratio of 𝑦 to 𝑧 is 3 : 2, and the ratio of 𝑧 to 𝑥 is 1 : 6. What is the ratio of 𝑤 to 𝑦?

(A) 4 : 3 (B) 3 : 2 (C) 8 : 3 (D) 4 : 1 (E) 16 : 3

Solution:

Let \( 𝑥 = 6𝑘 \) and \( 𝑧 = 𝑘 \) from \( 𝑧 : 𝑥 = 1 : 6. \)
From \( 𝑦 : 𝑧 = 3 : 2, \) we get

\[
𝑦 = \frac{3}{2}𝑘.
\]

From \( 𝑤 : 𝑥 = 4 : 3, \) we get

\[
𝑤 = \frac{4}{3} \cdot 6𝑘 = 8𝑘.
\]

Then

\[
𝑤 : 𝑦 = 8𝑘 : \frac{3}{2}𝑘 = 8 : 1.5 = 16 : 3.
\]

Answer:(E)

Common Mistakes:

• Mixing which variable is in the numerator of each ratio.
• Not linking all three ratios through a common parameter (𝑘).
• Reducing \( 8 : 1.5 \) incorrectly.

Number Theory Example – Problem 4

Question:

The acute angles of a right triangle are \( 𝑎^\circ \) and \( 𝑏^\circ, \) where \( 𝑎 > 𝑏 \) and both \( 𝑎 \) and \( 𝑏 \) are prime numbers. What is the least possible value of \( 𝑏? \)

(A) 2 (B) 3 (C) 5 (D) 7 (E) 11

Solution:

In a right triangle, \( 𝑎 + 𝑏 = 90. \)
Since 90 is even, either one angle is \( 2^\circ \) and the other \( 88^\circ \) (not prime), or both are odd primes.

To minimize \( 𝑏, \) maximize \( 𝑎 < 90 \) while keeping \( 𝑏 = 90 – 𝑎 \) prime.

Testing large primes below 90:
\( 𝑎 = 89 \Rightarrow 𝑏 = 1 \) (not prime)
\( 𝑎 = 83 \Rightarrow 𝑏 = 7 \) (prime)

Thus, the smallest possible \( 𝑏 \) is 7.

Answer:(D)

Common Mistakes:

• Trying \( 𝑏 = 2 \) (forces \( 𝑎 = 88, \) not prime).
• Forgetting both must be primes.
• Not checking valid prime pairs summing to 90.

Geometry Example – Problem 8

Question:

Points 𝑃 and 𝑄 lie in a plane with \( 𝑃𝑄 = 8. \) How many locations for point 𝑅 in this plane are there such that \( △𝑃𝑄𝑅 \) is a right triangle with area 12 square units?

(A) 2 (B) 4 (C) 6 (D) 8 (E) 12

Solution:

If \( △𝑃𝑄𝑅 \) is right at 𝑅, then 𝑃𝑄 is the hypotenuse (by Thales’ theorem).

Let the legs be 𝑃𝑅 = 𝑎 and 𝑄𝑅 = 𝑏, so

\[
𝑎^2 + 𝑏^2 = 8^2 = 64, \qquad \frac{1}{2}𝑎𝑏 = 12 \Rightarrow 𝑎𝑏 = 24.
\]

Then

\[
(𝑎 + 𝑏)^2 = 𝑎^2 + 𝑏^2 + 2𝑎𝑏 = 64 + 48 = 112 \Rightarrow 𝑎 + 𝑏 = 4\sqrt{7}.
\]

Solving

\[
𝑡^2 − (𝑎 + 𝑏)𝑡 + 𝑎𝑏 = 0 \Rightarrow 𝑡^2 − 4\sqrt{7}𝑡 + 24 = 0
\]

gives two distinct roots \( 𝑡 = 2(\sqrt{7} ± 1), \) so \( 𝑎 ≠ 𝑏. \)

For each ordered pair \( (𝑎, 𝑏), \) the circles \( 𝑃𝑅 = 𝑎 \) and \( 𝑄𝑅 = 𝑏 \) intersect in 2 points (mirror images across line 𝑃𝑄).
Swapping \( (𝑎, 𝑏) \) and \( (𝑏, 𝑎) \) gives another 2 points.
Total = 4.

Answer: (B)

Common Mistakes:

• Forgetting the right angle is at 𝑅 (so 𝑃𝑄 must be the hypotenuse).
• Counting only one side of line 𝑃𝑄.
• Assuming \( 𝑎 = 𝑏 \) (not true here).

Combinatorics Example – Problem 11

Question:

Ms. Carr asks her students to read any 5 of the 10 books on a reading list. Harold randomly selects 5 books from this list, and Betty does the same. What is the probability that there are exactly 2 books that they both select?

(A) \( \frac{1}{8} \) (B) \( \frac{5}{36} \) (C) \( \frac{14}{45} \) (D) \( \frac{25}{63} \) (E) \( \frac{1}{2} \)

Solution:

Fix Harold’s chosen set of 5 books.
Betty must choose exactly 2 of Harold’s 5 and 3 of the remaining 5.

Number of favorable outcomes:

\[
\frac{5!}{2!(5−2)!} \times \frac{5!}{3!(5−3)!} = 10 \times 10 = 100.
\]

Total possible selections for Betty:

\[
\frac{10!}{5!(10−5)!} = 252.
\]

Therefore,

\[
P = \frac{100}{252} = \frac{25}{63}.
\]

Answer:(D)

Common Mistakes:

• Multiplying by \( 2! \) unnecessarily (order doesn’t matter).
• Using \( \frac{10!}{2!(10−2)!} \times \frac{8!}{3!(8−3)!} \), which double-counts since Harold’s set is fixed.
• Forgetting to divide by the total \( \frac{10!}{5!(10−5)!} \).

2020 AMC 10B Answer Key

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

AMC 10 Past Exam Solutions & Knowledge Packs

Think Academy provides downloadable AMC 10 past exams, full worked solutions, and AR/HR Knowledge Point practice sets to help students learn deeply — not just memorize.

Why Past Exams Matter

• Reveal recurring patterns and concept frequency
• Strengthen logical reasoning + problem-solving skills
• Build confidence heading into real competition day

Don’t just read — start mastering the AMC 10 now!

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

2020 AMC 10B Topic Distribution

The 2020 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, 3, 12, 13, 15, 22	Algebraic operations, coordinate geometry
Number Theory	4, 6, 9, 19, 24	Congruence, factorization
Geometry	2, 8, 10, 14, 20, 21	Solids, triangles, polygons
Combinatorics / Probability	5, 7, 11, 16, 17, 18, 23, 25	Counting, logic, 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.

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