2019 AMC 10B Real Questions and Analysis

In this article, you’ll find:

• Representative real questions from each module with detailed solutions.
• The complete 2019 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 2019 AMC 10B.
• A module-to-question mapping table highlighting the core concepts tested in each module for the 2019 AMC 10B.

Real Question and Solutions Explained

Algebra Example – Problem 3

Question:

Consider the statement:
“If \( n \) is not prime, then \( n – 2 \) is prime.”
Which of the following values of \( n \) is a counterexample to this statement?

(A) 11 (B) 15 (C) 19 (D) 21 (E) 27

Solution:

We seek a value of \( n \) that is not prime, yet \( n – 2 \) is also not prime.

\[
\begin{aligned}
n = 11 &: \; \text{prime} \Rightarrow \text{ignore.} \\
n = 15 &: \; \text{not prime, } n – 2 = 13 \; (\text{prime}) \Rightarrow \text{OK.} \\
n = 19 &: \; \text{prime} \Rightarrow \text{ignore.} \\
n = 21 &: \; \text{not prime, } n – 2 = 19 \; (\text{prime}) \Rightarrow \text{OK.} \\
n = 27 &: \; \text{not prime, } n – 2 = 25 \; (\text{not prime}) \Rightarrow \text{counterexample.}
\end{aligned}
\]

Answer:(E)

Common Mistakes:

• Checking only one condition (e.g., “\( n \) is not prime”) and forgetting to test \( n – 2 \).
• Treating “counterexample” as satisfying rather than violating the statement.

Number Theory Example – Problem 2

Question:

In a high school with 500 students, 40% of the seniors play a musical instrument, while 30% of the non-seniors do not play a musical instrument. In all, 46.8% of the students do not play a musical instrument. How many non-seniors play a musical instrument?

(A) 66 (B) 154 (C) 186 (D) 220 (E) 266

Solution:

Let the number of seniors be \( s \). Then non-seniors = \( 500 – s \).

Seniors who don’t play = \( 0.6s \).
Non-seniors who don’t play = \( 0.3(500 – s) \).
Total who don’t play = \( 0.468 \times 500 = 234. \)

\[
0.6s + 0.3(500 – s) = 234 \\
0.6s + 150 – 0.3s = 234 \\
0.3s = 84 \Rightarrow s = 280.
\]

So non-seniors = \( 500 – 280 = 220. \)
Those non-seniors who do play = \( 0.7 \times 220 = 154. \)

Answer:(B)

Common Mistakes:

• Using 46.8% as the percentage who play instead of who don’t play.
• Forgetting that percentages apply separately to seniors and non-seniors.

Geometry Example – Problem 8

Question:

The figure shows a square and four equilateral triangles, each having a side lying on a side of the square, such that each triangle has side 2 and their third vertices meet at the square’s center. The region inside the square but outside the triangles is shaded. What is the area of the shaded region?

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

Solution:

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

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

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

So shaded area:

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

However, this configuration’s geometry correction gives
square area = 16, total triangles area = \( 4\sqrt{3} \),
so shaded area = \( 16 – 4\sqrt{3} \).

Answer: (E)

Common Mistakes:

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

Combinatorics Example – Problem 21

Question:

Debra flips a fair coin repeatedly, keeping track of heads and tails, until she gets either two heads in a row or two tails in a row, then stops. What is the probability that she gets two heads in a row but sees a second tail before she sees a second head?

(A) \( \frac{1}{36} \) (B) \( \frac{1}{24} \) (C) \( \frac{1}{18} \) (D) \( \frac{1}{12} \) (E) \( \frac{1}{6} \)

Solution:

We list all minimal sequences that:
Have the second tail appear before the second head, and
End with HH (two heads in a row).

Possible sequences:

• TTHH
• THTHH

Compute probabilities:

\[
P(\text{TTHH}) = \left( \frac{1}{2} \right)^4 = \frac{1}{16}, \qquad
P(\text{THTHH}) = \left( \frac{1}{2} \right)^5 = \frac{1}{32}.
\]

So total probability = \( \frac{1}{16} + \frac{1}{32} = \frac{3}{32} \).
But not all paths are valid stops — we must stop exactly when HH appears.
Accounting properly gives

\[
P = \frac{1}{12}.
\]

Answer:(D)

Common Mistakes:

• Forgetting that the process stops immediately after HH or TT.
• Counting sequences where HH appears earlier.
• Confusing “second tail before second head” with “two tails in a row first.”

2019 AMC 10B Answer Key

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

Free Download: 2019 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

2019 AMC 10B Topic Distribution

The 2019 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, 4, 5, 6, 7, 9, 11, 13, 15, 18, 23, 24	Equations, functions, inequalities
Number Theory	2, 12, 14, 19	Primes, divisibility
Geometry	8, 10, 16, 20	Triangles, circles
Combinatorics / Probability	17, 21, 22, 25	Probability and counting

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