2017 AMC 10B Real Questions and Analysis

In this article, you’ll find:

• Representative real questions from each module with detailed solutions.
• The complete 2017 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 2017 AMC 10B.
• A module-to-question mapping table highlighting the core concepts tested in each module for the 2017 AMC 10B.

Real Question and Solutions Explained

Algebra Example – Problem 2

Question:

Sofia ran 5 laps around the 400–meter track at her school. For each lap, she ran the first 100 meters at an average speed of 4 meters per second and the remaining 300 meters at an average speed of 5 meters per second. How much time did Sofia take running the 5 laps?

(A) 5 minutes and 35 seconds (B) 6 minutes and 40 seconds (C) 7 minutes and 5 seconds (D) 7 minutes and 25 seconds (E) 8 minutes and 10 seconds

Solution:

Time per lap:

\[
\frac{100}{4} + \frac{300}{5} = 25 + 60 = 85\text{ s.}
\]

For 5 laps:

\[
5 \times 85 = 425\text{ s.}
\]

Convert to minutes:

\[
425\text{ s} = 7\text{ minutes } 5\text{ seconds.}
\]

Answer:(C)

Common Mistakes:

• Averaging speeds instead of summing times.
• Mixing meters/seconds with minutes/hours.
• Forgetting the “per lap” structure and multiplying distances incorrectly.

Number Theory Example – Problem 14

Question:

An integer \( N \) is selected at random in the range \( 1 \le N \le 2020 \). What is the probability that the remainder when \( N^{16} \) is divided by 5 is 1?

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

Solution:

If 5 does not divide \( N \), then by Euler’s theorem:

\[
N^{\varphi(5)} = N^4 \equiv 1 \pmod{5}.
\]

Hence,

\[
N^{16} = (N^4)^4 \equiv 1 \pmod{5}.
\]

If \( 5 \mid N \), then \( N^{16} \equiv 0 \pmod{5} \).

Therefore, all numbers not divisible by 5 give remainder 1.

Count of multiples of 5 in \( 1..2020 \):

\[
\frac{2020}{5} = 404.
\]

Favorable outcomes:

\[
2020 – 404 = 1616.
\]

Probability:

\[
\frac{1616}{2020} = \frac{4}{5}.
\]

Answer:(D)

Common Mistakes:

• Checking only small bases (1–4) and assuming uniform distribution.
• Forgetting to exclude multiples of 5.
• Claiming \( N^{16} \equiv 1 \) for all \( N \).

Geometry Example – Problem 15

Question:

Rectangle \(ABCD\) has \(AB = 3\) and \(BC = 4\). Point \(E\) is the foot of the perpendicular from \(B\) to diagonal \(\overline{AC}\).
What is the area of \(\triangle AED\)?

(A) 1 (B) \( \frac{42}{25} \) (C) \( \frac{28}{15} \) (D) 2 (E) \( \frac{54}{25} \)

Solution:

Place \( A(0,4),\ B(3,4),\ C(3,0),\ D(0,0) \).
Vector along diagonal \( AC \):
\[
\vec{v} = (3, -4).
\]

From \( B \), foot of perpendicular onto \( AC \):
\[
t = \frac{(B – A) \cdot \vec{v}}{\vec{v} \cdot \vec{v}} = \frac{9}{25},
\quad
E = A + t \vec{v} = \left( \frac{27}{25}, \frac{64}{25} \right).
\]

Triangle \( \triangle AED \) has base \( AD = 4 \) (vertical) and height equal to the horizontal distance from \( E \) to \( AD \), which is \( \frac{27}{25} \).

Area:

\[
\text{Area} = \frac{1}{2} \times 4 \times \frac{27}{25} = \frac{54}{25}.
\]

Answer: (E)

Common Mistakes:

• Computing via slope–angle formulas instead of projection.
• Using distance from \(E\) to \(AB\) or \(BC\) instead of \(AD\).
• Arithmetic mistakes when calculating \(t\).

Combinatorics Example – Problem 9

Question:

A radio program has a quiz consisting of 3 multiple–choice questions, each with 3 choices. A contestant wins if he or she gets 2 or more of the questions right. The contestant answers randomly to each question. What is the probability of winning?

(A) \( \frac{1}{27} \) (B) \( \frac{1}{9} \) (C) \( \frac{2}{9} \) (D) \( \frac{7}{27} \) (E) \( \frac{1}{2} \)

Solution:

Let \( p = \frac{1}{3} \) be the probability of answering a question correctly.

\[
P(\text{win}) = P(\text{exactly 2 correct}) + P(\text{exactly 3 correct})
\]

There are 3 ways to get exactly 2 correct out of 3 questions,
so number of combinations \( = \frac{3!}{2!(3-2)!} = 3. \)

\[
P(\text{win}) = 3 \cdot p^2 (1 – p) + p^3
= 3 \cdot \frac{1}{9} \cdot \frac{2}{3} + \frac{1}{27}
= \frac{6}{27} + \frac{1}{27} = \frac{7}{27}.
\]

Answer:(D)

Common Mistakes:

• Counting only the “all 3 correct” case.
• Using \( p = \frac{1}{2} \) instead of \( \frac{1}{3} \).
• Forgetting the binomial coefficient for “exactly 2.”

2017 AMC 10B Answer Key

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

Free Download: 2017 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

2017 AMC 10B Topic Distribution

The 2017 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	1–5, 7, 8, 10–13, 17, 24	Equations, ratios, coordinate geometry
Number Theory	14, 20, 23, 25	Congruence, divisibility
Geometry	6, 15, 19, 21, 22	Triangles, solids
Combinatorics / Probability	9, 16, 17, 18	Counting and probability

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