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2021 AMC Fall 10A Problem 11

2021 AMC Fall 10A Problem 11AMC 10 Step by Step

Name: Date: September 13, 2026

Linear Equations & Word Problems · difficulty 3 of 5 · about 4 min · position P11-15

Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?

A)
$70$
B)
$84$
C)
$98$
D)
$105$
E)
$126$

Answer

Key insight

Measure in Emily's steps: walking with the ship she gains 210 - 210k, against it 42 + 42k, where k is the ship's speed ratio; both equal the length.

Solution

Use one of Emily's steps as the unit of length and the time of one step as the unit of time, so Emily's speed is $1$ . Let the ship's speed be $k$ (with $k < 1$ ) and its length $L$ .

Walking in the ship's direction, Emily takes $210$ steps. In that time she moves $210$ and the ship moves $210k$ , and she has gone from the stern to the bow, so she gained exactly one ship length:
$L = 210 - 210k.$

Walking against the ship, she takes $42$ steps; she moves $42$ and the ship moves $42k$ toward her, and together they cover one ship length:
$L = 42 + 42k.$

Equating: $210 - 210k = 42 + 42k$ , so $168 = 252k$ and $k = \tfrac{2}{3}$ . Then
$L = 42 + 42 \cdot \tfrac23 = 42 + 28 = 70.$
(Check with the other equation: $210 - 140 = 70$ .)

The answer is $\boxed{\textbf{(A)}\ 70}$ .

Why this works

This is the classic "walking past a moving train" setup: relative speed is $1 - k$ when moving with the object and $1 + k$ when moving against it, and the ship's length is relative speed times time in both cases. Choosing Emily's step as the unit of both distance and time removes all extra variables, leaving one linear equation in $k$ .

Alternative approach

Ratio form: the relative speeds are inversely proportional to the times, so $\dfrac{1 + k}{1 - k} = \dfrac{210}{42} = 5$ , giving $1 + k = 5 - 5k$ and $k = \tfrac23$ immediately; then $L = 42(1 + k) = 70$ .

The trap

Forgetting that the ship moves while Emily walks and taking the length as 210 steps, or as the average of 210 and 42.

Common mistakes

Forgetting that the ship moves while Emily walks and taking the length as 210 steps, or as the average of 210 and 42.
Setting up $L = 210 + 210k$ for the same-direction walk (adding instead of subtracting the ship's motion).

Techniques

Set up the equation/formula and compute; no special trick needed