2021 AMC Fall 10A Problem 11

Measure distance in Emily's steps, and time in "how long one step takes". Then Emily moves $1$ step per unit of time. Say the ship moves $k$ steps per unit of time. Emily is faster, so $k$ is less than $1$. Call the ship's length $L$.

1. Walking the same way as the ship ($210$ steps). In this time Emily moves $210$ steps forward, and the ship moves $210k$ steps forward too. She starts at the back and ends at the front, so she got ahead of the ship by exactly one ship length:
$$L = 210 - 210k.$$
2. Walking the opposite way ($42$ steps). Now Emily moves $42$ steps one way and the ship moves $42k$ steps toward her. Together they close up one ship length:
$$L = 42 + 42k.$$
3. Set the two expressions equal. Both equal $L$:
$$210 - 210k = 42 + 42k.$$
Add $210k$ to both sides and subtract $42$: $168 = 252k$, so $k = \frac{168}{252} = \frac23$.
4. Find $L$. $L = 42 + 42 \times \frac23 = 42 + 28 = 70$.
5. Check with the first equation. $210 - 210 \times \frac23 = 210 - 140 = 70$. It matches.

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

This is the "walking past a moving train" idea. When you walk the same way as the ship, you only gain on it by the difference of the speeds. When you walk toward it, the speeds add. In both cases, the distance gained or closed is the length of the ship. Measuring everything in Emily's steps means you only need one unknown speed.

Compare the two trips directly. Going with the ship, Emily gains $1 - k$ steps each unit of time. Going against, the gap closes by $1 + k$ each unit of time. The same length takes $210$ time units one way and $42$ the other, which is $5$ times as long. So the slower closing speed is $5$ times smaller: $1 + k = 5(1 - k)$. That gives $6k = 4$, $k = \frac23$, and $L = 42 \times (1 + \frac23) = 70$.