2021 Fall AMC 10A Problems/Problem 11

Problem

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 tha 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?

$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126$

Solution 1

Let the speed at which Emily walks be 42 steps per hour. Let the speed at which the ship is moving be $s$ . Walking in the direction of the ship, it takes her 210 steps, or 210/42 = 5 hours, to travel. We can create the equation: $d = (42-s)5$ Where d is the length of the ship. Walking in the opposite direction of the ship, it takes her 42 steps, or 42/42 = 1 hour. We can create a similar equation: $d = (42+s)1$ Now we have 2 variables and 2 equations, and we can solve for d. $210-5s = 42 + s$ $s = 28$ $d = 42 + s = \boxed{\textbf{(A) } 70}$ ~LucaszDuzMatz

Solution 2

Let $x$ be the length of the ship. Then, in the time that Emily walks $210$ steps, the ship moves $210-x$ steps. Also, in the time that Emily walks $42$ steps, the ship moves $x-42$ steps. Since the ship and Emily both travel at some constant rate, $\frac{210}{210-x} = \frac{42}{x-42}$ . Dividing both sides by $42$ and cross multiplying, we get $5(x-42) = 210-x$ , so $6x = 420$ , and $x = \boxed{70}$ . ~ihatemath123

2021 Fall AMC 10A (Problems • Answer Key • Resources)
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2021 Fall AMC 10A Problems/Problem 11

Contents

Problem

Solution 1

Solution 2

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