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 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?

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

Solution 1 (One Variable)

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 $x = \boxed{\textbf{(A) }70}$ .

~ihatemath123

Solution 2 (Two Variables)

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 $\frac {210}{42} = 5$ hours, to travel. We can create an equation: $d = 5(42-s),$ 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: $d = 1(42+s).$ Now we have two variables and two equations. We can equate the expressions for $d$ and solve for $s$ : $\begin{align*} 210-5s &= 42 + s \\ s &= 28. \\ \end{align*}$ Therefore, we have $d = 42 + s = \boxed{\textbf{(A) }70}$ .

~LucaszDuzMatz (Solution)

~Arcticturn (Minor $\LaTeX$ Edits)

Solution 3 (Three Variables)

Suppose that Emily and the ship take steps simultaneously such that Emily's steps cover a greater length than the ship's steps.

Let $L$ be the length of the ship, $E$ be Emily's step length, and $S$ be the ship's step length. We wish to find $\frac LE.$

When Emily walks from the back of the ship to the front, she walks a distance of $210E$ and the front of the ship moves a distance of $210S.$ We have $210E=L+210S$ for this scenario, which rearranges to $210E-210S=L. \hspace{15mm}(1)$ When Emily walks in the opposite direction, she walks a distance of $42E$ and the back of the ship moves a distance of $42S.$ We have $42E=L-42S$ for this scenario, which rearranges to $42E+42S=L. \hspace{19.125mm}(2)$ We multiply $(2)$ by $5$ and then add $(1)$ to the result: $420E=6L,$ from which $\frac LE = \boxed{\textbf{(A) }70}.$

~Steven Chen (www.professorchenedu.com)

~MRENTHUSIASM

Solution 4 (Relative Speeds)

Call the speed of the boat $v_s$ and the speed of Emily $v_e$ . Consider the scenario when Emily is walking along with the boat. Relative to an observer on the boat, her speed is $v_e-v_s$ . Consider the scenario when Emily is walking in the opposite direction. Relative to an observer on the boat, her speed is $v_e+v_s$ Since Emily takes 210 steps to walk along with the boat and 42 steps to walk opposite the boat, that means it takes her 5x longer to walk the length of a stationary boat at $v_e-v_s$ compared to $v_e+v_s$ . This means that $5(v_e-v_s)=v_e+v_s \rightarrow v_s = \frac{2v_e}{3}$ As Emily takes 210 steps to walk the length of the boat at a speed of $v_e- \frac{2v_e}{3}=\frac{v_e}{3}$ , she must take 1/3rd of the time to walk the length of the boat at a speed of $v_e$ , so our answer is 210/3=70 steps.

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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