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2021 Fall AMC 10A Problems/Problem 11

Revision as of 12:16, 26 November 2021 by MRENTHUSIASM (talk | contribs) (→‎Solution 2 (Two Variables))

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)

Denote by $L$ the length of the ship, $x$ and $y$ the rates of Emily and the ship (distance per walking step), respectively.

Hence, $L = 210 \left( x - y \right)$ and $L = 42 \left( x + y \right)$.

By solving these equations, we get $y = \frac{2}{3} x$. Plugging this result into the first equation, we get $\frac{L}{x} = 70$. This is exactly the length of the ship, measured in terms of Emily's equal steps.

Therefore, the answer is $\boxed{\textbf{(A) }70}$.

~Steven Chen (www.professorchenedu.com)

See Also

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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