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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 11"

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== Solution 3 (Three Variables) ==
 
== Solution 3 (Three Variables) ==
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Denote by <math>L</math> the length of the ship, <math>x</math> and <math>y</math> the rates of Emily and the ship (distance per walking step), respectively.
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Suppose that Emily and the ship take steps simultaneously such that Emily's steps cover a greater length than the ship's steps.
  
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Hence, <math>L = 210 \left( x - y \right)</math> and <math>L = 42 \left( x + y \right)</math>.
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Let <math>L</math> be the length of the ship, <math>E</math> be Emily's step length, and <math>S</math> be the ship's step length. We wish to find <math>\frac LE.</math>
  
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By solving these equations, we get <math>y = \frac{2}{3} x</math>.
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When Emily walks from the back of the ship to the front, she walks a distance of <math>210E</math> and the front of the ship moves a distance of <math>210S.</math> We have <math>210E=L+210S</math> for this scenario, which rearranges to <cmath>210E-210S=L. \hspace{15mm}(1)</cmath> When Emily walks in the opposite direction, she walks a distance of <math>42E</math> and the back of the ship moves a distance of <math>42S.</math> We have <math>42E=L-42S</math> for this scenario, which rearranges to <cmath>42E+42S=L. \hspace{19.125mm}(2)</cmath>
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Plugging this result into the first equation, we get <math>\frac{L}{x} = 70</math>.
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We multiply <math>(2)</math> by <math>5</math> and then add <math>(1)</math> to the result: <cmath>420E=6L,</cmath> from which <math>\frac LE = \boxed{\textbf{(A) }70}.</math>
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This is exactly the length of the ship, measured in terms of Emily's equal steps.
 
  
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Therefore, the answer is <math>\boxed{\textbf{(A) }70}</math>.
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~Steven Chen (www.professorchenedu.com)
  
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~Steven Chen (www.professorchenedu.com)
+
~MRENTHUSIASM
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=10|num-a=12}}
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:10, 27 November 2021

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

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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