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

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Then, in the time that Emily walks <math>210</math> steps, the ship moves <math>210-x</math> steps.
 
Then, in the time that Emily walks <math>210</math> steps, the ship moves <math>210-x</math> steps.
 
Also, in the time that Emily walks <math>42</math> steps, the ship moves <math>x-42</math> steps.
 
Also, in the time that Emily walks <math>42</math> steps, the ship moves <math>x-42</math> steps.
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Since the ship and Emily both travel at some constant rate, <math>\frac{210}{210-x} = \frac{42}{x-42}</math>. Dividing both sides by <math>42</math> and cross multiplying, we get <math>5(x-42) = 210-x</math>, so <math>6x = 420</math>, and <math>x = \boxed{\textbf{(A) }70}</math>.
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Since the ship and Emily have the same ratio of absolute speeds in either direction, <math>\frac{210}{210-x} = \frac{42}{x-42}</math>. Dividing both sides by <math>42</math> and cross multiplying, we get <math>5(x-42) = 210-x</math>, so <math>6x = 420</math>, and <math>x = \boxed{\textbf{(A) }70}</math>.
  
 
~ihatemath123
 
~ihatemath123
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~LucaszDuzMatz (Solution)
 
~LucaszDuzMatz (Solution)
  
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~Arcticturn (Minor <math>\LaTeX</math> Edits)
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== Solution 3 (Three Variables) ==
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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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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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 +
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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We multiply <math>(2)</math> by <math>5</math> and then add <math>(1)</math> to get <math>420E=6L,</math> from which <math>\frac LE = \boxed{\textbf{(A) }70}.</math>
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 +
~Steven Chen (www.professorchenedu.com)
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~MRENTHUSIASM
  
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== Solution 3 (Three Variables) ==
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==Solution 4 (Using the Boat's "Step")==
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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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Every time Emily takes a step, the boat also "takes a step". Call the length of the boats step <math>s</math>. Call the length of the boat <math>x</math>.
 +
 +
When Emily is walking in the same direction as the boat, every time she takes a step the boat moves an additional distance of <math>s</math>. This means that she travels a total distance of <math>x + 210 s</math> to reach the other end of the boat.
 +
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When Emily is walking in the opposite direction of the boat, every time she takes a step the distance till the end of the boat reduces by <math>s</math> (since the boat is coming towards her and moves a distance of <math>s</math>). This means that she travels a total distance of <math>x - 42 s</math> to reach the other end of the boat.
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Taking Emily's step as a unit of distance, we now have two equations
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<cmath>\begin{align*}
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210 &= x + 210 s, \\
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42 &= x - 42s.
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\end{align*}</cmath>
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Solving for <math>x</math> you get <math>\boxed{\textbf{(A) }70}</math>.
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~zeeshan12
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== Solution 5 (Relative Speeds) ==
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Call the speed of the boat <math>v_s</math> and the speed of Emily <math>v_e</math>.
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Consider the scenario when Emily is walking along with the boat. Relative to an observer on the boat, her speed is <math>v_e-v_s</math>.
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Consider the scenario when Emily is walking in the opposite direction. Relative to an observer on the boat, her speed is <math>v_e+v_s</math>
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Since Emily takes <math>210</math> steps to walk along with the boat and <math>42</math> steps to walk opposite the boat, that means it takes her <math>5</math> times longer to walk the length of a stationary boat at <math>v_e-v_s</math> compared to <math>v_e+v_s</math>.
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This means that <math>5(v_e-v_s)=v_e+v_s</math>, so <math>v_s = \frac{2v_e}{3}</math>.
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As Emily takes <math>210</math> steps to walk the length of the boat at a speed of <math>v_e- \frac{2v_e}{3}=\frac{v_e}{3}</math>, she must take <math>\frac13</math> of the time to walk the length of the boat at a speed of <math>v_e</math>, so our answer is <math>\frac{210}{3} = \boxed{\textbf{(A) }70}</math>.
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==Video Solution by TheBeautyofMath==
 +
https://youtu.be/zq3UPu4nwsE
  
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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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~IceMatrix
  
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By solving these equations, we get <math>y = \frac{2}{3} x</math>.
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==Video Solution by WhyMath==
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Plugging this result into the first equation, we get <math>\frac{L}{x} = 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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https://youtu.be/0JOj_fbCB40
  
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~Steven Chen (www.professorchenedu.com)
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~savannahsolver
  
 
==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}}

Latest revision as of 04:47, 30 October 2024

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 have the same ratio of absolute speeds in either direction, $\frac{210}{210-x} = \frac{42}{x-42}$. 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)

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 get $420E=6L,$ from which $\frac LE = \boxed{\textbf{(A) }70}.$

~Steven Chen (www.professorchenedu.com)

~MRENTHUSIASM

Solution 4 (Using the Boat's "Step")

Every time Emily takes a step, the boat also "takes a step". Call the length of the boats step $s$. Call the length of the boat $x$.

When Emily is walking in the same direction as the boat, every time she takes a step the boat moves an additional distance of $s$. This means that she travels a total distance of $x + 210 s$ to reach the other end of the boat.

When Emily is walking in the opposite direction of the boat, every time she takes a step the distance till the end of the boat reduces by $s$ (since the boat is coming towards her and moves a distance of $s$). This means that she travels a total distance of $x - 42 s$ to reach the other end of the boat.

Taking Emily's step as a unit of distance, we now have two equations \begin{align*} 210 &= x + 210 s, \\ 42 &= x - 42s. \end{align*} Solving for $x$ you get $\boxed{\textbf{(A) }70}$.

~zeeshan12

Solution 5 (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 $5$ times longer to walk the length of a stationary boat at $v_e-v_s$ compared to $v_e+v_s$.

This means that $5(v_e-v_s)=v_e+v_s$, so $v_s = \frac{2v_e}{3}$.

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 $\frac13$ of the time to walk the length of the boat at a speed of $v_e$, so our answer is $\frac{210}{3} = \boxed{\textbf{(A) }70}$.

Video Solution by TheBeautyofMath

https://youtu.be/zq3UPu4nwsE

~IceMatrix

Video Solution by WhyMath

https://youtu.be/0JOj_fbCB40

~savannahsolver

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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