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

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s &= 28. \\
s &= 28. \\
\end{align*}</cmath>
\end{align*}</cmath>
Therefore, we have <math>d + </math>s = 42 + s = \boxed{\textbf{(A) }<math>70}</math>.
Therefore, we have <math>d + </math>s = 42 + s = \boxed{\textbf{(A) }70}<math>.


~LucaszDuzMatz (Solution)
~LucaszDuzMatz (Solution)
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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.
Suppose that Emily and the ship take steps simultaneously such that Emily's steps cover a greater length than the ship's steps.


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>
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>


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>
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>
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>
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>


~Steven Chen (www.professorchenedu.com)
~Steven Chen (www.professorchenedu.com)
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==Solution 4 (Using the Boat's "Step")==
==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 <math>s</math>. Call the length of the boat <math>x</math>.
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.
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.
   
   
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.  
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.  


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


~zeeshan12
~zeeshan12


== Solution 5 (Relative Speeds) ==
== Solution 5 (Relative Speeds) ==
Call the speed of the boat <math>v_s</math> and the speed of Emily <math>v_e</math>.
Call the speed of the boat </math>v_s<math> and the speed of Emily </math>v_e<math>.


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>.
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>.


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>
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>


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>.
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>.


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


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>.
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}$.


==Video Solution by TheBeautyofMath==
==Video Solution by TheBeautyofMath==

Revision as of 17:46, 1 November 2025

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 $d +$s = 42 + s = \boxed{\textbf{(A) }70}$.

~LucaszDuzMatz (Solution)


~Minor edits by Yvz2900

== 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$ (Error compiling LaTeX. Unknown error_msg)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 <cmath>210E-210S=L. \hspace{15mm}(1)</cmath> 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 <cmath>42E+42S=L. \hspace{19.125mm}(2)</cmath> 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$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)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 <cmath>\begin{align*} 210 &= x + 210 s, \\ 42 &= x - 42s. \end{align*}</cmath> Solving for$ (Error compiling LaTeX. Unknown error_msg)x$you get$\boxed{\textbf{(A) }70}$.

~zeeshan12

== Solution 5 (Relative Speeds) == Call the speed of the boat$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)5(v_e-v_s)=v_e+v_s$, so$v_s = \frac{2v_e}{3}$.

As Emily takes$ (Error compiling LaTeX. Unknown error_msg)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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