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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 11"
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<math>\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126</math>
 
<math>\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126</math>
  
−
==Solution 1==
+
==Solution 1 (One Variable)==
−
Let the speed at which Emily walks be <math>42</math> steps per hour. Let the speed at which the ship is moving be <math>s</math>. Walking in the direction of the ship, it takes her <math>210</math> steps, or <math>\frac {210}{42} = 5</math> hours, to travel. We can create the equation: <cmath>d = 5(42-s)</cmath> 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: <cmath>d = 1(42+s)</cmath> Now we have 2 variables and 2 equations, and we can solve for d.
 
−
<cmath>210-5s = 42 + s</cmath>
 
−
<cmath>s = 28</cmath>
 
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<cmath>d = 42 + s = \boxed{\textbf{(A) } 70} </cmath>
 
−
~LucaszDuzMatz
 
−
 
 
−
~Minor LaTeX edits by Arcticturn
 
−
 
 
−
==Solution 2==
 
 
Let <math>x</math> be the length of the ship.
 
Let <math>x</math> be the length of the ship.
 
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.
−
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{70}</math>.
+
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>.
  
 
~ihatemath123
 
~ihatemath123
  
−
== Solution 3 ==
+
==Solution 2 (Two Variables)==
 +
Let the speed at which Emily walks be <math>42</math> steps per hour. Let the speed at which the ship is moving be <math>s</math>. Walking in the direction of the ship, it takes her <math>210</math> steps, or <math>\frac {210}{42} = 5</math> hours, to travel. We can create an equation: <cmath>d = 5(42-s),</cmath> where <math>d</math> is the length of the ship. Walking in the opposite direction of the ship, it takes her <math>42</math> steps, or <math>42/42 = 1</math> hour. We can create a similar equation: <cmath>d = 1(42+s).</cmath> Now we have two variables and two equations. We can equate the expressions for <math>d</math> and solve for <math>s</math>:
 +
<cmath>\begin{align*}
 +
210-5s &= 42 + s \\
 +
s &= 28 \\
 +
\end{align*}</cmath>
 +
Therefore, we have <math>d = 42 + s = \boxed{\textbf{(A) }70}</math>.
 +
 
 +
~LucaszDuzMatz (Solution)
 +
 
 +
~Arcticturn (Minor <math>\LaTeX</math> Edits)
 +
 
 +
== Solution 3 (Three Variables) ==
 
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.
 
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.
  

Revision as of 04:56, 26 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)

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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