Difference between revisions of "2015 AMC 8 Problems/Problem 3"

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==Problem==
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Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of <math>10</math> miles per hour. Jack walks to the pool at a constant speed of <math>4</math> miles per hour. How many minutes before Jack does Jill arrive?
 
Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of <math>10</math> miles per hour. Jack walks to the pool at a constant speed of <math>4</math> miles per hour. How many minutes before Jack does Jill arrive?
  
 
<math>\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10</math>
 
<math>\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10</math>
  
===Solution===
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==Solution==
Jill arrives in <math>\dfrac{1}{10}</math> of an hour, which is <math>6</math> minutes. Jack arrives in <math>\dfrac{1}{4}</math> of an hour which is <math>15</math> minutes. Thus, the time difference is <math>\boxed{\textbf{(D)}~9}</math> minutes.
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Using <math>d=rt</math>, we can set up an equation for when Jill arrives at the swimming pool:
 
 
==Solution 2==
 
Using <math>d=rt</math>, we can set up an equation for when Jill arrives at swimming:
 
  
 
<math>1=10t</math>
 
<math>1=10t</math>
  
Solving for t, we get that Jill gets to the pool in <math>\frac{1}{10}</math> of on hour, which translates to <math>6</math> minutes.  Doing the same for Jack, we get that  
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Solving for <math>t</math>, we get that Jill gets to the pool in <math>\frac{1}{10}</math> of an hour, which is <math>6</math> minutes.  Doing the same for Jack, we get that  
  
Jack arrives at the pool in <math>\frac{1}{4}</math> of an hour, which in turn translates to <math>15</math> minutes.  Thus, Jill has to wait <math>15-6=\boxed{\textbf{9}}</math>  
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Jack arrives at the pool in <math>\frac{1}{4}</math> of an hour, which in turn is <math>15</math> minutes.  Thus, Jill has to wait <math>15-6=\boxed{\textbf{(D)}~9}</math>  
  
minutes for Jack to arrive at the pool: Therefore, answer <math>\boxed{{\textbf{(D) 9} }}</math>.
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minutes for Jack to arrive at the pool.
  
 
==See Also==
 
==See Also==

Latest revision as of 15:56, 16 January 2021

Problem

Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of $10$ miles per hour. Jack walks to the pool at a constant speed of $4$ miles per hour. How many minutes before Jack does Jill arrive?

$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10$

Solution

Using $d=rt$, we can set up an equation for when Jill arrives at the swimming pool:

$1=10t$

Solving for $t$, we get that Jill gets to the pool in $\frac{1}{10}$ of an hour, which is $6$ minutes. Doing the same for Jack, we get that

Jack arrives at the pool in $\frac{1}{4}$ of an hour, which in turn is $15$ minutes. Thus, Jill has to wait $15-6=\boxed{\textbf{(D)}~9}$

minutes for Jack to arrive at the pool.

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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 AJHSME/AMC 8 Problems and Solutions

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