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

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Solving for <math>t</math>, 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  
 
Solving for <math>t</math>, 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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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 translates to <math>15</math> minutes.  Thus, Jill has to wait <math>15-6=9</math>  
  
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minutes for Jack to arrive at the pool: Therefore, the answer is <math>\boxed{{\textbf{(D) 9} }}</math>.
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minutes for Jack to arrive at the pool: Therefore, the answer is <math>\boxed{{\textbf{(D)}~9}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 11:39, 30 November 2015

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

Jill arrives in $\dfrac{1}{10}$ of an hour, which is $6$ minutes. Jack arrives in $\dfrac{1}{4}$ of an hour which is $15$ minutes. Thus, the time difference is $\boxed{\textbf{(D)}~9}$ minutes.

Solution 2

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

$1=10t$

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

Jack arrives at the pool in $\frac{1}{4}$ of an hour, which in turn translates to $15$ minutes. Thus, Jill has to wait $15-6=9$

minutes for Jack to arrive at the pool: Therefore, the answer is $\boxed{{\textbf{(D)}~9}}$.

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
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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