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

(Solution)
(Solution2)
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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>  
 
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>  
  
minutes for Jack to arrive at the pool: answer <math>\boxed{textbf{(D)}}</math>.
+
minutes for Jack to arrive at the pool: answer <math>\boxed{{(D)}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 22:19, 25 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.

Solution2

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 $15-6=\boxed{\textbf{9}}$

minutes for Jack to arrive at the pool: answer $\boxed{{(D)}}$.

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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