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

(Solution 2)
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<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>
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.
==Solution 2==
==Solution 2==

Revision as of 10:19, 16 December 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 2

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


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 $15-6=\boxed{\textbf{(D)}~9}$

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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