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

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

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