2015 AMC 8 Problems/Problem 3

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.

Solution 2

T=D/s Jill: (1/10)x60 because in minutes, is equal to 6 min Jack:(1/4)x60 is 15 minutes. 15-6 is 9, so our answer is $15-6=\boxed{\textbf{(D)}~9}$ - TheNerdWhoIsNerdy.

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/srGyMofBMsE

~Education, the Study of Everything

Video Solution

https://youtu.be/YvYq3iM4jP8

~savannahsolver

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search