# Difference between revisions of "2015 AMC 8 Problems/Problem 17"

Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?

$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$

## Contents

### Solution 1

So $\frac{d}{v}=\frac{1}{3}$ and $\frac{d}{v+18}=\frac{1}{5}$.

This gives $d=\frac{1}{5}v+3.6=\frac{1}{3}v$, which gives $v=27$, which then gives $d=\boxed{\textbf{(D)}~9}$

### Solution 2

$d = rt$, $d$ is obviously constant

$\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)$

$\frac{r}{3} = \frac{r}{5} + \frac{18}{5}$

$\frac{2r}{15} = \frac{18}{5}$

$10r = 270$ so $r = 27$, plug into the first one and it's $\boxed{\textbf{(D)}~9}$ miles to school

### Solution 3

We set up an equation in terms of $d$ the distance and $x$ the speed In miles per hour. We have $\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)$ $d=(5)(x)=(3)(x+18)$ $5x=3x+54$ $2x=54$ $x=27$

So $d=\dfrac{27}{3}=\boxed{\textbf{(D)},~9}$