# 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

[INCOMPLETE]

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

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

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

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

$10r = 270$ 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 $d=\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)$ $$d=(5)(x)=(3)(x+18)$$ $$5x=3x+54$$ $$2x=54$$ $$x=27$$

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

### Solution 4

Since it takes 3/5 of the original time for him to get to school when there is no traffic, the speed must be 5/3 of the speed in traffic or 2/3 more. Letting x be the distance he can drive with traffic in 1 hour, we have 2x/3 = 18 miles per hour. Solving for x gives us 27 miles per hour. Because 20 minutes is a third of an hour, the distance would then be 9 miles. (D)9