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

Revision as of 20:09, 10 January 2022

Problem

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?

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

Video Solution

https://www.youtube.com/watch?v=TFm1jNgB4QM

Solutions

Solution 1

Somehow(yes no maybe so?) we get $\frac{d}{v}=\frac{1}{3}$ 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

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

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

$10r = 270$ 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 $d=\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)$ , giving $(5)(x)=(3)(x+18)$ $5x=3x+54$ $2x=54$ $x=27$ Hence, $d=\dfrac{27}{3}=\boxed{\textbf{(D)}~9}$ .

Solution 4

Since it takes $\frac{3}{5}$ of the original time for him to get to school when there is no traffic, the speed must be $\frac{5}{3}$ of the speed in no traffic or $\frac{2}{3}$ more. Letting $x$ be the rate and we know that $\frac{5}{3}x = x + 18$ , so we have $\frac{2x}{3} = 18$ 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 $\boxed{\textbf{(D)}~9}$ .

Solution 5

When driving in rush hour traffic, he drives $20$ minutes for one distance ( $1d$ ) to the school. It means he drives $60$ minutes for $3$ distances ( $3d$ ) to the school. When driving in no traffic hours, he drives $12$ minutes for one distance ( $1d$ ) to the school. It means he drives $60$ minutes for $5$ distances ( $5d$ ) to the school. Subtracting these two situations, it gives us $5d-3d = 18 = 2d$ , then $d=\frac{18}{2}=9$ . So the distance to the school would be $\boxed{\textbf{(D)}~9}$ miles. ----LarryFlora

@@ Line 19: / Line 19: @@
 ===Solution 1===
-Somehow(?) we get <math>\frac{d}{v}=\frac{1}{3}</math> and <math>\frac{d}{v+18}=\frac{1}{5}</math>.
+Somehow(yes no maybe so?) we get <math>\frac{d}{v}=\frac{1}{3}</math> and <math>\frac{d}{v+18}=\frac{1}{5}</math>.
 This gives <math>d=\frac{1}{5}v+3.6=\frac{1}{3}v</math>, which gives <math>v=27</math>, which then gives <math>d=\boxed{\textbf{(D)}~9}</math>.

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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