2015 AMC 8 Problems/Problem 17: Difference between revisions

Revision as of 17:06, 25 November 2015

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?

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

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

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

@@ Line 23: / Line 23: @@
 <math>\frac{2r}{15} = \frac{18}{5}</math>
-<math>10r = 270</math> so <math>r = 27</math>, plug into the first one and it's <math>\boxed{\textbf{B}~9}</math> miles to school
+<math>10r = 270</math> so <math>r = 27</math>, plug into the first one and it's <math>\boxed{\textbf{(D)}~9}</math> miles to school
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2015 AMC 8 Problems/Problem 17: Difference between revisions

Revision as of 17:06, 25 November 2015

Solution 1

Solution 2

See Also