Art of Problem Solving

2015 AMC 8 Problems/Problem 17: Difference between revisions

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===Solution 1===
So <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>
==See Also==
==See Also==


{{AMC8 box|year=2015|num-b=16|num-a=18}}
{{AMC8 box|year=2015|num-b=16|num-a=18}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 17:04, 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 $d=\boxed{\textbf{(D)}~9}$

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
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

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