Difference between revisions of "2013 AMC 12B Problems/Problem 4"

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==Solution==
 
==Solution==
 
Let both Ray and Tom drive 40 miles. Ray's car would require <math>\frac{40}{40}=1</math> gallon of gas and Tom's car would require <math>\frac{40}{10}=4</math> gallons of gas. They would have driven a total of <math>40+40=80</math> miles, on <math>1+4=5</math> gallons of gas, for a combined rate of <math>\frac{80}{5}=</math> <math>\boxed{\textbf{(B) }16}</math>
 
Let both Ray and Tom drive 40 miles. Ray's car would require <math>\frac{40}{40}=1</math> gallon of gas and Tom's car would require <math>\frac{40}{10}=4</math> gallons of gas. They would have driven a total of <math>40+40=80</math> miles, on <math>1+4=5</math> gallons of gas, for a combined rate of <math>\frac{80}{5}=</math> <math>\boxed{\textbf{(B) }16}</math>
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== See also ==
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{{AMC12 box|year=2013|ab=B|num-b=3|num-a=5}}

Revision as of 18:03, 22 February 2013

Problem

Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?
$\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40$

Solution

Let both Ray and Tom drive 40 miles. Ray's car would require $\frac{40}{40}=1$ gallon of gas and Tom's car would require $\frac{40}{10}=4$ gallons of gas. They would have driven a total of $40+40=80$ miles, on $1+4=5$ gallons of gas, for a combined rate of $\frac{80}{5}=$ $\boxed{\textbf{(B) }16}$


See also

2013 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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 AMC 12 Problems and Solutions