Difference between revisions of "2013 AMC 10B Problems/Problem 8"

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==Problem==
 
==Problem==
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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?
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<math> \textbf{(A) }10\qquad\textbf{(B) }16\qquad\textbf{(C) }25\qquad\textbf{(D) }30\qquad\textbf{(E) }40 </math>
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==Solution==
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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>

Revision as of 19:49, 21 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 $\frac{80}{5}=$ $\boxed{\textbf{(B) }16}$