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 == | ||
+ | {{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 miles per gallon of gasoline, and Tom's car averages 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?
Solution
Let both Ray and Tom drive 40 miles. Ray's car would require gallon of gas and Tom's car would require gallons of gas. They would have driven a total of miles, on gallons of gas, for a combined rate of
See also
2013 AMC 12B (Problems • Answer Key • Resources) | |
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 |