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 17:03, 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 | |
