Difference between revisions of "2013 AMC 10B Problems/Problem 8"
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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> | ||
| + | == See also == | ||
| + | {{AMC10 box|year=2013|ab=B|num-b=7|num-a=9}} | ||
Revision as of 17:00, 27 March 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?
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 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 10 Problems and Solutions | ||
