Difference between revisions of "2013 AMC 12B Problems/Problem 4"
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− | Taking the harmonic mean of the two rates, we get <cmath>\left(\frac{40^{-1} + 10^{-1}}{2}\right)^{-1} = \frac{2}{\frac{1}{40}+\frac{1}{10}} = \frac{2}{\frac{5}{40}} = \frac{2}{\frac{1}{8}} = \boxed{\textbf{(B) }16}</cmath> | + | Taking the harmonic mean of the two rates, we get <cmath>\left(\frac{40^{-1} + 10^{-1}}{2}\right)^{-1} = \frac{2}{\frac{1}{40}+\frac{1}{10}} = \frac{2}{\frac{5}{40}} = \frac{2}{\frac{1}{8}} = \boxed{\textbf{(B) }16}.</cmath> |
-Solution by Joeya | -Solution by Joeya |
Revision as of 00:03, 18 January 2021
- The following problem is from both the 2013 AMC 12B #4 and 2013 AMC 10B #8, so both problems redirect to this page.
Contents
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 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
Solution 2
Taking the harmonic mean of the two rates, we get
-Solution by Joeya
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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.