Difference between revisions of "2013 AMC 12B Problems/Problem 4"
m (→Solution) |
|||
| (9 intermediate revisions by 4 users not shown) | |||
| Line 5: | Line 5: | ||
<math>\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40</math> | <math>\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | + | Let 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> | |
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | 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 3== | ||
| + | |||
| + | Let the number of miles that Ray and Tom each drive be denoted by <math>m</math>. Thus the number of gallons Ray's car uses can be represented by <math>\frac{m}{40}</math> and the number of gallons that Tom's car uses can likewise be expressed as <math>\frac{m}{10}</math>. Thus the total amount of gallons used by both cars can be expressed as <math>\frac{m}{40} + \frac{m}{10} = \frac{m}{8}</math>. The total distance that both drive is equal to <math>2m</math> so the total miles per gallon can be expressed as <math>\frac{2m}{\frac{m}{8}} = \boxed{\textbf{(B) }16}</math> | ||
| + | |||
| + | ~AnkitAMC | ||
== See also == | == See also == | ||
Latest revision as of 11:40, 3 August 2022
- The following problem is from both the 2013 AMC 12B #4 and 2013 AMC 10B #8, so both problems redirect to this page.
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 1
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 ![\[\left(\frac{40^{-1} + 10^{-1}}{2}\right)^{-1} = \frac{2}{\frac{1}{40}+\frac{1}{10}} = \frac{2}{\frac{5}{40}} = \frac{2}{\frac{1}{8}} = \boxed{\textbf{(B) }16}.\]](http://latex.artofproblemsolving.com/9/b/6/9b6db99b489317a9225179391b994358e160d481.png)
-Solution by Joeya
Solution 3
Let the number of miles that Ray and Tom each drive be denoted by 
. Thus the number of gallons Ray's car uses can be represented by 
and the number of gallons that Tom's car uses can likewise be expressed as 
. Thus the total amount of gallons used by both cars can be expressed as 
. The total distance that both drive is equal to 
so the total miles per gallon can be expressed as 
~AnkitAMC
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. 
