Difference between revisions of "2022 AMC 10A Problems/Problem 4"
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<cmath>100\cdot\frac{m}{(\frac{x}{l})} = \frac{100lm}{x} \rightarrow \boxed{\textbf{(E) } \frac{100lm}{x}}.</cmath> | <cmath>100\cdot\frac{m}{(\frac{x}{l})} = \frac{100lm}{x} \rightarrow \boxed{\textbf{(E) } \frac{100lm}{x}}.</cmath> | ||
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+ | -Benedict T (countmath1) | ||
==Video Solution 1 (Quick and Easy)== | ==Video Solution 1 (Quick and Easy)== |
Revision as of 22:55, 13 November 2022
Problem
In some countries, automobile fuel efficiency is measured in liters per kilometers while other countries use miles per gallon. Suppose that 1 kilometer equals miles, and gallon equals liters. Which of the following gives the fuel efficiency in liters per kilometers for a car that gets miles per gallon?
Solution 1
The formula for fuel efficiency is Note that mile equals kilometers. We have Therefore, the answer is
~MRENTHUSIASM
Solution 2
Since it can be a bit odd to think of "liters per 100 km", this statement's numerical value is equivalent to .
so the numerator is simply . Since liters gallon, and miles = gallon, .
Therefore, the requested expression is
-Benedict T (countmath1)
Video Solution 1 (Quick and Easy)
~Education, the Study of Everything
See Also
2022 AMC 10A (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 10 Problems and Solutions |
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