Difference between revisions of "2022 AMC 10A Problems/Problem 4"
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<math>\textbf{(A) } \frac{x}{100lm} \qquad \textbf{(B) } \frac{xlm}{100} \qquad \textbf{(C) } \frac{lm}{100x} \qquad \textbf{(D) } \frac{100}{xlm} \qquad \textbf{(E) } \frac{100lm}{x}</math> | <math>\textbf{(A) } \frac{x}{100lm} \qquad \textbf{(B) } \frac{xlm}{100} \qquad \textbf{(C) } \frac{lm}{100x} \qquad \textbf{(D) } \frac{100}{xlm} \qquad \textbf{(E) } \frac{100lm}{x}</math> | ||
− | == Solution == | + | == Solution 1 == |
The formula for fuel efficiency is <cmath>\frac{\text{Distance}}{\text{Gas Consumption}}.</cmath> | The formula for fuel efficiency is <cmath>\frac{\text{Distance}}{\text{Gas Consumption}}.</cmath> | ||
Note that <math>1</math> mile equals <math>\frac 1m</math> kilometers. We have <cmath>\frac{x\text{ miles}}{1\text{ gallon}} = \frac{\frac{x}{m}\text{ kilometers}}{l\text{ liters}} = \frac{1\text{ kilometer}}{\frac{lm}{x}\text{ liters}} = \frac{100\text{ kilometers}}{\frac{100lm}{x}\text{ liters}}.</cmath> | Note that <math>1</math> mile equals <math>\frac 1m</math> kilometers. We have <cmath>\frac{x\text{ miles}}{1\text{ gallon}} = \frac{\frac{x}{m}\text{ kilometers}}{l\text{ liters}} = \frac{1\text{ kilometer}}{\frac{lm}{x}\text{ liters}} = \frac{100\text{ kilometers}}{\frac{100lm}{x}\text{ liters}}.</cmath> | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
+ | |||
+ | == Solution 2 == | ||
+ | Since it can be a bit odd to think of "liters per <math>100</math> km", this statement's numerical value is equivalent to <math>100</math> km per <math>1</math> liter: | ||
+ | |||
+ | <math>1</math> km requires <math>l</math> liters, so the numerator is simply <math>l</math>. Since <math>l</math> liters is <math>1</math> gallon, and <math>x</math> miles is <math>1</math> gallon, we have <math>1\text{ liter} = \frac{x}{l}</math>. | ||
+ | |||
+ | Therefore, the requested expression is <cmath>100\cdot\frac{m}{(\frac{x}{l})} = \boxed{\textbf{(E) } \frac{100lm}{x}}.</cmath> | ||
+ | -Benedict T (countmath1) | ||
==Video Solution 1 (Quick and Easy)== | ==Video Solution 1 (Quick and Easy)== | ||
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~Education, the Study of Everything | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://youtu.be/qACAjp1HSxA | ||
== See Also == | == See Also == |
Latest revision as of 01:32, 25 January 2024
Contents
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 km", this statement's numerical value is equivalent to
km per
liter:
km requires
liters, so the numerator is simply
. Since
liters is
gallon, and
miles is
gallon, we have
.
Therefore, the requested expression is
-Benedict T (countmath1)
Video Solution 1 (Quick and Easy)
~Education, the Study of Everything
Video Solution 2
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.