2022 AMC 10A Problems/Problem 4

Problem

In some countries, automobile fuel efficiency is measured in liters per $100$ kilometers while other countries use miles per gallon. Suppose that 1 kilometer equals $m$ miles, and $1$ gallon equals $l$ liters. Which of the following gives the fuel efficiency in liters per $100$ kilometers for a car that gets $x$ miles per gallon?

$\textbf{(A) } \frac{x}{100lm} \qquad \textbf{(B) } \frac{xlm}{100} \qquad \textbf{(C) } \frac{lm}{100x} \qquad \textbf{(D) } \frac{100}{xlm} \qquad \textbf{(E) } \frac{100lm}{x}$

Solution 1

The formula for fuel efficiency is $\frac{\text{Distance}}{\text{Gas Consumption}}.$ Note that $1$ mile equals $\frac 1m$ kilometers. We have $\frac{x\text{ miles}}{1\text{ gallon}} = \frac{\frac{x}{m}\text{ kilometers}}{l\text{ liters}} = \frac{1\text{ kilometer}}{\frac{lm}{x}\text{ liters}} = \frac{100\text{ kilometers}}{\frac{100lm}{x}\text{ liters}}.$ Therefore, the answer is $\boxed{\textbf{(E) } \frac{100lm}{x}}.$

~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 $100$ km per $1$ liter:

$1$ km requires $l$ liters, so the numerator is simply $l$ . Since $l$ liters is $1$ gallon, and $x$ miles is $1$ gallon, we have $1\text{ liter} = \frac{x}{l}$ .

Therefore, the requested expression is $100\cdot\frac{m}{(\frac{x}{l})} = \boxed{\textbf{(E) } \frac{100lm}{x}}.$ -Benedict T (countmath1)

Video Solution 1 (Quick and Easy)

https://youtu.be/JX4u3V2IqY0

~Education, the Study of Everything

Video Solution 2

https://youtu.be/qACAjp1HSxA

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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