# 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)

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