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2022 AMC 10A Problems/Problem 4: Difference between revisions

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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)
== Solution 3 ==
A car that gets a gallon for <math>x</math> miles has a fuel efficiency of <math>\frac{1 \text{ gallon}}{x\text{ miles}}.</math> We want to convert these units to liters and kilometers. Since one gallon is <math>l</math> liters, that can be rewritten as <math>\frac{l \text{ liters}}{x\text{ miles}}</math>.
Now to convert the miles into kilometers. <math>1</math> km is <math>m</math> miles, so <math>1</math> mile is <math>\frac{1}{m}</math> km. Plug that in to get <math>\frac{l \text{ liter}}{x * \frac{1}{m} \text{ km}}.</math> Simplifying, we get <math>\frac{lm \text{ liters}}{x \text{ km}}</math>.
But the problem asked for the fuel efficiency of liters per <math>100</math> km, not just <math>1</math> km. So instead of <math>m</math>, we put in <math>100 m</math>, getting
<math>\boxed{\textbf{(E) } \frac{100lm}{x}}</math> as the answer.
~mihikamishra
==Video Solution 1 (Quick and Easy)==
https://youtu.be/JX4u3V2IqY0
~Education, the Study of Everything
==Video Solution 2==
https://youtu.be/qACAjp1HSxA


== See Also ==
== See Also ==

Latest revision as of 21:06, 31 May 2025

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)

Solution 3

A car that gets a gallon for $x$ miles has a fuel efficiency of $\frac{1 \text{ gallon}}{x\text{ miles}}.$ We want to convert these units to liters and kilometers. Since one gallon is $l$ liters, that can be rewritten as $\frac{l \text{ liters}}{x\text{ miles}}$.

Now to convert the miles into kilometers. $1$ km is $m$ miles, so $1$ mile is $\frac{1}{m}$ km. Plug that in to get $\frac{l \text{ liter}}{x * \frac{1}{m} \text{ km}}.$ Simplifying, we get $\frac{lm \text{ liters}}{x \text{ km}}$.

But the problem asked for the fuel efficiency of liters per $100$ km, not just $1$ km. So instead of $m$, we put in $100 m$, getting $\boxed{\textbf{(E) } \frac{100lm}{x}}$ as the answer.

~mihikamishra

Video Solution 1 (Quick and Easy)

https://youtu.be/JX4u3V2IqY0

~Education, the Study of Everything

Video Solution 2

https://youtu.be/qACAjp1HSxA

See Also

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

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