Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AMC 10A Problems/Problem 4"

(→‎Solution 2)
(→‎Solution 2)
Line 15: Line 15:
 
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:
 
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\text{km}</math> requires <math>l\text{ liters},</math> 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 liter = \frac{x}{l}</math>.
+
<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 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>
 
Therefore, the requested expression is <cmath>100\cdot\frac{m}{(\frac{x}{l})} = \boxed{\textbf{(E) } \frac{100lm}{x}}.</cmath>

Revision as of 12:27, 14 November 2022

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 $1 liter = \frac{x}{l}$.

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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems/Problem_4&oldid=181424"