Difference between revisions of "2022 AMC 10A Problems/Problem 4"
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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</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>. | + | <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> | Therefore, the requested expression is <cmath>100\cdot\frac{m}{(\frac{x}{l})} = \boxed{\textbf{(E) } \frac{100lm}{x}}.</cmath> | ||
-Benedict T (countmath1) | -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>. | ||
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+ | 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{ miles}}.</math> Simplifying, we get <math>\frac{lm \text{ liters}}{x \text{ miles}}</math>. | ||
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+ | 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)== | ==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 11:37, 9 November 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)
Solution 3
A car that gets a gallon for miles has a fuel efficiency of We want to convert these units to liters and kilometers. Since one gallon is liters, that can be rewritten as .
Now to convert the miles into kilometers. km is miles, so mile is km. Plug that in to get Simplifying, we get .
But the problem asked for the fuel efficiency of liters per km, not just km. So instead of , we put in , getting as the answer.
~mihikamishra
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 |
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