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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 3"

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== Solution 2 ==
 
== Solution 2 ==
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We use the equation <math>d=st</math> to solve this problem. Recall that <math>1\text{ mile per hour}=\frac{1}{60}\text{ mile per minute}.</math>
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We use the equation <math>d=st</math> to solve this problem. Recall that <math>1</math> mile per hour is equal to <math>\frac{1}{60}</math> mile per minute.
  
 
For Route A, the distance is <math>6</math> miles and the speed to travel this distance is <math>\frac{1}{2}</math> mile per minute. Thus, the time it takes on Route A is <math>12</math> minutes.
 
For Route A, the distance is <math>6</math> miles and the speed to travel this distance is <math>\frac{1}{2}</math> mile per minute. Thus, the time it takes on Route A is <math>12</math> minutes.
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~NH14
 
~NH14
 +
 +
==Video Solution (Simple and Quick)==
 +
https://youtu.be/jE8kjbkcrwI
 +
 +
~Education, the Study of Everything
  
 
==Video Solution==
 
==Video Solution==
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~savannahsolver
 
~savannahsolver
 +
 +
==Video Solution==
 +
https://youtu.be/F4om1fIfYYw
 +
 +
~Charles3829
 +
 +
==Video Solution by TheBeautyofMath==
 +
for AMC 10: https://youtu.be/o98vGHAUYjM?t=389
 +
 +
for AMC 12: https://youtu.be/jY-17W6dA3c?t=158
 +
 +
~IceMatrix
 +
 +
==Video Solution==
 +
https://youtu.be/Si1AMjHYxGE
 +
 +
~Lucas
 +
 +
==Video Solution==
 +
https://youtu.be/jE8kjbkcrwI
 +
 +
~Education, the Study of Everything
  
 
==See Also==
 
==See Also==

Latest revision as of 10:04, 16 June 2023

The following problem is from both the 2021 Fall AMC 10A #4 and 2021 Fall AMC 12A #3, so both problems redirect to this page.

Problem

Mr. Lopez has a choice of two routes to get to work. Route A is $6$ miles long, and his average speed along this route is $30$ miles per hour. Route B is $5$ miles long, and his average speed along this route is $40$ miles per hour, except for a $\frac{1}{2}$-mile stretch in a school zone where his average speed is $20$ miles per hour. By how many minutes is Route B quicker than Route A?

$\textbf{(A)}\ 2 \frac{3}{4} \qquad\textbf{(B)}\ 3 \frac{3}{4} \qquad\textbf{(C)}\ 4 \frac{1}{2} \qquad\textbf{(D)}\ 5 \frac{1}{2} \qquad\textbf{(E)}\ 6 \frac{3}{4}$

Solution 1

If Mr. Lopez chooses Route A, then he will spend $\frac{6}{30}=\frac{1}{5}$ hour, or $\frac{1}{5}\cdot60=12$ minutes.

If Mr. Lopez chooses Route B, then he will spend $\frac{9/2}{40}+\frac{1/2}{20}=\frac{11}{80}$ hour, or $\frac{11}{80}\cdot60=8\frac14$ minutes.

Therefore, Route B is quicker than Route A by $12-8\frac14=\boxed{\textbf{(B)}\ 3 \frac{3}{4}}$ minutes.

~MRENTHUSIASM

Solution 2

We use the equation $d=st$ to solve this problem. Recall that $1$ mile per hour is equal to $\frac{1}{60}$ mile per minute.

For Route A, the distance is $6$ miles and the speed to travel this distance is $\frac{1}{2}$ mile per minute. Thus, the time it takes on Route A is $12$ minutes.

For Route B, we have to use the equation twice: once for the distance of $5- \frac{1}{2} = \frac{9}{2}$ miles with a speed of $\frac{2}{3}$ mile per minute and a distance of $\frac{1}{2}$ miles at a speed of $\frac{1}{3}$ mile per minute. Thus, the time it takes to go on Route B is $\frac{9}{2} \cdot \frac{3}{2} + \frac{1}{2} \cdot 3 = \frac{27}{4} + \frac{3}{2} = \frac{33}{4}$ minutes. Thus, Route B is $12 - \frac{33}{4} = \frac{15}{4} = 3\frac{3}{4}$ faster than Route A. Thus, the answer is $\boxed{\textbf{(B)}\ 3 \frac{3}{4}}.$

~NH14

Video Solution (Simple and Quick)

https://youtu.be/jE8kjbkcrwI

~Education, the Study of Everything

Video Solution

https://youtu.be/MIsp9QysU7Y

~savannahsolver

Video Solution

https://youtu.be/F4om1fIfYYw

~Charles3829

Video Solution by TheBeautyofMath

for AMC 10: https://youtu.be/o98vGHAUYjM?t=389

for AMC 12: https://youtu.be/jY-17W6dA3c?t=158

~IceMatrix

Video Solution

https://youtu.be/Si1AMjHYxGE

~Lucas

Video Solution

https://youtu.be/jE8kjbkcrwI

~Education, the Study of Everything

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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 12 Problems and Solutions
2021 Fall 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

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