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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 4"

(Created page with "==See Also== {{AMC10 box|year=2021 Fall|ab=A|num-b=3|num-a=5}} {{MAA Notice}}")
 
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==Problem==
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Mr. Lopez has a choice of two routes to get to work. Route A is <math>6</math> miles long, and his average speed along this route is <math>30</math> miles per hour. Route B is <math>5</math> miles long, and his average speed along this route is <math>40</math> miles per hour, except for a <math>\frac{1}{2}</math>-mile stretch in a school zone where his average speed is <math>20</math> miles per hour. By how many minutes is Route B quicker than Route A?
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<math>\textbf{(A)}\ 2 \frac{3}{4}  \qquad\textbf{(B)}\  3 \frac{3}{4} \qquad\textbf{(C)}\  4 \frac{1}{2} \qquad\textbf{(D)}\
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5 \frac{1}{2} \qquad\textbf{(E)}\ 6 \frac{3}{4}</math>
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==Solution==
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If Mr. Lopez chooses Route A, then he will spend <math>\frac{6}{30}=\frac{1}{5}</math> hour, or <math>12</math> minutes.
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If Mr. Lopez chooses Route B, then he will spend <math>\frac{9/2}{40}+\frac{1/2}{20}=\frac{11}{80}</math> hour, or <math>8\frac14</math> minutes.
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Therefore, Route B is quicker than Route A by <math>12-8\frac14=\boxed{\textbf{(B)}\  3 \frac{3}{4}}</math> minutes.
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~MRENTHUSIASM
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==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=3|num-a=5}}
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:32, 22 November 2021

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

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

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

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

~MRENTHUSIASM

See Also

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

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