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

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==Problem==
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#REDIRECT [[2021_Fall_AMC_12A_Problems/Problem_3]]
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?
 
 
 
<math>\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}</math>
 
 
 
==Solution==
 
If Mr. Lopez chooses Route A, then he will spend <math>\frac{6}{30}=\frac{1}{5}</math> hour, or <math>12</math> minutes.
 
 
 
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.
 
 
 
Therefore, Route B is quicker than Route A by <math>12-8\frac14=\boxed{\textbf{(B)}\  3 \frac{3}{4}}</math> minutes.
 
 
 
~MRENTHUSIASM
 
 
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=3|num-a=5}}
 
{{MAA Notice}}
 

Latest revision as of 17:44, 23 November 2021