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| − | ==Problem==
| + | #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?
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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 1==
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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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| − | == 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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| − | On route <math>A,</math> 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 <math>A</math> is <math>12</math> minutes. For route <math>B</math> we have to use the equation twice, once for the distance of <math>5- \frac{1}{2} = \frac{9}{2}</math> miles with a speed of <math>\frac{2}{3}</math> mile per minute and a distance of <math>\frac{1}{2}</math> miles at a speed of <math>\frac{1}{3}</math> mile per minute. Thus, the time it takes to go on Route <math>B</math> is <math>\frac{9}{2} \cdot \frac{3}{2} + \frac{1}{2} \cdot 3 = \frac{27}{4} + \frac{3}{2} = \frac{33}{4}</math> minutes. Thus, Route B is <math>12 - \frac{33}{4} = \frac{15}{4} = 3\frac{3}{4}</math> faster than Route <math>A.</math> Thus, the answer is <math>\boxed{\textbf{(B)}\ 3 \frac{3}{4}}.</math>
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| − | ~NH14
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| − | ==See Also==
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| − | {{AMC10 box|year=2021 Fall|ab=A|num-b=3|num-a=5}}
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| − | {{MAA Notice}}
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