2021 Fall AMC 12A Problems/Problem 3

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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 $12$ 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

Solution 2

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

On 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

Solution 3

The time spent on Route $A$ is \begin{align*} \frac{6 \mbox{ miles}}{30 \mbox{ miles per hour}} & = \frac{1}{5} \mbox{ hours} \\ & = \frac{1}{5} \cdot 60 \mbox{ minutes} \\ & = 12 \mbox{ minutes} . \end{align*}

The time spent on Route $B$ is \begin{align*} \frac{\left( 5 - \frac{1}{2} \right) \mbox{ miles}}{40 \mbox{ miles per hour}} + \frac{\frac{1}{2} \mbox{ miles}}{20 \mbox{ miles per hour}} & = \frac{9}{80} \mbox{ hours} + \frac{1}{40} \mbox{ hours}  \\ & = \frac{11}{80} \mbox{ hours} \\ & = \frac{11}{80} \cdot 60 \mbox{ minutes} \\ & = \frac{33}{4} \mbox{ minutes} . \end{align*}

Therefore, the number of minutes Route $B$ is quicker than Route $A$ is $12 - \frac{33}{4} = 3 \frac{3}{4}$.

Therefore, the answer is $\boxed{\textbf{(B) }3 \frac{3}{4}}$.

~Steven Chen (www.professorchenedu.com)

See Also

2021 Fall AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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