2021 AMC 10 A Fall

Complete problem set with solutions and individual problem pages

Problem 4 Easy

Mr. Lopez has a choice of two routes to get to work. Route \text{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?(2021 AMC Fall 10A, Question #4)

  • A.

    2 \frac{3}{4}

  • B.

    3 \frac{3}{4}

  • C.

    4 \frac{1}{2}

  • D.

    5 \frac{1}{2}

  • E.

    6 \frac{3}{4}

Answer:B

Solution 1:

If Mr. Lopez chooses Route A, then he will spend \frac{6}{30}=\frac{1}{5} hour, or \frac{1}{5} \cdot 60=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} \cdot 60=8 \frac{1}{4} minutes. Therefore, Route B is quicker than Route A by 12-8 \frac{1}{4}= (B) 3 \frac{3}{4} minutes.

Solution 2:

We use the equation d=s t 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 \text{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 (B) 3 \frac{3}{4}.