2021 Fall AMC 12A Problems/Problem 3
- 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.
Contents
Problem
Mr. Lopez has a choice of two routes to get to work. Route A is miles long, and his average speed along this route is miles per hour. Route B is miles long, and his average speed along this route is miles per hour, except for a -mile stretch in a school zone where his average speed is miles per hour. By how many minutes is Route B quicker than Route A?
Solution 1
If Mr. Lopez chooses Route A, then he will spend hour, or minutes.
If Mr. Lopez chooses Route B, then he will spend hour, or minutes.
Therefore, Route B is quicker than Route A by minutes.
~MRENTHUSIASM
Solution 2
We use the equation to solve this problem. Recall that mile per hour is equal to mile per minute.
For Route A, the distance is miles and the speed to travel this distance is mile per minute. Thus, the time it takes on Route A is minutes.
For Route B, we have to use the equation twice: once for the distance of miles with a speed of mile per minute and a distance of miles at a speed of mile per minute. Thus, the time it takes to go on Route B is minutes. Thus, Route B is faster than Route A. Thus, the answer is
~NH14
Video Solution (Simple and Quick)
~Education, the Study of Everything
Video Solution
~savannahsolver
Video Solution
~Charles3829
Video Solution by TheBeautyofMath
for AMC 10: https://youtu.be/o98vGHAUYjM?t=389
for AMC 12: https://youtu.be/jY-17W6dA3c?t=158
~IceMatrix
Video Solution
~Lucas
Video Solution
~Education, the Study of Everything
See Also
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
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 (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.