2019 AMC 8 Problems/Problem 16
Contents
Problem 16
Qiang drives miles at an average speed of miles per hour. How many additional miles will he have to drive at miles per hour to average miles per hour for the entire trip?
Solution 1
The only option that is easily divisible by is , which gives 2 hours of travel. And, the formula is .
And, = .
Thus, .
Both are equal and thus our answer is
Solution 2
To calculate the average speed, simply evaluate the total distance over the total time. Let the number of additional miles he has to drive be Therefore, the total distance is and the total time (in hours) is We can set up the following equation: Simplifying the equation, we get Solving the equation yields so our answer is .
Solution 3
If he travels miles at a speed of miles per hour, he travels for 30 min. Average rate is total distance over total time so , where d is the distance left to travel and t is the time to travel that distance. Solve for to get . You also know that he has to travel miles per hour for some time, so . Plug that in for d to get and and since , , the answer is .
Solution 4
Let be the amount of hours Qiang drives after his first 15 miles. Average speed, which we know is mph, means total distance over total time. For 15 miles at 30 mph, the time taken is hour, so the total time for this trip would be hours. For the total distance, 15 miles are traveled in the first part and miles in the second. This gives the following equation:
Cross multiplying, we get that , and simple algebra gives . In 2 hours traveling at 55 mph, the distance traveled is , which is choice .
~TaeKim
Video Solution
https://www.youtube.com/watch?v=OC1KdFeZFeE
Associated Video
- happytwin
https://www.youtube.com/watch?v=0rcDe2bDRug
Video Solution
Solution detailing how to solve the problem:
https://www.youtube.com/watch?v=sEZ0sM-d1FA&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=17
Video Solution
- Soo, DRMS, NM
Video Solution
~savannahsolver
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.