Difference between revisions of "2013 AMC 8 Problems/Problem 11"
(→Solution) |
(→Solution) |
||
| Line 7: | Line 7: | ||
You have to find the hours spent on each day. For people who do not know how to do this, here is a slower way to do it (you will be able to do quickly once you learn it): | You have to find the hours spent on each day. For people who do not know how to do this, here is a slower way to do it (you will be able to do quickly once you learn it): | ||
| − | On Monday, he spent 5 hours. So, 5x = 2 miles. x = <math>\frac{2}{5}</math> | + | On Monday, he spent 5 hours. So, 5x = 2 miles. x = <math>\frac{2}{5} hours</math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=10|num-a=12}} | {{AMC8 box|year=2013|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:54, 27 November 2013
Problem
Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less?
Solution
You have to find the hours spent on each day. For people who do not know how to do this, here is a slower way to do it (you will be able to do quickly once you learn it):
On Monday, he spent 5 hours. So, 5x = 2 miles. x = 
.
See Also
| 2013 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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. 
