Difference between revisions of "2013 AMC 8 Problems/Problem 11"
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Thus, the solution to this problem is <math>\dfrac{4}{60}\times 60=\boxed{\textbf{(D)}\ 4}</math> | Thus, the solution to this problem is <math>\dfrac{4}{60}\times 60=\boxed{\textbf{(D)}\ 4}</math> | ||
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==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 11:31, 4 January 2018
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
We use that fact that . Let d= distance, r= rate or speed, and t=time. In this case, let represent the time.
On Monday, he was at a rate of . So, .
For Wednesday, he walked at a rate of . Therefore, .
On Friday, he walked at a rate of . So, .
Adding up the hours yields + + = .
We now find the amount of time Grandfather would have taken if he walked at per day. Set up the equation, .
To find the amount of time saved, subtract the two amounts: - = . To convert this to minutes, we multiply by .
Thus, the solution to this problem is
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.