Difference between revisions of "2013 AMC 8 Problems/Problem 11"
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Thus, the solution to this problem is <math>\dfrac{1}{15}\times 60=\boxed{\textbf{(D)}\ 4}</math> | Thus, the solution to this problem is <math>\dfrac{1}{15}\times 60=\boxed{\textbf{(D)}\ 4}</math> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/b3z2bfTLk4M ~savannahsolver | ||
==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}} | ||
Latest revision as of 16:57, 5 May 2022
Contents
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 
Video Solution
https://youtu.be/b3z2bfTLk4M ~savannahsolver
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. 
