Difference between revisions of "2013 AMC 8 Problems/Problem 11"
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To find the amount of time saved, subtract the two amounts: <math>\frac{94}{60} \text {hours}</math> - <math>\frac{3}{2} \text {hours}</math> = <math>\frac{4}{60} \text {hours}</math>. To convert this to minutes, use the conversion rate; multiply by 60. | To find the amount of time saved, subtract the two amounts: <math>\frac{94}{60} \text {hours}</math> - <math>\frac{3}{2} \text {hours}</math> = <math>\frac{4}{60} \text {hours}</math>. To convert this to minutes, use the conversion rate; multiply by 60. | ||
− | Thus, the solution to this problem | + | Thus, the solution to this problem is <math>\dfrac{4}{60}\times 60=\boxed{\textbf{(D)}\ 4}</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 20:06, 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
On Monday, he was at a rate of 5 mph. So, 5x = 2 miles. x = . For Wednesday, he jogged at a rate of 3 mph. Therefore, 3x = 2 miles. x = . On Friday, he jogged at a rate of 4 hours. So, 4x = 2 miles. x=.
Add the hours = + + = .
Once you find this, answer the actual question by finding the amount of time Grandfather would take by jogging at 4 mph per day. Set up the equation, 4x = 2 miles 3 days. x = .
To find the amount of time saved, subtract the two amounts: - = . To convert this to minutes, use the conversion rate; multiply by 60.
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 |
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