2013 AMC 8 Problem 11

Below is the video solution and professionally curated solution for Problem 11 of the 2013 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 8 solutions, or check the answer key.

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Concepts:distance rate and timeunit conversion

Difficulty rating: 1280

11.

Ted's grandfather used his treadmill on 33 days this week. He went 22 miles each day.

On Monday he jogged at a speed of 55 miles per hour. He walked at the rate of 33 miles per hour on Wednesday and at 44 miles per hour on Friday.

If Grandfather had always walked at 44 miles per hour, he would have spent less time on the treadmill. How many minutes less?

11

22

33

44

55

Video solution:
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Written solution:

Grandfather spent 25\frac{2}{5} hours jogging on Monday. He spent 23\frac{2}{3} hours walking on Wednesday and 24=12\frac{2}{4}=\frac{1}{2} hours on Friday. Converting these times to minutes, we get 2560=24 minutes\dfrac{2}{5} \cdot 60 = 24~\mathrm{minutes} 2360=40 minutes\dfrac{2}{3} \cdot 60 = 40~\mathrm{ minutes} 1260=30 minutes\dfrac{1}{2} \cdot 60 = 30~\mathrm{ minutes} for Monday, Wednesday, and Friday respectively. Therefore, Grandfather totaled 9494 minutes of exercise throughout the week.

If he walked at a pace of 44 miles per hour each day, he would have spent 24=12\frac{2}{4}=\frac{1}{2} hours each day walking. This equals 2460=30\frac{2}{4} \cdot 60 = 30 minutes every day. If he did this for 33 days, he would have totaled 330=903 \cdot 30 = 90 minutes of exercise.

Therefore, Grandfather would have walked for 9490=494 - 90 = 4 less minutes.

Thus, D is the correct answer.

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