Question 77941This question is from textbook Addison-Wesley Algebra
: A motorcycle breaks down and the rider has to walk the rest of the way to work.The motorcycle was traveling at 45 mi/h , and the rider walks at a speed of 6 mi/h. The distance from home is 25 miles, and the total time for the trip was 2 hours. How far did the motorcycle go before it broke down?
This question is from textbook Addison-Wesley Algebra
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A motorcycle breaks down and the rider has to walk the rest of the way to work.The motorcycle was traveling at 45 mi/h , and the rider walks at a speed of 6 mi/h. The distance from home is 25 miles, and the total time for the trip was 2 hours. How far did the motorcycle go before it broke down?
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Let x = distance the motorcycle traveled
Then (25-x) = distance he walked
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Write a time equation; Time = distance/speed
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Motor cycle time + walking time = 2 hrs
x/45 + (25-x)/6 = 2
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Multiply equation by 90 to get rid of the denominators, you then have
2x + 15(25-x) = 90(2)
:
2x + 375 - 15x = 180
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2x - 15x = 180 - 375
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-13x = -195
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x = -195/-13
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x = +15 miles by motorcycle
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Then we know the poor guy walk 25 - 15 = 10 miles!
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Check using the time:
15/45 + 10/6 =
1/3 + 5/3 = 6/3 = 2 hrs
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Did this make sense to you?
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