SOLUTION: Suppose that Arlene can mow the entire lawn in 40 minutes less time with the power mower than she can with the push mower. One day the power mower broke down after she had been mow
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-> SOLUTION: Suppose that Arlene can mow the entire lawn in 40 minutes less time with the power mower than she can with the push mower. One day the power mower broke down after she had been mow
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Question 60965This question is from textbook Algebra for College Students
: Suppose that Arlene can mow the entire lawn in 40 minutes less time with the power mower than she can with the push mower. One day the power mower broke down after she had been mowing for 30 minutes. She finished the lawn with the push mower in 20minutes. How long does it take Arlene to mow the entire lawn the power mower? This question is from textbook Algebra for College Students
You can put this solution on YOUR website! Suppose that Arlene can mow the entire lawn in 40 minutes less time with the power mower than she can with the push mower. One day the power mower broke down after she had been mowing for 30 minutes. She finished the lawn with the push mower in 20minutes. How long does it take Arlene to mow the entire lawn the power mower?
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Let the completed job = 1
Let t = time required to mow the lawn with power mower (minutes)
Let (t+40) = time required with a push mower
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Using both mowers: = 1
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Eliminate denominators, mult equation by t(t+40), resulting in:
30(t+40) + 20t = t(t+40)
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30t + 1200 + 20t = t^2 + 40t; mult what's in brackets
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-t^2 + 30t + 20t - 40t + 1200 = 0; arrange as a quadratic eq on the left
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-t^2 + 10t + 1200 = 0
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t^2 - 10t - 1200 = 0; mult by -1 to get t^2 positive, easier to factor
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(t-40)(t+30) = 0
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Positive solution
t = +40 min for power mower to do the job
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Check: = 1
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