SOLUTION: Rowing upstream for 5 mi against a strong current, your speed is 20% less than your normal rowing speed. On the return trip downstream, your speed is 20% greater than your normal r

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Question 616651: Rowing upstream for 5 mi against a strong current, your speed is 20% less than your normal rowing speed. On the return trip downstream, your speed is 20% greater than your normal rowing speed. Will you finish the whole trip in the same time as if you had rowed at normal speed throughout? How much longer than usual will your total time be?
Answer by lwsshak3(11628) About Me  (Show Source):
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Rowing upstream for 5 mi against a strong current, your speed is 20% less than your normal rowing speed. On the return trip downstream, your speed is 20% greater than your normal rowing speed. Will you finish the whole trip in the same time as if you had rowed at normal speed throughout? How much longer than usual will your total time be?
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let x=normal rowing speed
Travel time=distance/speed
Travel time downstream =5/1.2x
Travel time upstream =5/.8x
Total travel time=5/1.2x+5/.8x
=5(1/1.2x+1/.8x)
=5(2/.96x)
=10/.96x
..
Total travel time at normal rowing speed=10/x
It takes less time at normal rowing speed
difference=(10/.96x)-(10/x)
=10(1/.96x-1/x)
=10(1-.96)/.96x
=10(.04/.96x
=4/9.6x