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by charlie33 » Sat Jun 27, 2009 9:03 am
A boater travels 21 miles upstream against a 2 mi/h current, then returns downstream to the starting point. If the entire trip took 10 hours, what is the rate of the boat in still water?

A. 4 mi/h
B. 7 mi/h
C. 5 mi/h
D. 6 mi/h
E. 3 mi/h
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Answer

by chintanjadwani » Sat Jun 27, 2009 9:10 am
IMO C[/spoiler]
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by Morgoth » Sat Jun 27, 2009 9:35 am
Let speed of the boat in still water = x

Upstream

Since, the boat is going upstream, more speed is required. The speed would be x+2

D/S = T

T= 21/(x+2)


Downstream

Since, the boat is going downstream, less speed is required. The speed would be x-2

D/S = T

T= 21/(x-2)


Total time = 21/(x+2) + 21/(x-2) = 10

Solve the equation, you'll get

5x^2 -21x -20 = 0

(5x+4)(x-5) = 0

discard 5x+4, coz speed can never be in negative

x-5 = 0

x = 5 mph

Hence C.


Hope this helps.
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by truplayer256 » Sat Jun 27, 2009 9:39 am
Answer should be C. I had the same explanation as Morgoth.
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by mehravikas » Tue Jun 30, 2009 9:37 pm
An easy solution would be to plug in the answer choices. Pick C

U = 3
D = 7

downstream travel time - 21 = T * 7 -> T = 3

upstream travel time - 21 = 3 (10 - T)
21 = 3 * (10-3) -> 21 = 21

Hence answer C.
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