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Work Problem....a little help please

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Work Problem....a little help please

by smoore320 » Thu Dec 11, 2008 9:27 pm
Working alone, a small pump takes twice as long as a large pump takes to fill an empty tank. Working together at their respective constant rates, the pumps can fill the tank in 6 hours. How many hours would it take the small pump to fill the tank working alone?


(A) 8
(B) 9
(C) 12
(D) 15
(E) 18



OA is E.
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by anayeri » Thu Dec 11, 2008 9:43 pm
plug and chug the work problem formula, where S is small, and L is large.

(1/S)+(1/L) = (1/6).

S=2L, replace that in the formula.

(1/2L)+(1/L)=(1/6)

isolating L we get L=[6*(2+1)]/2, so L=9, and since S=2L, S=18.
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by cramya » Thu Dec 11, 2008 9:45 pm
Let the large pump take t hours to fill a tank
Let the small pump take 2t hours

Rate of large pump = 1/t
rate of smaller pump = 1/2t

Rate * time = work (working together add the rates)

(1/t+1/2t) * 6 = 1 (1 job)

t = 9

2t = 18 (smaller pump takes 18 hours to fill the tank)

E)
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