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Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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30 Nov 2010, 07:03

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Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?

Re: Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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30 Nov 2010, 07:30

Inlet 1 fills x/6 in 1 hour inlet 2 fills (2x/3)/6 = x/9 in 1 hour

so together in 1 hour they will fill x/6 + x/9 = 5x/18 which means to fill 5x/18 it takes 1 hour => to fill x it will take 18/5=3.6 hrs
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Re: Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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30 Nov 2015, 19:30

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Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?

A. 3.25 B. 3.6 C. 4.2 D. 4.4 E. 5.5

Assume the volume of the container to be 6 Lts We have assumed 6 because 6 is the LCM of 2 and 3, the denominators of the capacity filled.

Inlet 1: 1/2 of capacity in 3 hours i.e. 3 lts in 3 hours Hence 1 lts/hour

Inlet 2: 2/3 of capacity in 6 hours i.e. 4 lts in 6 hours Hence 2/3 lts/hour

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