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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

Concentration: International Business, General Management

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

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03 Jun 2017, 01:08

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The first pipe fills the empty pool to half in 3 hours --> meaning it will take the first pipe double the time, 6h to fill the whole pool. Now we have the rate of the first pipe: 1/6 The second pipe fills the pool to 2/3 in 6 hours. Multiply capacity and and time by 3/2 to obtain the rate it will take pipe 2 to fill to full capacity: 1/9 Now as we have two machines, two entities working together we can use the work formula: A*B/(A+B)= 6*9/(6+9) =54/15= 3.6 Answer B!

This question can be solved in a couple of different ways, but it's essentially just a Work Formula question.

Work = (A)(B)/(A+B) where A and B are the respective times it takes for two entities to individually complete a task.

To start, we have to figure out how long it takes each pipe to fill the FULL tank....

First pipe = fills 1/2 the tank in 3 hours... so it fills the FULL tank in 6 hours Second pipe = fills 2/3 the tank in 6 hours... so it fills the FULL tank in 9 hours

Working together, the two pipes will fill the tank in (6)(9)/(6+9) = 54/15 = 3 9/15 hours = 3.6 hours

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

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10 Sep 2017, 20:54

rite2deepti wrote:

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?

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

The rate for the first inlet pipe is (1/2)/3 = 1/6 and the rate for the second pipe is (2/3)/6 = 2/18 = 1/9.

If we let t = the number of hours they work together to fill the empty tank, we have:

(1/6)t + (1/9)t = 1

t/6 + t/9 = 1

Multiplying the entire equation by 18, we have:

3t + 2t = 18

5t = 18

t = 18/5 = 3 ⅗ = 3.6

Answer: B
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