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Pumping alone at their respective constant rates, one inlet pipe fills

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

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New post 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?

A. 3.25
B. 3.6
C. 4.2
D. 4.4
E. 5.5
[Reveal] Spoiler: OA
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Re: Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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New post 30 Nov 2010, 07:19
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1/2 of capacity in 3 hours

full capacity in 6 hours

---> pump A.


2/3 capacity in 6 hours

(2/3 +1/3) in 6+3 = 9 hours

---> pump B

in 1 hour filled 1/6 +1/9 = 5/18 parts
full in 18/5 hours 3.6 hrs.

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

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New post 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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New post 30 Nov 2015, 19:30
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Re: Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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New post 01 Dec 2015, 02:51
Expert's post
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?

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

Inlet 1 + Inlet 2 combined rate = 6/(1 + 2/3) hours= 6/ (5/3) hours = 18/5 = 3.6 hours
Option B
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Re: Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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New post 01 Dec 2015, 13:04
rite2deepti wrote:
one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours

Time required to fill the full tank is 6 hours
rite2deepti wrote:
a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours.

Time required to fill the full tank is 9 hours
rite2deepti wrote:
How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?

Use the formula - \(\frac{AB}{(A + B)}\)

Or, \(\frac{6*9}{(6 + 9)}\)

Or, \(\frac{54}{15}\)

Or, \(\frac{18}{5}\)

Or, \(3.6\)

Hence answer is (B) 3.6
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Re: Pumping alone at their respective constant rates, one inlet pipe fills [#permalink]

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New post 10 May 2016, 18:21
Attached is a visual that should help.
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Screen Shot 2016-05-10 at 5.50.37 PM.png
Screen Shot 2016-05-10 at 5.50.37 PM.png [ 100.66 KiB | Viewed 209 times ]

Re: Pumping alone at their respective constant rates, one inlet pipe fills   [#permalink] 10 May 2016, 18:21
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Pumping alone at their respective constant rates, one inlet pipe fills

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