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PS: Pool 2

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Director
Joined: 01 Apr 2008
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Schools: IIM Lucknow (IPMX) - Class of 2014
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PS: Pool 2 [#permalink]  01 Apr 2009, 11:07
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100% (02:38) correct 0% (00:00) wrong based on 4 sessions
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With both valves open, the pool will be filled with water in 48 minutes. The first valve alone would fill the pool in 2 hours. What is the capacity of the pool if every minute the second valve admits 50 cubic meters of water more than the first?
* 9000 cubic meters
* 10500 cubic meters
* 11750 cubic meters
* 12000 cubic meters
* 12500 cubic meters
Current Student
Joined: 13 Jan 2009
Posts: 372
Location: India
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Kudos [?]: 89 [0], given: 1

Re: PS: Pool 2 [#permalink]  01 Apr 2009, 15:12
A fills tank in 120 mins.
A+B fill in 48 mins
So, B fills in 80 mins. ( A*T/A-T)

Let x be the capacity.

A will in one min x/120
B will in one min x/80

Now X/80 - x/120 = 50

So x is 12000

D.
Intern
Joined: 28 Feb 2010
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Re: PS: Pool 2 [#permalink]  30 Mar 2010, 20:30
can you Explain how did you get the min of B?
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Re: PS: Pool 2 [#permalink]  31 Mar 2010, 13:30
First valve allows=x cubic meter/min
Second valve allows=x+50 cubic meter/min
total capacity=48(x+x+50)
and total capacity as well=120x

so 48(2x+50)=120x
x=100
and capacity 120*100=12000 cubic meter
Intern
Joined: 28 Feb 2010
Posts: 5
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Re: PS: Pool 2 [#permalink]  31 Mar 2010, 13:33
saqibbaig wrote:
First valve allows=x cubic meter/min
Second valve allows=x+50 cubic meter/min
total capacity=48(x+x+50)
and total capacity as well=120x

so 48(2x+50)=120x
x=100
and capacity 120*100=12000 cubic meter

ThankSSSS.perfect
Re: PS: Pool 2   [#permalink] 31 Mar 2010, 13:33
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