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# work ps pool

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Manager
Joined: 05 Oct 2005
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work ps pool [#permalink]  12 Jul 2006, 01:51
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PLZ Explain ur approach on this one and show the equation used to come to the answer
thanks

A pool has two inlet pipes. The first inlet can fill the pool in 6 hours, the other can fill it in 6 hours. If the first pipe is open for 2 hours and then the second is open. How long will take for the pool to be filled after the second pipe is open.
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Manager
Joined: 05 Oct 2005
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my trial

c the pool capacity
so the first will do 2* 1/6C

so 1/12 OF C SO 11/12 OF W LEFT

(11/12 W)/ 2/6

= 11/4

Is it right what do i miss there?

thanks
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Manager
Joined: 01 Jun 2006
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[quote="mand-y"]:? my trial

c the pool capacity
so the first will do 2* 1/6C

so 1/12 OF C SO 11/12 OF W LEFT

(11/12 W)/ 2/6

= 11/4

Is it right what do i miss there?

thanks[/quote]
I don't think 2*1/6c=1/12c
Because both inlet number 1 and 2 can fill the pool in 6 hours so there must be 6-2=4 hours left till the number 2 makes the pool full of water.
SVP
Joined: 30 Mar 2006
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Pipe A can fill a pool in 6 hrs
Hence Pipe A can fill in 2 hrs 1/3 of the pool.

Remaining pool to be filled = 2/3

Pipe A can fill a pool in 6 hrs
Hence Pipe A can fill 2/3 of pool in 6*2/3 = 4 hrs

Pipe B can fill the pool in 6 hrs
Hence Pipe B cab fill 2/3 of the pool in 6*2/3 = 4 hrs

Let time taken by both the pipes be X
Hence,

1/X = 1/4 + 1/4 = 2/4 =1/2

Hence 2 hrs ,

Hence after the second pipe is opened they both can fill the pool in 2 hrs.
Total time taken = 4
GMAT Instructor
Joined: 04 Jul 2006
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As the two pipes fill the pool at the same rate, we can say that 6 pipe-hours are needed to fill the pool.

If pipe A is left open for 2 hours before pipe B opens, the pipes will have filled the pool after 6-2=4 pipe-hours (i.e the two pipes open for two hours).

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If the first pipe is open for 2 hours the pool will be filled 1/6*2=1/3 of its capacity
so 2/3 of the work left the work rate of two pipes combined is (1/6)*2=1/3
(2/3)/(1/3)=2
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