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Flow Rida (m05q09)

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Joined: 15 Jun 2012
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Re: Flow Rida (m05q09) [#permalink] New post 27 Feb 2013, 12:45
Very nice question. My answer is D:

Inflow volume in 1 hour = number of pipes x rate = 12 x 1 = 12
Outflow volume in 1 hour = number of pipes x rate = Q x 1.5 = 1.5Q
-----------------------------------------------------------------------------
Remaining volume = Inflow volume - outflow volume = 12 - 1.5Q

Because rates are constant, after 12 hour we have --> 12 x (12 - 1.5Q) = 54
--> Q = 5
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"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."

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Re: Flow Rida (m05q09) [#permalink] New post 01 Mar 2013, 19:45
Mashuri, MKParris...both have used a different approach...good ones...
vipinktyagi...simplified approach...i should say easy to understand...

+1 kudos for all...
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Re: Flow Rida (m05q09) [#permalink] New post 25 Feb 2014, 05:18
it can be done in 1 step and less than 45 sec

54=1212*1-x*1.5}
54=total work to be done
12=time taken
12*1 =nos of pipe multiplied by work done by each pipe
1.5*x where X is the nos of pipes , On solvin We get 5 D
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Re: Flow Rida (m05q09) [#permalink] New post 21 Apr 2014, 05:49
a: number of big pipes. After 12hrs the tank is full -> 12(12*1-1.5a)=54-> 1.5a=7.5-> a=5

Choose D
Re: Flow Rida (m05q09)   [#permalink] 21 Apr 2014, 05:49

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Flow Rida (m05q09)

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