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Flow Rida (m05q09)

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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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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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