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Two pipes P and Q would fill an empty cistern in 24 and 32 minutes respectively.Both the pipes being opened together, find when the first pipe must be

turned off so that the empty cistern may be just filled in 16 minutes.


After 12 minutes switch off P See , Work done by Q in 16 minutes -1/32×16=1/2 That means P has to do 1/2 of the work left Therefore Time taken by P,=24×1/2=12 minutes. So after 12minutes P can be closed
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Let volume of cistern be V
Flow rate through pipe P is V/24
Flow rate through pipe Q is  V/32
Flow rate through both pipes = V/32+V/24= 7V/96
let first pipe be turn off after x minute
Cistern will be filled by 7V*x/96 volume
Rest of volume be filled by pipe Q for 16-x minutes
Volume filled by Q for 16-x minute is V*(16-x)/32
So we have 7Vx/96+ V(16-x)/32 = V
             ⇒ 7x/96 + (16-x)/32 = 1
             ⇒ 7x +48 - 3x = 96
             ⇒ 4x = 48
             ⇒   x = 12
First pipe should be closed after 12 minutes.
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