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Two taps P and Q can fill a cistern in 24 min and 32 min respectively. Both the taps are opened together for a certain time but due to some obstruction the flow of water was restricted to 7/4 of full flow in tap P and 5/3 of full in tap Q. This obstruction is removed after some time and cistern is now filled in 6min from that moment. How long was it before the full flow. A) 8 min B) 3 min C) 5.6 min D) 4.5 min

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Two taps P and Q can fill a cistern in 24 min and 32 min respectively. Both the taps are opened together for a certain time but due to some obstruction the flow of water was restricted to 7/4 of full flow in tap P and 5/3 of full in tap Q. This obstruction is removed after some time and cistern is now filled in 6min from that moment. How long was it before the full flow. A) 8 min B) 3 min C) 5.6 min D) 4.5 min
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D

Let the obstruction remain for X min. 

Hence,

Part of cistern filled in X min + part of cistern filled in 6 min = full cistern

[(7X/4*24)+(5X/3*32)]+[(6/24)+(6/32)] = 1

(12X/96)+(7/16) = 1

12x/96=9/16

Thus,

X = 4.5 min.
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