Two pipes A&B can fill a cistern in 32 mins and 48 mins respectively. Both pipes are opened together for some times and then pipe B is turned off. Find when pipe B must be turned off so that the cistern get filled in 24 mins..
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Solution:
Let the total amount of work be 192 units -------(L.C.M of 32, 48, and 64)
Amount of work done by pipe A in one hour = 192/32 = 6 units
Amount of work done by pipe B in one hour = 192/48 = 4 units
Amount of work done by pipe C in one hour = -192/64 = -3 units
Let Pipe A and Pipe C operate for x hours. After x hours, Pipe A is turned off and Pipe B is turned on. Now Pipe B and Pipe C will operate for remaining 112 – x hours.
Work done by Pipe A and C in x hours = (6 – 3) × x units
Work done by Pipe B and C in (112 – x) hours = (4 – 3) × (112 – x) units.
Total work done should be 192 units.
⇒ (6 – 3) × x + (4 – 3) × (112 – x) = 192
⇒ 3x + 112 – x = 192
⇒ 2x = 80
⇒ x = 40
∴ Pipe B was turned on for 112 – 40 = 72 hours.
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