A tank can be filled by a tap in 10 minutes and by

Solution

Part filled by Tap A in 1 minute = 1/10 Part filled by Tap B in 1 minute = 1/30 (A+B)’s 5 minute work = 5 × (1/10+1/30) = 5 × ((3+1)/30) = 5 × (4/30) = 2/3 Remaining work = 1 - 2/3 = 1/3 1/30 Part filled by Tap B in = 1 minute 1/3 Part will be filled in = (1/3)/(1/30) = 30/3 = 10 minute Alternate method: Let total capacity of tank = LCM(10, 30) = 30 L So A can fill in 1 minute = 30/10 = 3L & B can fill in 1 minute = 30/30 = 1L So Both taps together filled in 5 minutes =( 3+1)×5 = 4×5 = 20 L So remaining tank = 30 – 20 = 10 So this 10L is completed filled by B tap = 10/1 =10 minutes