PIPE can fill or empty a tank in x hours then in 1 hour it can fill or empty 1/x part of the tankCalculation:Let, the total tank is 1 partTap A can fill the tank in 6 hoursThat means in 6 hours tap A can fill 1 part of the tank⇒ In 1 hour tap A can fill 1/6 part of the tankTap B can empty the tank in 10 hoursThat means in 10 hours tap B can empty 1 part of the tank⇒ In 1 hour tap B can empty 1/10 part of the tankIn 1 hour part of the tank will be filled when both taps are opened together is (1/6 – 1/10)⇒ (5 - 3)/30⇒ 2/30⇒ 1/15To fill the tank time taken is 1/(1/15) hours⇒ 15 hours∴ If both taps are opened together the tank will be filled in 15 hours. ALTERNATIVE Solution:Tap A can fill the tank in 6 hours and tap B can empty the same tank in 10 hoursLCM of 6 and 10 is 30Let, the tank be of 30 UNITSIN 1 hour tap A can fill (30/6) = 5 unit of the tankIn 1 hour tap B can empty (30/10) = 3 unit of the tankWhen both the taps are opened together in 1 hour tank will be fill (5 - 3) = 2 unit∴ To fill the tank total time needs (30/2) = 15 hours