Time taken by pipe A and pipe B to fill up the cistern is 12 minutes whereas time taken by pipe C to empty the whole cistern is twice that of time taken by A and B together to fill the tank and ratio of time taken by pipe A and B to fill up the tank when operating alone is 2 : 3. If pipe A and B are alternatively opened every 6 minutes and C is opened with pipe B but closed when pipe A is opened then calculate total time taken to fill up the whole cistern. Correct Answer 48 minutes Given: Pipe A + Pipe B = 12 minutes Pipe C = 24/2 minutes Pipe A and B alternatively opening = 6 minutes Pipe A: Pipe B = 2 : 3 Formula: If time taken by pipe A to fill tank is ‘a’ minutes and by pipe B is ‘b’ minutes to empty the tank then total time taken to fill the tank t ⇒ (1/ t) = (1/a) - (1/b) If time taken by pipe A to fill tank is ‘a’ minutes and by pipe B is ‘b’ minutes then total time taken to fill the tank t ⇒ (1/ t) = (1/ a) + (1/b) Calculation: Let time taken by A = X ⇒ Time taken by B = 3 × X/2 Placing the assumptions in equation, ⇒ 1/ 12 = (1/X) + (2/3X) ⇒ 1/ 12 = (3 + 2)/3X ⇒ 3X = 60 ⇒ X = 20 Time taken by B to fill up the tank = 3 × 20/2 ⇒ Time taken by B = 30 minutes Now, time taken by A = 20 minutes, time taken by B = 30 minutes and time taken by C = 24 minutes (To drain the water) ⇒ LCM of 20, 30 and 24 = 600 ⇒ Efficiency of Pipe A = 30 units/minute (∵ 600/20 = 30) ⇒ Efficiency of pipe B = 20 units/minute ⇒ Efficiency of pipe C = 25 units/minute ⇒ after every 6 minutes number of units filled by pipe A = 180 units ⇒ after every 6 minutes number of units filled by pipe B = 120 units ⇒ Number units removed by pipe C when opened for 6 minutes = 150 units Hence, after every 12 minutes number of units filled = (180 + 120 – 150) ⇒ after every 12 minutes number of units filled = 150 ⇒ Total pairs of such adjustment = 600/150 ⇒ Pairs of adjustment = 4 ⇒ One pair takes 12 minutes so, total time taken to fill the cistern = 12 × 4 ∴ it takes 48 minutes to fill up the cistern