There are four pipes A, B, C and D where A and B are

Solution

Let’s assume the total capacity of the tank is 72 units. The efficiency of pipe A is 50% more than that of pipe B. Let’s assume the efficiency of pipe B is ‘4y’. efficiency of pipe A = 4y of (100+50)% = 4y of 150% = (4y x 150)/100 = 600y/100 = 6y units/hour The efficiency of pipe C is 25% less than that of pipe B. efficiency of pipe C = (100-25)% of (efficiency of pipe B) efficiency of pipe C = 75% of (efficiency of pipe B) efficiency of pipe C = 75% of 4y = -3y units/hour [Here the negative sign represents the pipe used to empty the tank.] Pipe A and C together can fill the tank in (a+9) hours. (a+9)x(6y-3y) = 72 (a+9)x3y = 72    Eq.(i) Pipe B alone can fill 75% of the tank in (a-1.5) hours. Time taken by Pipe B alone to fill 100% of the tank = ((a-1.5)/75)x100 = ((a-1.5)/3)x4 So ((a-1.5)/3)x4x4y = 72    Eq.(ii) We can say that Eq.(i) = Eq.(ii). (a+9)x9 = (a-1.5)x16 9a+81 = 16a-24 16a-9a = 81+24 7a = 105 a = 15 Pipe D alone can empty half of the tank in (a-3) hours. Time taken by Pipe D alone to empty the tank = 2x(a-3) So 2x(a-3)x(efficiency of Pipe D) = 72 Put the value of ‘a’ in the above equation. 2x(15-3)x(efficiency of Pipe D) = 72 2x12x(efficiency of Pipe D) = 72 24x(efficiency of Pipe D) = 72 efficiency of Pipe D = -3 units/hour [Here the negative sign represents the pipe used to empty the tank.] Put the value of ‘a’ in Eq.(i) to obtain the value of ‘y’. (15+9)x3y = 72 24x3y = 72 y = 1 Time taken by pipe A and D together to fill the empty tank completely = capacity of tank/(efficiency of pipe A and D together) = 72/(6y-3) Put the value of ‘y’ in the above equation. = 72/(6-3) = 72/3 = 24 hours