Three pipes A, B and C can fill a tank in t, (t+12) & (t-12) hours respectively. Pipe A opened for d hours in the tank and after that B and C also opened in the tank, while all three pipes opened for (d+6) hours and after that pipes A and C closed. If the remaining tank filled by pipe B in (d+2) hours and tank filled by pipe A in three hours equal to tank filled by B in four hours, then find in (t+3d) hours what portion of tank will be filled by pipe B. 

Solution

Given, tank filled by pipe A in three hours equal to tank filled by B in four hours Let efficiency of pipe A and B be a & b respectively So, a×3=b×4 a/b=4/3   And ratio of time taken by pipe A and B will be 3 : 4 ATQ.  t/(t+12)=3/4    t=36 So, time taken by pipe A to fill the tank = 36 hours Time taken by pipe B to fill the tank = (36 + 12) = 48 hours Time taken by pipe C to fill the tank = (36 – 12) = 24 hours So, total capacity of tank = 144 units (L.C.M. of 36, 48 and 24)  Efficiency of pipe A = 144/36=4 units/hour Efficiency of pipe B = 144/48=3 units/hour Efficiency of pipe C = 144/24= 6 units/hour  ATQ. 4×d+13(d+6)+3(d+2)=144   4d+13d+78+3d+6=144 20d = 60 d = 3 Required portion = [((36+3×3)×3)/144] = 15/16