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Let’s assume the capacity of the tank is 600 units. Pipe T can empty a completely filled tank in 50 hours. Efficiency of Pipe T = 600/50 = -12 units/hour [Here negative sign is used to represent that the pipe is used to empty the tank.] The efficiency of pipe R is 50% more than the efficiency of pipe S. Let’s assume the efficiency of pipe S is ‘2a’. efficiency of pipe R = 2a of (100+50)% = 2a of 150% = 3a Pipe R and S together can fill the empty tank together in 24 hours. (3a+2a)x24 = 600 5a = 25 a = 5 Pipe R and T together can fill an empty tank completely in (z+5) hours. (3a-12)x(z+5) = 600 Put the value of ‘a’ in the above equation. (3x5-12)x(z+5) = 600 (15-12)x(z+5) = 600 3x(z+5) = 600 (z+5) = 200 z = 200-5 Value of ‘z’ = 195
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