Question
A tank is filled by three pipes with uniform flow. The
first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 45 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is?Solution
Let time taken by third pipe to fill the tank = x hrs so time taken by 2nd pipe to fill the tank = x + 4 hrs & time taken by 1st pipe to fill the tank = x + 4 + 45 = x + 49 hrs; Hence 1/(x+4) +1/(x+49) = 1/x (x + 4 + x + 49)/ {(x + 4)(x + 49)} = 1/x (2x + 53)/(x2 + 53x + 196) = 1/x 2x2 + 53x = x2 + 53x + 196 x2 = 196 x = 14 hrs So time taken by 1st pipe to fill the tank = x + 49 = 14 + 49 = 63 hrs Alternate Method :- Let time taken by third pipe to fill the tank = x hrs so time taken by 2nd pipe to fill the tank = x + 4 hrs & time taken by 1st pipe to fill the tank = x + 4 + 45 = x + 49 hrs; Then x = root of (4 × 49) = root of 196 = 14 hrs So time taken by 1st pipe to fill the tank = x + 49 = 14 + 49 = 63 hrs.
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