A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in 6 hours when 6 filling and 5 draining pipes are on, but this time becomes 60 hours when 5 filling and 6 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on?
Explanation:
Let a filling pipe fills the tank at ‘a’ liters per hour and a draining pipe drains at ‘b’ liters per hour.
Work done by 6 filling and 5 emptying pipes in 6 hours = 6(6a - 5b) Work done by 5 filling and 6 emptying pipes in 60 hours = 60(5a - 6b)
6(6a – 5b) = 60(5a – 6b) ∴ 6a – 5b = 50a – 60b ∴ 44a = 55b ∴ a = 5b/4
Let the tank gets filled completely in ‘m’ hours when one draining pipe and two filling pipes are on. Work done by 2 filling and 1 emptying pipes in m hours = m(2a - b)
∴ m(2a – b) = 6(6a – 5b)
∴ m2×54b-b = 66×54b-5b
∴ m6b4 = 610b4
Solving this, we get m = 10
Hence, 10.
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