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?
Answers
Answer:
Step-by-step explanation:
♦ Let the rate of each filling pipe be x litres/hr and the rate of each draining pipe be y litres/hr.
Now,
Capacity of tank = (6x - 5y) × 6 = (5x - 6y) × 60
6x - 5y = 50x - 60y
44x = 55y
4x = 5y
x = 1.25y
Therefore, the capacity of the tank = (6x - 5y) × 6 = (7.5y - 5y) × 6 = 15y litres
Effective rate of 2 filling pipes and 1 draining pipe = (2x - y) = (2.5y - y) = 1.5y
♠ Hence, the required time = 15y/1.5y = 10 hours
Answer:
10 hours
Step-by-step explanation:
Let, x be the rate at which a single filling pipe fills the tanks.
Let y be the rate at which a single draining pipe drains the pipe.
Tank gets completely filled in 6 hours when 6 filling and 5 draining pipes.
=> 6x - 5y = 1/6 ---- (1)
Time becomes 60 hours when 5 filling and 6 draining pipes are on.
=> 5x - 6y = 1/60 ---- (2)
Multiply 1 by 5 and 2 by 6.
30x - 25y - 30x + 36y = 5/6 - 1/10
=> 11y = 11/15
=> y = 1/15
Place y = 1/15 in (1).
=> 6x - 5y = 1/6
=> 6x - 5(1/15) = 1/6
=> 6x - 1/3 = 1/6
=> 6x = 1/2
=> x = 1/12
Thus, time taken when one draining and two filling pipes are on is,
=> 1/(2x - y)
=> 1/[2(1/12) - 1/15]
=> 1/1/6 - 1/15
=> 1/10
Hence, the tank get completely filled in 10 hours.
#Hope my answer helped you!