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01 Aug 2019, 00:02
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Official CAT 2018 Questions; Section: QA
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?
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?
10
03 Aug 2019, 03:43
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Narenn
Official CAT 2018 Questions; Section: QA
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?
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?
10
If x=efficiency of input pipe and y=efficiency of output pipe, then
36x-30y=1
300x-360y=1
t*2x-t*y=1
=> x= 1/12, and y=1/15
Thus, time taken when one draining and two filling pipes are on = t = 1/(2x-y) = 1/(1/6-1/15) = 10 hours.
07 Aug 2019, 02:33
Kudos
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Alternative approach:
If x = rate of input and y = rate of output [original].
Then:
(x-y) = 10*(5x/6 - 6y/5) [as the time taken is 10 times in the second case, so the rate in the first case must be 10 times the second]
=> y =(2/3)x
Put this in 6*(x-y)=1, to get x=1/2 and y=1/3
Then to find the required time t, we solve t*(x/3 - y/5) = 1, to get t=10 hours.
If x = rate of input and y = rate of output [original].
Then:
(x-y) = 10*(5x/6 - 6y/5) [as the time taken is 10 times in the second case, so the rate in the first case must be 10 times the second]
=> y =(2/3)x
Put this in 6*(x-y)=1, to get x=1/2 and y=1/3
Then to find the required time t, we solve t*(x/3 - y/5) = 1, to get t=10 hours.
