Time taken by the pipe to empty the cistern = 3 h
Then, time taken by the pipe to empty the 2/3 part = 3 x 2/3 = 2 h
We know that, when a pipe fills 1/m part of a tank in 1 h, then the pipe takes m h to fill the tank.
Here, 1/m = 1/8 ? m = 8
? Required time to fill the tank = 8 h
Here x = 8 hrs. and y = 8 + 2 = 10 hrs.
Now, applying the given rule, we have the
Required answer = (8 x 10) /(10 - 8) = 40 hrs.
Required answer = (12 x 16) /(12 + 16) = 48 / 7 = 66/7 minutes.
Work done by leak in 1 hour = (1/8 - 1/10) = 1/40
? The leak will empty the cistern in 40 hours.
We know that, when a pipe fills a tank in m h, then the part of tank filled in 1 h = 1/m
Here, m = 6
? Required part of the tank to be filled in 1h = 1/6 part
Let the leak empties the full tank in x h, then
Part emptied in 1 by leak = 1/x
Part filled by inlet in 1 h = 1/20
According to the question,
1/20 + 1/x = 1/40
? 1/x = 1/40 - 1/20 = 1- 2/40 = -1/40 [-ve sign Indicates emptying.]
Clearly, leak will empty the full tank in 40 h.
We know that, when a pipe fills a tank in m h, then the part of tank filled in 1 h = 1/m
Here, m = 6
? Required part of the tank to be filled in 1 h = 1/6 part
We know that, when a pipe empties a cistern in n min, then the part emptied by the pipe in 1 min = 1/n
Here, n=30
? Required part of the tank emptied in 1 min = 1/30 part
Part filled by A in 1 h = 1/8
Part filled by B in 1 h = 1/10
Part filled by C in 1 h = 1/20
Part filled by (A + B + C) in 1 h = 1/8 + 1/10 + 1/20
= (5 + 4 + 2)/40 = 11/40
? Required time to fill the tank = 40/11 h
= 37/11 h
Part emptied in 1 min. = (1/8 - 1/16) = 1/16
? Time taken to empty the full tank = 16 min.
Hence, time taken to empty the half tank = 8 min.
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