Net filling in 1 hour = (1/4 + 1/6 - 1/8) = 7/24
? Time taken to fill the cistern = (24/7) hrs. = 3 hrs. 26 min.
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.
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
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
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
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.
Net filling in 1 hour = (1/2 - 1/3) = 1/6
? Time taken to fill the cistern = 6 hours
Net filling in 1 hour = (1/x - 1/y) = (y - x) / xy
? Time taken to fill the tank = xy / (y - x) hrs.
Part filled by intel in 1 hour = 1/8
Part emptied by outlet in 1 hour = 1/16
Net filling in 1 hour = (1/8 - 1/16) = 1/16
? Time taken to fill the tank = 16 hours
Part filled by A in 1 hour = 1/10
Part filled by B in 1 hours = 1/15
Part filled by (A + B) in 1 hour = (1/10 + 1/15) = 5/30 = 1/6
? Both pipes together can fill the tank in 6 hours.
Part emptied by the third pipe in 1 min (1/10 + 1/12) - 1/25 = 7/60
So, the full tank will be emptied by third pipe in = (60/7) min. = 8 min. 34 ec.
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