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.
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 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
Net filling in 1 hour = (1/x - 1/y) = (y - x) / xy
? Time taken to fill the tank = xy / (y - x) hrs.
Net filling in 1 hour = (1/2 - 1/3) = 1/6
? Time taken to fill the cistern = 6 hours
Net filling in 1 hour = (1/4 + 1/6 - 1/8) = 7/24
? Time taken to fill the cistern = (24/7) hrs. = 3 hrs. 26 min.
T = (25 x 50) / (50 - 25) = + 50 minutes
+ ve sign shows that tank is filled up in 50 minutes.
Part filled by A in 1 h = 1/18
Part filled by B in by 1 h = 1/6
Part filled (A + B) in 1 h =1/18 + 1/6 = (1 + 3)/18 = 4/18 = 2/9
Hence, both the pipes together will fill the tank in 9/2 h or 41/2 h
Part filled by A in 1 h = 1/4
Part emptied by B in 1 h = 1/8
Part filled by (A + B) In 1 h = 1/4+ ( -1/8 ) = 1/4 - 1/8 = (2 - 1)/8 = 1/8
? Required time to fill the cistern = 8
Part filled by A in 1 h = 1/5
Part filled by B in 1 h = 1/6
Part filled by C in 1 h = 1/30
Net part filled by ( A + B + C ) in h 1 = ( 1/5 + 1/6 + 1/30 )
= (6 + 5 + 1)/30 = 12/30 = 2/5
? Required time to fill the tank = 5/2 = 21/2 h
Part filled by tap in 1 h = 1/12
Part emptied by leak in 1 h = 1/20
Net part filled in 1 h when both (tap and leakage) work = 1/12 - 1/20
= (5 - 3)/60 = 2/60 = 1/30
? Required time to fill the tank = 30 h
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