A cistern has 3 pipes A, B and C, A and B can fill it in 2 and 3 hours respectively. C is a waste pipe. If all the 3 pipes be opened at once, 7/24 of the cistern will be filled up in 30 minutes. In what time can C empty the full cistern?

Let the leak empty the full cistern in N hours
Now, applying the given rule (9 x N) / (N - 9) = 9 + 1
or N = 90 hours.

Part filled in 10 hours = 10 x (1/15 + 1/20 - 1/25) = 23/30
Remaining part = (1 - 23/30) = 7/30
Now, (1/15 + 1/20) part is filled by A and B in 1 hr.
7/30 part be filled them in (60/7) x (7/30) = 2 hrs.
? Total time taken to fill the tank = (10 + 2) hrs. = 12 hrs.

Suppose that one pipe takes N hours to fill the reservoir. Then another pipe takes (N - 10) hours.
? 1/N + 1/(N - 10) = 1/12
? 12(N - 10 + N) = N(N - 10)
or N² - 34N + 120 = 0
or (N - 30)(N - 4) = 0
? N = 30 or N = 4
So, the faster pipe takes 30 hours to fill the reservoir.

Part filled in 2 hours = 2 x 1/6 = 1/3
Remaining part = (1 - 1/3) = 2/3
(A + B)'s 7 hour's work = 2/3
? (A + B)'s 1 hour's work = (2/3 x 1/7) = 2/21
(A + B + C)'s 1 hour's work = 1/6
C's 1 hours work = (1/6 - 2/21) = 1/14
Hence, C alone can fill the cistern in 14 hours.

Net filling in 1 min. = (1/32 + 1/36 - 1/20) = 13/1440
? Time taken to fill the tank = (1440/13) min.
Time taken to fill half of the tank = (1440/13) x 2 min. = (720/13) min.
= 55⁵/₁₃ min.

12(1 - t/25) = 3
? t = 45/4 = 11¹/₄ minutes
? Required answer = 11¹/₄ - 3
= 8¹/₄ minutes

[(A's 1 hour work) + (A + B)'s 1 hour work]
= (1/10) + [(1/10) + (1/12)] = 17/60
Remaining part = 1 - (17/60) = 43/60
Now, (A + B + C)'s 1 hour work = (1/10) + (1/12) + (1/15) = 1/4

1/4 part is filled by 3 pipes in 1 hour.
43/60 part will be filled by them in 4 x (43/60) hrs. = 2 hours 52 min.

? Total time taken to fill the cistern = 4 hours 52 min.

Part filled in 5 min. = 5 x (1/12 + 1/15) = 5 x 9/60 = 3/4

Part emptied in 1 min. (when all the pipes are opened) = 1/6 - (1/12 + 1/15) = (1/6 - 3/20) = 1/60

Now, 1/60 part is emptied in 1 min. 3/4 part will be emptied in (60 x 3/4) = 45 min.

Part filled by inlet in 1 hour = (1/6 - 1/8) = 1/24
So, the inlet can fill the tank in 24 hours
? Capacity of the tank = Water that flows in 24 hours = (4 x 24 x 60) liters = 5760 liters

Let Y be closed after T min
Then, T x (1/ 24 + 1/32) + (18 - T) x 1/24 = 1
? (7 x T)/96 + (18 - T)/24 = 1
or 7T + 72 - 4T = 96.
? 3T = 24
or T = 8 min.