A cistern has three pipes A, B and C. Pipes A and B can fill it in 3 and 4 h, respectively, while pipe C can empty the completely filled cistern in 1 h. If the pipes are opened in order at 3:00 pm, 4:00 pm and 5:00 pm, respectively , at what time will the cistern be empty?

Part of the tank filled by both taps in 5 min = 5 x (1/20 + 1/60)
= 5 x (6 +2 )/120 = 8/24 = 1/3
? Remaining part = (1 - 1/3) = 2/3
? 1/60 Part is now filled in 1 min.
? 2/3 Part is now filled in 60 x 2/3 = 40 min.

Part filled by 1st pipe in 1 min = 1/20
Part filled by 2nd pipe in 1 min = 1/24
Part filled by all the pipes in 1 min = 1/15
Work done by the waste pipe 1 min
= 1/15 - (1/20 + 1/24 )
= 1/15 - 11/120
= (8 - 11)/120 = ( - 3/120 ) = ( -1/40 )
[-ve sign indicates emptying]
Now, volume of 1/40 part = 6 gallon
? Volume of whole tank = 40 x 6 = 240 gallon

1st pipe takes 1 h to fill 1/2 part of the tank.
So, time taken to fill the whole tank (m) = 2 h
2nd pipe takes 1 h to 1/3 part of the tank
So, time taken to fill the whole tank (n) = 3 h
Let 3rd pipe takes P h to empty the tank = x
? 1/m + 1/n - 1/x = 7/12 ? 1/2 + 1/3 - 1/x = 7/12
? 1/x = (6 + 4 - 7)/12 = 3/12 = 1/4
? x = 4 h

Here, a = 16 and = 9
Required time = ?ab
?16 x 9 = 4 x 3 = 12 min

Part filled by pipe P in 1 min = 1/12
Part filled by pipe Q in 1 min = 1/15
Part filled by both pipes in 1 min = 1/12 + 1/15 + = (5 + 4)/60 = 9/60
Now, Part filled by both pipes in 3 min = (3 x 9)/60 = 27/60 = 9/20
? Remaining part = 1 - 9/20 = 11/20
Let the remaining part is filled by pipe Q in T min.
Then, T/15 = 11/20
T = (15 x 11)/20 = (3 x 11)/4
= 33/4 = 8⁴/₄ min

Time taken to fill the tank by the tap having 2 cm diameter = 61 min

? Time taken to fill the tank by the tap having 1 cm diameter
= 61 x (2/1)² = 244 min

Similarly, time taken to fill the tank by the tap having 4/3 cm diameter
= 61 x [(2 x 3)/4]² = 61 x 9/4 = 549/4 min.

? Part of the tank filled by all the three pipes in 1 min
= 1/61 + 1/244 + 1/(549/4)
= (36 + 9 + 16)/2196 = 61/2196 = 1/36
Hence, required time taken = 36 min

Part filled by 1st pipe in 1 h = 1/14
Part filled by 2nd pipe in 1 h = 1/16
Part filled by the two pipes in 1 h
= ( 1/14 + 1/16 ) = (8 + 7)/112 = 15/112
? Time taken by these two pipes to fill the cistern = 112/15 h = 7 h 28 min
Due to leakage, the time taken 7 h 28 min + 92 min = 9 h
&thee4; Work done by ( two pipes + leak ) in 1 h = 1/9
Work done by the leak in 1 h = 1/9 - 15/112 = (112 - 135)/1008
= - 23/1008
? Time taken by leak to empty the full cistern
= 1008/23 = 43¹⁹/₂₃ h

T = (10 x 15 x 20) / (10 x 15 + 10 x 20 + 15 x 20) = 60/13 minutes
Now, applying the given rule, we have [60/13 x y]/ [y - 60/13 + 3] = 20
or y = 21/10 = 2 min. 6 seconds.

Net filling in last 15¹/₂ minutes
= 31/2 (1/30 + 1/36) = 341/360
Now, suppose they remained clogged for x minutes.
Net filling in these x minutes
= (x/30 x 5/6 + x/36 x 9/10) = 19 x/360
Remaining part = (1 - 19x/360) = (360 - 19x/360)
360 - 19x/360 = 341/360 or x = 1.
Hence, the pipes remained clogged for 1 minutes.

Let the leak empties it in T hours
From the given rule, we have (T x 30) / (T - 30) = 40
? T = 120 minutes = 2 hours.

Now, from the question, applying the rule, we have time taken by B to fill the tank when crack in the bottom develops
= (120 x 40) / (120 - 40) = 60 minutes = 1 hour