There are three taps of diameter 1 cm, 4/3 cm and 2 cm, respectively. The ratio of the water flowing through them is equal to the ratio of the square of their diameters. The biggest tap can fill the tank alone in 61 min. If all the taps are opened simultaneously, how long will the tank take to be filled?

Let the cistern gets emptied in m h after 3:pm
Work done by A in m h, by B in (m - 1) h and by c in (m - 2) h= 0
? m/3 + (m - 1)/4 - (m - 2) = 0
? 4m + 3(m - 1) - 12(m - 2) = 0
? 5m = 21
? m = 21/5 = 4.2
? m = 4 h 12 min
? Required time = 7 : 12 pm.

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 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

Suppose they will meet after T hours.
Distance = Speed x Time
Sum of distance traveled by them after T hours
6T + 4T = 20 km
T = 2 hours.
So they will meet at 7:00 AM + 2 hours = 9:00 AM