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

Part filled in 4 minutes =4(1/15+1/20) = 7/15

Remaining part =(1-7/15) = 8/15

Part filled by B in 1 minute =1/20 : 8/15 :: 1:x 

x = (8/15*1*20) =  10 2 3 m i n = 10 m i n 40 s e c

The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec

If the total area of pump=1 part

The pumop take 2 hrs to fill 1 part 

The pumop take1 hour to fill 1/2 portion 

Due to lickage  

The pumop take 7/3 hrs to fill 1 part 

The pumop take1 hour to fill 3/7 portion 

Now the difference of area = (1/2-3/7)=1/14 

This 1/14 part of water drains in 1 hour 

Total area=1 part of water drains in (1x14/1)hours= 14 hours 

So the leak can drain all the water of the tank in 14 hours. 

Leak will empty the tank in 14 hrs

Time taken by one tap to fill half of the tank = 3 hrs.

  Part filled by the four taps in 1 hour =4*1/6 =2/3

Remaining part = 1 - 1 2 = 1 2

2 3 : 1 2 : : 1 : x

=> x =  1 2 * 1 * 3 2 = 3 4

So, total time taken = 3 hrs. 45 mins.

Part filled in 2 hours = 2/6=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 = 2/21

C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }

=1/6-2/21 = 1/14

C alone can fill the tank in 14 hours.

A + B ' s 1 h o u r w o r k = 1 12 + 1 15 = 9 60 = 3 20

A + C ' s 1 h o u r w o r k = 1 12 + 1 20 = 8 60 = 2 15  

p a r t f i l l e d i n 2 h r s = 3 20 + 2 15 = 17 60  

p a r t f i l l e d i n 6 h r s = 3 * 17 60 = 17 20  

  r e m a i n i n g p a r t = 1 - 17 20 = 3 20  

Now, it is  the turn of A and B (3/20) part is filled by A and B in 1 hour.

Therefore, Total  time taken to fill the tank =(6+1)hrs= 7 hrs

1/8 - 1/x = 1/10

 => x = 40 hrs

Clearly,pipe B is faster than pipe A and so,the tank will be emptied.

part to be emptied = 2/5

part emptied by (A+B) in 1 minute= 1 6 - 1 10 = 1 15

1 15 : 2 5 : : 1 : x

2 5 * 15 = 6 m i n s

so, the tank will be emptied in 6 min