Let the filling capacity of the pump be x m 3/min. 

Then, emptying capacity of the pump=(x+10) m 3/min.  

so, 2400 x - 2400 x + 10 = 8 ? x 2 + 10 x - 3000 = 0 

? x - 50 + x + 60 = 0 ? x = 50

1/10 + 1/15 = 1/6 x 3 = 1/2

1 - 1/2 = 1/2

1/10 + 1/15 - 1/x = 1/2

x = 8.57 min

Work done by the waste pipe in 1 minute = 1 15 - 1 20 + 1 24 = 1 15 - 11 120 = - 1 40 [-ve sign means emptying]

Volume of  1 40part = 3 gallons

Volume of whole = (3 x 40) gallons = 120 gallons.

The work to be done = Capacity of reservoir  = 30 litres.

 1^st Minute ---> inlet pipe opened ---> 5 lit filled

 2^nd minute ---> inlet pipe closed; outlet pipe opened ---> 4 lit emptied

In 2 minutes (5 litres - 4 litres = 1lit) is filled into the reservoir.

It takes 2 minutes to fill 1lit ---> it takes 50 minutes to fill 25 litres into the reservoir.  

In the 51^st minute inlet pipe is opened and the reservoir is filled.

Volume of water collected in the tank in 1 hour

? (0.3 × 0.2 × 20km × 1000mts) = 1200 m cubic

If after t hours, the water is at height of 12m,

1200t=200×150×12 

? t = 300 Hours.

Capacity of the tank =(12 x 13.5) liters =162 liters.

Capacity of each bucket =9 liters

Number of buckets needed = 162/9 =18.

Let the slower pipe alone fill the tank in x minutes. 

Then,faster pipe will fill it in x/3 minutes. 

= > 1 x + 3 x = 1 36 = > 4 x = 1 36 = > x = 144 m i n s

Given A alone can fill the tank of capacity 240 lit in 16 hrs.

=> A can fill in 1 hr = 240/16 = 15 lit

=> B alone can fill the tank of capacity 240 lit in 12 hrs.

=> B can fill in 1 hr = 240/12 = 20 lit

Now, (A + B) in 1 hr = 15 + 20 = 35 lit

But they are opened for 2 hrs

=> 2 x 35 = 70 lit rae filled

Remaining water to be filled in tank of 240 lit = 240 - 70 = 170 lit.

Part filled by (A + B) in 1 minute = (1/60 + 1/40) = 1/24
Suppose the tank is filled in x minutes.
Then, x/2(1/24 + 1/40) = 1
(x/2) * (1/15) = 1 => x = 30 min.