Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively

Quantity of A in mixture left = 7 x - 7 12 × 9 = 7 x - 21 4 litres.

Quantity of B in mixture left =   5 x - 5 12 × 9 = 5 x - 15 4litres.

7 x - 21 4 5 x - 15 4 + 9 = 7 9

? 28 x - 21 20 x + 21 = 7 9

? x = 3

 So, the can contained 21 litres of A.

Suppose the vessel initially contains 8 litres of liquid.

Let x litres of this liquid be replaced with water.
Quantity of water in new mixture =   3 - 3 x 8 + xlitres.

Quantity of syrup in new mixture =   5 - 5 x 8 litres.

   3 - 3 x 8 + x = 5 - 5 x 8

=>   5x + 24 = 40 - 5x 

=>   10x = 16    => x = 8/5

So, part of the mixture replaced =  8 5 × 1 8 = 1/5.

Number of liters of water in 125 liters of the mixture = 20% of 150 = 1/5 of 150 = 30 liters

Let us Assume that another 'P' liters of water are added to the mixture to make water 25% of the new mixture. So, the total amount of water becomes (30 + P) and the total volume of the mixture becomes (150 + P)

Thus, (30 + P) = 25% of (150 + P)

Solving, we get P = 10 liters

           Wine                         Water

            8L                              32L

            1                :               4

           20 %                           80% (original ratio)

           30 %                           70% (required ratio)

In  ths case, the percentage of water being reduced when the mixture is being replaced with wine.

so the ratio of left quantity to the initial quantity is 7:8

Therefore ,  7 8 = 1 - K 40

=> K = 5 Lit

pool : kerosene 

   3  :  2(initially) 

   2  :  3(after replacement)

  R e m a i n i n g Q u a n t i t y I n i t i a l Q u a n t i t y = 1 - R e p l a c e d Q u a n t i t y T o t a l Q u a n t i t y

(for petrol)    2 3 = 1 - 10 k

 => K = 30

 Therefore the total quantity of the mixture in the container is 30 liters.

Amount of milk left after 3 operations

40 1 - 4 40 3 = 40 * 9 10 * 9 10 * 9 10 = 29 . 16 litres