According to question,
Quantity of Milk/Quantity of Water = 3/2
Let us assume the product ratio = n.
Quantity of Milk = 3n and Quantity of Water = 2n liters
Quantity of Milk + Quantity of Water = 20 liters
3n + 2n = 20 liters
5n = 20 liters
n = 20/5
n = 4
Quantity of Milk = 3n liters
Put the value of n,
Quantity of Milk = 3 x 4 = 12 liters
Quantity of Water = 2n
Quantity of Water = 2 x 4 = 8 liters
If 10 liters of mixture are removed first time, we will find how much milk and water contain in mixture.
? 20 liters of mixture contains 12 liter of milk.
? 1 liters of mixture contains 12/20 liter of milk.
? 10 liters of mixture contains 10 x 12/20 liter of milk.
? 10 liters of mixture contains 6 liter of milk.
? 10 liters of mixture contains 4 liter of water.
If 10 liters of mixture are removed, then amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 - 6 = 6 liters
Remaining water = 8 - 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters and water = 4 liters.
If the process is repeated one more time and 10 liters of the mixture are removed second time, then
If 10 liters of mixture are removed, Again we will find how much milk and water contain in the mixture.
? 20 liters of mixture contains 16 liter of milk.
? 1 liters of mixture contains 16/20 liter of milk.
? 10 liters of mixture contains 10 x 16/20 liter of milk.
? 10 liters of mixture contains 8 liter of milk.
? 10 liters of mixture contains 2 liter of water.
If 10 liters of mixture are removed, then amount of milk removed = 8 liters and amount of water removed = 2 liters.
Remaining milk = (16 - 8) = 8 liters.
Remaining water = (4 - 2) = 2 liters.
Now 10 liters milk is added => total milk = 18 liters and water will be 2 liters.
The required ratio of milk and water in the final mixture obtained
Quantity of milk/Quantity of Water= 18:2 = 9:1

           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

Let the capacity of the pot be 'P' litres.
Quantity of milk in the mixture before adding milk = 4/9 (P - 8)
After adding milk, quantity of milk in the mixture = 6/11 P.
6P/11 - 8 = 4/9(P - 8)
10P = 792 - 352 => P = 44.

The capacity of the pot is 44 liters.

From the given data,

let the initial quantity of the mixture = 5x

Then,

$$\begin{gathered} 2 x - 16 3 x - 24 + 40 = 1 4 \\ 8 x - 64 = 3 x + 16 \\ 5 x = 80 \\ x = 16 lit \end{gathered}$$

Then the initial quantity of the mixture = 5x = 5 x 16 = 80 lit.

Milk = 3/5 x 20 = 12 liters, water = 8 liters

If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.

Remaining milk = 12 - 6 = 6 liters

Remaining water = 8 - 4 = 4 liters

10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.

The ratio of milk and water in the new mixture = 16:4 = 4:1

If the process is repeated one more time and 10 liters of the mixture are removed,
then amount of milk removed = 4/5 x 10 = 8 liters.

Amount of water removed = 2 liters.

Remaining milk = (16 - 8) = 8 liters.

Remaining water = (4 -2) = 2 liters.

Now 10 lts milk is added => total milk = 18 lts

The required ratio of milk and water in the final mixture obtained

= (8 + 10):2 = 18:2 = 9:1.

w i n e ( l e f t ) w i n e ( a d d e d ) = 343 169

It means  w i n e ( l e f t ) w i n e ( i n i t i a l a m o u n t ) = 343 512    (since 343 + 169 = 512)

Thus,   343 x = 512 x 1 - 15 k 3

  343 512 = 7 8 3 = 1 - 15 k 3

  1 - 15 k = 7 8 = 1 - 1 8

Thus the initial amount of wine was 120 liters.

Let the tin contain 5x litres of liquids

5 x - 45 x 4 5 5 x - 45 x 1 5 + 45 = 2 5

=> 5(4x - 36) = 2(x + 36)

=> 20x - 180 = 2x + 72

=> x = 14 litres

Hence, the initial quantity of mixture = 70l

Quantity of liquid B

=  70 - 45 x 1 5 + 45

= 50 litres.

Water in 60 gm mixture=60 x 75/100 = 45 gm. and Milk = 15 gm.

After adding 15 gm. of water in mixture, total water = 45 + 15 = 60 gm and weight of a mixture = 60 + 15 = 75 gm.

So % of water = 100 x 60/75 = 80%.