Consider volume of milk = M and volume of water = W

Situation 1:
M/W=2/1
Let x denote the common factor: 2x+x=30
M = 20 and W=10

Situation 2:
Let us consider that "a" litres of solution has been replaced.
In "a" liters --> 2a/3 = milk and a/3 = water
M1= 20-2a/3
W1=10-(a/3)+a=10+(2a/3)
Note that we have replaced "a" liters of solution with water, hence "a" liters of water should be added
M1/W1=8/7
Solving this equation we get a=6
M1 = 16 and W1=14

Situation 3:
M2/W2=3/1
Let y be common factor: 3y+y=20
M2= 15 and W2=5

Total M = 16+15 = 31
Total W=14+5=19

Total M/Total(m+W) = 31/50 =62%

drakell thank you for noticing the mistake.

From first two sentences we can say that 30 ltr of water has 8:7 ratio.

8+7 =15 and 30/15= 2

Therefore final mixture has 16 ltr milk and 14 ltr water

In last sentence, we will follow the same process.

3+1=4, 20/4= 5
Therefore, new mixture contains 15 ltr of milk and 5 ltr of water

Total: 31 ltr milk and 19 ltr water
Total mixture 50 ltrs

31/50=62/100 ie 62 %

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1). we have 20 liters of milk & 10 liters of water
2). 20-x/10+x = 8/7 , we can find x = 4. so now we have 16 liters of milk and 14 liters of water
3). we pour another 20 liters in a ratio 3:1, so we have another 15 liters milk and 5 liters water. Combining with the previous numbers:

16+15/ 14+5 = 31/19 -> looking for % of total: 31/50 or 62/ 100 -> 62%