How to solve this problem on alligations using successive technique?

Solution 1:

The initial mixture is $8/40=20$% wine. So with every 5 litres you remove, you remove a litre of wine. Leaving 7 litres of wine. If you now add 5 litres of pure wine, then you will have 12 litres of wine in 40 litres of mixture, which is your 30% solution.

Solution 2:

The diluted wine contains only 8 litres of wine and the rest is water. A new mixture whose concentration is 30 % is to be formed by replacing wine. How many litres of mixture shall be replaced with pure wine if there was initially 32 litres of water?

Initially, the concentration is $20$% wine, since there are 8 liters of wine, mixed with $32$ liters of water. You will remove $x$ liters of the concentration, and then add back $x$ liters of pure wine. The result must be $30$% wine (instead of $20$%).

The final volume will be unchanged at $(40)$ liters, since you are removing $x$ liters of the mixture, and replacing it with $x$ liters of pure wine.

When you remove $x$ liters of the mixture, since the mixture is $20$% wine, you are (in effect) removing $(0.2)x$ liters of wine.

Then, you are adding back $x$ liters of (pure) wine.

So, the new amount of wine is $(8) - (0.2)x + x = 8 + (0.8)x$.

The requirement is that the new amount of wine must equal $30$% of the new volume, which is the same as the old volume. Therefore, the new amount of wine must equal $$0.3 \times 40 = 12.$$

So, $x$ must satisfy:

$$8 + (0.8)x = 12 \implies (0.8)x = 4 \implies x = \frac{4}{0.8} = 5.$$