Aptitude - Alligation or Mixture - Discussion

If this process done with 5 litres only one time, and the given ratio is 361:39. What will be the answer?

Suppose, the initial amount of mixture is X.

Remaining quantity of wine after four operations ( P suppose)= X*(1-8/X)^4 = P(Suppose)
The total quantity of mixture will always remain X because how much mixture we are removing same we are adding as water and quantity which is not of wine that is water so after 4 operations quantity of water in the mixture = X-P.

Given: remaining wine/water=16/65= P/(X-P)...................(1).
By solving equation (1) we will get P/X=16/81, that is the ratio of remaining wine and present water in the mixture.

So this is the explanation why he divided by X.
Further is simple as we supposed P = X * (1-8/X) ^4.
So, P/X = (1-8/X)^4 = 16/81.............(2).
Solving equation (2) we get X = 24.

Thanks @Ayush.

Thanks, @Pravu.

You explained the derivation of the formula is very well.

Let initial quantity of wine = x.
After, 4 operations, quantity of wine left = x(1 - 8/x)^4.
Also the quantity of wine in the jar = (16/81)x.
Therefore, x(1-8/x)^4 = (16/81)x.
On solving, x = 24.

Good explanation @Ayush. You cleared the doubt everyone had.

Let wine left be x,

4 times 8 liter wine removed = 4*8 = 32.

Therefore wine left = (x-32).

Water filled 4 times = (4*8) = 32.

Therefore ratio of (Quantity of wine left/Quantity of water).

= (16/65).

Therefore ( (x-32)/32) = (16/65).

On solving we get x = 39.87~40.

Hi, anyone clears my doubt.

Logically, How can you remove only wine again after mixing 8 liters of water to the wine already present in the cask?

Does anybody know the formula for calculating no of liters of wine present if the solution removed from second time onwards contain both wine and the water?

Step 1: (x - 8)^4/x = 16/81.
Step 2: (x - 8)^4/x = (2/3)^4.
Therefore, x - 8/x = 2/3.
So, x=24.

If 16x/81 is written based on the LHS it converted to 2/3 then what about x.