Aptitude - Alligation or Mixture - Discussion

Equation should be equated with 16/97.

Sine wine to water ratio is 16:81.

So wine to total ratio is 16:97.

Thats what we finding out actually by dividing it by x.

Since each time we are adding 8 litres of water and talking away 8 litres. So total composition is same only the proportions of individual componnts will change.

So talking water as x litre is wrong.

He has calculated the ratio of wine to whole soln. !

@Prajakt
the soln. assumes in the 1st line that the original quantity of wine be x. the question says the wine removed is replaced with water(obviously, in equal amount). Since we assume x as wine quantity, this replaced water quantity is also x.

Can any one please explain ....How the quantity of water become x?

Awesome rachit.

Intial wine = x.
After 8 units are pulled out, (x-8) units of pure wine remains.
In fractions, ((x-8)/8) = (1-(8/x)) of the initial content x remains after the first pullout.
After the second pullout, (1-(8/x)) of the previous amount is left, i.e.- (1-(8/x)) of (1-(8/x)) of x = (1-(8/x))^2 of x.
In the same way, after n such pullouts, the fraction of wine left will be (1-(8/x))^n.
And the amount of pure wine left will be x*(1-(8/x))^n.

This part is well understood. But why was the above quantity divided by x, while the question reported the ratio as that of leftover pure wine to water? I am finding it hard to understand the logic behind this. Or is it wrong?

Well explained Bappa.

Lets wine x
after 1st time wine remain x-8
now ratio of wine:water x-8:8
so 2nd time wine removed=(x-8)*(8/x)
after 2nd time wine remain=(x-8)-(x-8)*(8/x)
=(x-8){1-8/x}
=x(1-8/x)(1-8/x)
=x(1-8/x)^2
.........
for n time removaline remain=x(1-8/x)^n.

How come it get divided by x, While it should be quantity of water ? Alternate way can be dividing by x and putting in 16/97 instead of 16/81.

Its 3 more times not "more then three times" so we used mul.