Aptitude - Alligation or Mixture - Discussion

I think most of you understood the solution but for those who didn't let me explain:

First of all, There is a formula which says:

Suppose container contains x of liquid from which if y units are taken out (of the whole liquid) and then replaced with water, and this process is simply repeated n no of times, then the quantity of original liquid present in the mixture now is:

x[(1- y/x )^n].

Ok , now as the question says here we need to find the original quantity of the wine, let it be x.

Now the quantity of liquid (not wine) removed is = 8 which is 'y' for the formula.

This thing is done total of 4 times.

So now the quantity of the wine present in the final mixture will be:

x[(1 - 8/x) ^4]. No problem till here.

Now as the problem says, the ratio of wine to water in the final mixture, we have calculated wine but we can't calculate the amount of water in the final mixture.
So, what to do next.

We are given the ratio of wine to water in the mixture and remember the water was being replaced each time, so total quantity of the mixture is still 'x'.

Thus, if 16/65 was the given ratio of wine to water then, 16/(16+65)

will be the ratio of wine to total mixture ie:

x[(1-8/x)^n] / x = 16/81

I hope everyone get it now.

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.

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?

If you guys really do this maths then see this technique other than wrong methods.

The question is - "How much wine did the cask hold originally?".

So the method is :

(liquid A left after n'th operation) / (liquid B left after n'th operation) =[ (1- (b/a))^n]/[1- (1- (b/a) ^n].

Here, a= total amount of vessel(a.k.a wine);
b = amount of other liquid poured(a.k.a. water);
n = no of operations had done.

In this question they also mentioned that 3 more times it had done after it had changed with 8 ltr of water.

So the n will b 4.

Mind the tricks of the questions.

Do a very sharp calculation you will get it? It is much easier than its look.

Here, x is just used for simplification.

Suppose we write the equation as:
[x({1-(8/x)}^4)]/32 = 16/65 it is correct.

But, it will be very difficult to simplify because of the fifth power in the resulting equation that we have to solve.

So instead, we take the ratio of wine to total liquid in the end. We can find the total liquid by simply adding 16+65, ie, wine=water, so this is not an issue. This way, we can eliminate the x outside bracket in the numerator and find the answer very quickly.

Hope it's clear.

Lets,
The volume of the drum is x.

In 1st operation 8 liter is drawn.
Remaining pure wine is x-8.

Now. If we draw whole volume = the volume of wine is drawn is x-8.

""""""""""''' 1 unit volume="""""""""""""""""""""""""" (x-8) /x.

""""""""""'''8 unit volume=""""""""""""""""""""""""""8*(x-8) /x.

The volume of wine remaining after 1st operation is (x-8) -{8* (x-8) /x} = { (x-8) ^2}/x...

Similarly in 4th draw pure wine is x (1-8/x) ^4.

There can be no doubt about the fact that they got the last ratio wrong in the question.

@rattle, sorry but you are right if the replacement stopped after the first round. After that when on the second round 8 litres is taken from the cask, it includes some wine and some water and then 8 litres of water in put in. This continues twice more. Sorry but we cannot assume x to be the amount of water in the cask after mixing is complete.

After n operations the wine left is {(x-8)^n}/x^n-1.

However, at the end of every cycle the total quantity is x. Therefore quantity of water left after n operations is x-quantity of wine left= x-{(x-8)^n}/x^n-1.

Now, dividing quantity of wine left by quantity of water left we get: {(x-8)^n}/{x^n-(x-8)^n} = 16/65. Substitute each one of the options n see which one matches rather than going to solve this complex expression.

Explanation for (x(1 - 8/x)pow 4)/x =16/81.

In question it's given water/wine = 16/65,
We assumed x= total wine in cask,
That means total capacity of cask is x,

So looking at ratio 16/65 we can see that total
Capacity is (16+65) = 81.

(x(1 - 8/x)pow 4) is nothing but the wine replaced by water, In other word, it is the quantity of water in a final mixture of water and wine.

So, (x(1 - 8/x)pow 4)/x must be 16/81.

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.

After 3rd time wine remain=[x(1-8/x)^2]-[x(1-8/x)^2]x/8.
= [x(1-8/x)^2]{1-8/x}
= [x(1-8/x)^2](1-8/x)
= x(1-8/x)^3 and so on;

For n operation remain=x(1-8/x)^n.