Aptitude - Alligation or Mixture - Discussion

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

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.

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.

How can did wine left ratio is 16:65 to change 16:81 please explain anybody?

See the formula is well understood a(1-b/a)^n. Now even after adding and removing water we have the total amount o be the same that is x.

So we have 16/81 of x = This is the part of wine left that should be equal to the formula.

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.

Given ratio i.e. 16/65 is wine/water.

Hence wine/(wine+water) is 16/81.

Total quantity of wine+water will remain fixed i.e. equal to initial quantity of wine i.e. x.

Hence final ratio of wine/(wine+water) will be (wine left/initial wine).

Hence it will be x (1-8/x)^n/x equal to 16/81.

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.