Aptitude - Alligation or Mixture - Discussion

Thanks priya.

@sanjay

Its a formula....

[a(1-X/a)n]

a=initial quantity, X=quantity taken out, n=raised power on small bracket(no. of action done)

How did quantity of water become x?

It is like this:

x(1-(8/x)^4)-(1) gives quantity of wine, whereas final ratio is 16/81. So final wine quantity is (16/81)*x which is equal to (1).

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.

Let "x" liters of wine was present.

This wine was replaced by 8 liters of water for 4 times. So, wine at present is x[1-(8/x)]^4.

Also given that wine to water ratio at present is 16:65. So wine at present = 16 x/16+65.

So x[1-(8/x)]^4 = 16 x/16+65.

x = 24.

It is possible that drawn out 8l wine from the mixture the answer can be 24l, because whenever the wine is drawn from the liquid it is again filled with the water in the same amount.

Is another easy method for this problem?

Hello anybody explain of to me.

81 is how its coming?