Aptitude - Alligation or Mixture - Discussion

40-4 == 36.

Now mixture contain 36 part milk and 4 part water.
36-4 = 32.

But all 4 part is not milk, it is mixture of water and milk.
In 4 part amount of water was=4*4/40=1*4/10=0.4.

So now milk = 32+0.4 = 32.4.

Process repeats one more time as follows:

Milk = 32.4-4 = 28.4.
This 4 is mixture of water and milk.
Water in this 4 lit is = 7.6*4/40 = 0.76.

So finally milk = 28.4+0.76 = 29.16.

40 lit milk.
4 lit taken out.

36 lit milk and 4 lit water.
Again 4 lit mixture remove from total mixture.

4/ (total mixture = 40) --- 1 eqn.

36*4/40 = 3.6 milk and 4*4/40 = 0.4 water are remove.

Remaining milk = 32.4 (36-3.6) and water = 7.6 (8-0.4) lit.

Again same process as equation (1).

32.40-3.24 = 29.16 lit milk remains in mixture.

Simplest approach. !

First the milk had removed and replace by water it is 1 time later it had repeated for 2 times.

So total 1+2 = 3 times.

So the value of n is taken as 3.

It has given repeated twice, so three is taken as a power? Still not getting.

For this type of questions the formula is a(1-b/a) whole power n.

a = Initial quantity.
b = Taken out quantity.
n = Nor of times.

If the same amount of milk is drawn the formula is applicable but when different amounts of milk is drawn in different cases what have to do I have a problem here please solve this and explain.

There was 40 litres of pure milk in a vessel 16 litres of pure milk was taken out and replaced with equal amount of water again 10 litres of mixture was taken out and replaced with equal amount of water, what is the amount of water in the vessel now? (in litres).

@Ramesh.

40(1-16/40)(1-10/40) = 18 litres milk and 22 litres of water.

If you are taking out 4 liters out of a 40 liter solution and replacing it with water, you are effectively replacing 1/10th of the solution with water.

=> Fraction of milk will become (9/10) of the original.

If the process is repeated 'n' times, fraction of milk will become (9/10)^n of the original.

We are given the process was done thrice.

Can it it solved by x and why quantity? like:

Currently, we have 36(M), 4(W). Now 4 litres is taken out.

So let milk remains: 36-x.

Water 4-(4-x).

Now add 4 litres water: so, milk = 36-x, water = [4-(4-x)]+4.

Again 4 litres taken out. Now suppose in that 4 lit why lit was milk. So milk remains: (36-x) -y and water = [[4-(4-x)]+4]-(4-y).

Now 4 lit water mixed so, milk = 36-x-y.

Water = [[4-(4-x)]+4]-(4-y)+4.

= [[4-4+x]+4]-4+y+4.

= [x+4+y].

But where is the problem in this approach? I just want to know that.

Initial Status 1 => Milk = 40 {Volume = 40}.

Water = 0.

Iteration 1 => Milk = 40-4 = 36 {Volume = 40}.

Water = 4.

Fraction of water = 4/40 = 1/10.

Fraction of milk = 36/40 = 9/10.

Iteration 2 => Milk = 36-(9/10) 4 = 324/10 {Volume = 40}.

[Since 4 litres mixture has been removed].

Water = 4+(4-(1/10)4) = 76/10.

[Since 4 litres water has been added].

Fraction of water = (76/10)/40 = 76/400.

Fraction of milk = (324/10)/40 = 324/400.

Iteration 3 => Milk = 324/10-(324/400)4 = 729/25 = 29.16 {Vol = 40}.

[Since 4 litres mixture has been removed].

Water = 4+(76/10-(76/400) 4) = 271/25 = 10.84.

[Since 4 llitres water has been added].