Aptitude - Alligation or Mixture - Discussion

@Shobana.

Step1: 40-(10% of 40) = 36.
Step2: 36-(10% of 36) = 32.4.
Step3: 32.4-(10% of 32.4) = 29.16.

This is the simplest method. But if you want to understand the method clearly, go through my first comment.

Please explain with simple steps!

@Munna: Here the key is every time we are removing same 4 L from the 40 L of mixture which constitutes to 10% of entire solution.

Hence the percentage remains constant.

If mixture has two types of liquids say A and B. If you remove x% of the entire mixture it is obvious that same x% is removed from both the types of liquids from their respective quantities.

In the above example whenever we remove 4 lit of solution out of 40 lit. We are removing 10% of entire solution which means we are removing 10% of milk and also 10% of water.

After Step 1: We have 36 lit of milk and 4 lit of water.

Step 2: Here we are removing 4 lit of entire solution, which means we are removing (10% of 36 = 3.6 L of milk) and (10% of 4 = 0.4 of water). So now we are left with (36 - 3.6 = 32.4L of milk) and (4 - 0.4 = 3.6 L of water). But again the mixture is added with 4 lit of water which makes (32.4L of milk + 7.6L of water) = 40 L of solution.

Step 3: Here we are again removing 4 lit of entire solution, which means we are removing (10% of 32.4 = 3.24L of milk) and (10% of 7.6 = 0.76 of water). So now we are left with (32.4 - 3.24 = 29.16L of milk) and (7.6 - 0.76 = 6.84L of water)

But again the mixture is added with 4 lit of water which makes (29.16L of milk + 10.84L of water) = 40 L of solution.

And this process goes on. Here using Percentage method is more effective rather going with ratios.

I don't understand why milk is getting reduced 10% each time while every time ratio of milk and water is getting changed.

Milk, water.

40, 0.

36, 4 10% of 40 is 4.

32.4, 7.6 10% of 36 is 3.6.

19.16 10% of 32.4 is 3.24.

Somebody please explain why percentage is constant?

It's just taking 10% away each time.

First extract & refill. 36l milk, 4l water.

Milk in solution = 9/10; water = 1/10.

2nd extract: 4l taken out.

Milk = 9/10*4 = 3.6 water taken out = 0.4.

Milk remaining = 32.4, water remaining = 7.6.

Milk in solution = 32.4/40; water in solution = 7.6/40.

3rd extract: Milk extracted = 4*81/100.

Milk left will be 32.4-3.24 = 29.16.

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].

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.

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.