Aptitude - Alligation or Mixture - Discussion

@ Pankaj:
The explanation states that the normal cost of milk is actually Re. 1. Now this dishonest milkman is buying some of that milk, adding some water and selling the new mixture for Re. 1 (when the actual value should be lower). So since he is selling the mixture at Re. 1, that becomes his selling price. This is because he has made a profit due to having used some water to dilute it. Now we need to figure out how much of the actual milk has gone into the final product and since water is free, the cost of the milk becomes the cost price.
So:
Cost price of milk + 25% profit = Selling price of milk mixture
Lets assume the cost price of milk is x.
x + 0.25x = 1.00 (since the selling price of mixture is 1 and the profit is 25% of cost price)
1.25x = 1.00
- we all hate decimals, so lets make it easier:
125x = 100
x = 100/125 = 4/5(thats where it comes from). Now that multiplication by 1 is really useless in this problem. Unless the problem, specifically states that he sold something like 10 liters, there is no need multiply by total liters sold. Then do alligation to figure out how much milk and how much water.

Hope that helps.

@Vikas Rawat.

Thank you you explained very well.

Suppose Milk man have 100 litre he want sell same 100 litre by adding some water.

Milk...................100 litre(initially)
Milk+ water.......100 litre(after addition of milk)
So, milk = 100-water.

Now, cost price Rs/-100.
Because of 25% profit selling price Rs/-125
100 * 100 = (100-water) * 125.
Water = 20.
Out of 100 litre water quantity is 20 litre.
So, answer is 20%.

Is 25% gain means he must have added 25% of water. So if 100lts of milk is there then you have 25lts of water. So part of solution which is water is 25/125. Then convert this into percentage.

Here gain%/100 + gain%*100 it means 25/100 + 25*100 = 20%.

Suppose 1 litre is 1rs.
Then 0.5 litres is 0.5 right.

x litres cost x rupee,
So gain = ((s.p-c.p)/c.p)*100;
So he actually sells it for 1 rupee as 1 litre of milk but he has less than 1 litre of milk, say x,
So x litres cost x rupee,
So 25=((1-x)/x)*100;,
x=4/5;,
So water is 1/5 which is 20%,

Hope it helps.

Formula is; (R/R+100)*100 So, ( 25/25+100)*100=20%.

let 100 litres of milk costs ---> x rs.

Now, they say how much water needs to be added to milk, we shall get 25 % gain
so,
100 lit milk+ y litres water costs --> x + 25 * X/100.

re-arranging the above equations

100 lit --> x rs.
100 lit + y litres water ---> x + 0.25x (adding the gain).

Cross multiply.

(x+0.25x) * 100 = 100*x + y*x.
solving we get;
25x = x * y.

y = 25 litres of water.

So now total mixture ( 100 lit milk + 25 lit of water ) = 125 lit of mixture.
to find out the percentage of water in 125 litres.

25*100/125 = 20 litre of water.

The solution is;

Water/milk * 100 is the formula for solving this.

Initially, water is 25% so remaining milk is 75% which makes W:M=1:4.
%water = w/total*100 and %milk= m/total*100 so %water = 1/5*100 = 20%.

I am unable to understand. Please, anybody explain me clearly.