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.

If ingredient worth no value (water) is added to another ingredient of some value(milk), when the mixture (solution) is sold at cost of ingredient with some value and profit earned is 'P' percent, then amount of ingredient with no value(water) added = (P/100) x Amount of Milk.
Ratio of Water to Milk is P:100

If Solution is sold at some other rate than rate of milk, say K% higher or lower rate than milk, than;
Actual rate of solution = Rate of milk x (100+k)/100, when rate is K percent higher than rate of milk; and,
Actual rate of solution = Rate of milk x (100-k)/100, when rate is K percent lower than rate of milk.
Let the actual Rate be S,
Let P = S-100.
Ratio of Milk to Water is, P:100.

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.

Let the amount of pure milk in the mixture be x litres.
Since the mixture is sold at the cost price of 1 litre, but he gains 25%, the amount of milk he is effectively selling is:
𝑥=1/1.25 = 0.8 liters of milk
This means that in 1 litre of the mixture, there are 0.8 litres of milk.

Since the total volume of the mixture is 1 litre, the amount of water in the mixture is:
1 − 0.8 = 0.2 liters of water.
Now, the percentage of water in the mixture is:
0.2/1 × 100 = 20%.

There's a formula for this in Profit and Loss:

Gain% = Error/(True Value - Error).

Since Gain% = 25% (or 1/4), simply take any True Value and you'll find the Error amount for that.

Now that Error is how much of the True Value (in %), is your answer.

Normally we take True Value as 100 for a direct answer but you can take any value and then find the answer.

Eg. In this case,

1/4 = Error / (100-Error) => Error = 100/5 => 20%.

That's it.

Let the original amount of milk be 100 and the cost price is Rs. 1/L.

CP1 = 100.
SP1 = 100.

Case 2 : When the milk is milk with x liters of water milk remaining is 100-x.

But now the SP is 100.

Given gain% = 25.

Thus, (100-(100-x))/100-x *100 = 25.

x = 20 liters.

Therefore, 20 L of water is added to 80 liters of milk so that the quantities remain the same however the selling price now makes all the difference!

Let the original amount of milk be 100 and the cost price is Rs. 1/L.

CP1 = 100.
SP1 = 100.

Case 2 : When the milk is milk with x liters of water milk remaining is 100-x.

But now the SP is 100.

Given gain% = 25.

Thus, (100-(100-x))/100-x *100 = 25.

x = 20 liters.

Therefore, 20 L of water is added to 80 liters of milk so that the quantities remain the same however the selling price now makes all the difference!

CP of 1 liter of milk = Re1.
SP of 1 liter of milk = Re1(given).

Gain = 25%(for the mixture [M+W]).

CP of milk(mixture) = (SP*(100/100+gain)) = 1*100/125 = 4/5.

Hence by allegation rule,

Milk(old) water
CP = Re1(dearer) CP = 0(for milk)(cheaper)
Mixture
CP = 4/5(mean).

Cheaper:Dearer = (1-4/5)/(4/5-0) = 1:4.
Water : milk = 1:4.
Water% = 1/5*100 = 20%.

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%.

Assume he sells 100 litres of milk.

In those 100 litres, he mixes 25 %of water that means he mix 25 litres of water to 100 litres of Milk.
So , the total mixture is 125 litres.

The profit gained by the seller is nothing but the amount of water added to milk. i.e 25 litres

Now,profit% = amount of water added/total amount of solution * 100.
25/125 * 100 = 20%.