Mixture and Alligation

Mixture and Alligation

In mixture problems, substances with different characteristics are combined and it is necessary to determine the characteristics of the resulting mixture. To solve such problems, we can take the help of Alligation rule. This rule can also be used to find the average value of mixture when the price of two or more ingredients is mixed together.

Alligation basically means “Linking”. The rule states that when different quantities of different ingredients are mixed together to produce a mixture of a mean value, the ratio of their quantities is inversely proportional to the differences in their cost from the mean value.

Rule of Alligation: 

If 2 ingredients are mixed in a ratio, then we use the following ratio formula:

(Quantity of cheaper/ Quantity of dearer) = (Cost Price of dearer – Mean Price) / (Mean Price – Cost Price of Cheaper), Cheaper Quantity : Dearer Quantity = (d – m) : (m – c)

• m = mean price
• d = Cost price of dearer
• c = Cost Price of cheaper 

Example 1:

In what proportion must rice at Rs. 3.25 per kg be mixed with rice at Rs. 3.80 per kg, so that the mixture would be worth Rs. 3.50 per kg? 

Solution:

Here, d = Rs. 3.80, c = Rs. 3.25, m = Rs. 3.50

Quantity of cheaper / Quantity of dearer = (Cost Price of dearer – Mean Price) / (Mean Price – Cost Price of Cheaper)

Q_c / Q_d = (3.80 – 3.50) / (3.50 – 3.25) = 6/5

So, they must be mixed in the ratio 6 : 5. 

Repeated Dilution of a Mixture:

There’s a special case in mixtures in which there is repeated dilution of a mixture with one of the ingredients, by removing, say n litres of the mixture and replacing it with n litres of one of the ingredients.

Say, there are x litres of water initially. Now, n litres of the water are replaced by n litres of wine. This operation is repeated t times.

Quantity of water left in the vessel = x (1 – [n/x])^t

Here, 

• x = total quantity
• n = quantity drawn every time, and,
• t = no. of times 

Example 2:

In a vessel, there are 100 litres of water. 10 litres of water are withdrawn and replaced with milk. Again 10 litres of water are withdrawn and replaced with milk. What is the quantity of water left in the mixture?

Solution:

Here, x = total quantity = 100 litres, n = quantity withdrawn every time = 10 litres, t = no. of times = 2

Quantity of water left in the vessel = 100 (1-[10/100])² = 100 * 9 * 9/ (10*10)

Thus quantity of water left = 81 litres