Quants 14 - Alligations

Key Points to Remember:

If the average of the measured quantity was

A1 for group 1 containing n1 elements

A2 for group 2 containing n2 elements

A3 for group 3 containing n3 elements

Ak for group k containing nk elements

We say that the weighted average, Aw is given by:

Aw = (n1A1 + n2 A2 + n3 A3 + ……. + nkAk) / (n1 + n2 + n3 … + nk)

That is, the weighted average

Sum total of all groups / Total number of elements in all groups together

In the case of the situation where just two groups are being mixed, we can write this as:

Aw = (n1A1 + n2 A2 )/(n1 + n2)

Rewriting this equation we get: (n1 + n2) Aw = n1A1 + n2A2

n1(Aw – A1) = n2 (A2 – Aw) or n1/n2 = (A2 – Aw)/(Aw – A1) Æ The alligation equation.

The Alligation Situation

Two groups of elements are mixed together to form a third group containing the elements of both the  groups.

If the average of the first group is A1 and the number of elements is n1 and the average of the second group is A2 and the number of elements is n2, then to find the average of the new group formed, we can use either the weighted average equation or the alligation equation.

As a convenient convention, we take A1 < A2. Then, by the principal of averages, we get A1 < Aw < A2.

Graphical Representation of Alligation

The formula illustrated above can be represented by the following cross diagram:

As we have seen, there are five variables embedded inside the alligation equation. 

These being: the three averages Æ A1, A2 and Aw and the two weightsÆ n1 and n2 

Based on the problem situation, one of the following cases may occur with respect to the knowns and the unknown, in the problem.

Case I

Known

(a) A1, A2, Aw 

(b) A1, A2, Aw, n1 

Unknown

(a) n1 : n2

(b) n2 and n1 : n2

Case II 

Known - A1, A2, n1, n2 

Unknown - Aw

Case III 

Known - A1, Aw, n1, n2 

Unknown - A2

Now, let us try to evaluate the effectiveness of the cross method for each of the three cases illustrated

above:

Case 1: A1, A2, Aw are known; may be one of n1 or n2 is known.

To find: n1 : n2 and n2 if n1 is known OR n1 if n2 is known.

Practice Questions:

Question 1:

On mixing two classes of students having average marks 25 and 40 respectively, the overall average obtained is 30 marks. Find

(a) The ratio of students in the classes

(b) The number of students in the first class if the second class had 30 students.

(a) Hence, solution is 2 : 1.

(b) If the ratio is 2 : 1 and the second class has 30 students, then the first class has 60 students.

Question 2:

4 kg of rice at ` 5 per kg is mixed with 8 kg of rice at ` 6 per kg. Find the average price of the mixture.

The average of 4 kg rice at Rs. 5 and 8 kg rice at Rs. 6 is,

(6 - Aw) : (Aw - 5 ) :: 4 : 8

6 - Aw / Aw - 5 = 4 / 8

2 ( 6 - Aw ) = 1 ( Aw - 5 )

12 - 2Aw = Aw - 5

17 = 3Aw

Aw = 17 / 3 = ` 5.67 

Question 3:

5 kg of rice at ` 6 per kg is mixed with 4 kg of rice to get a mixture costing ` 7 per kg. Find the price of the costlier rice.

The price of the costlier rice can be calculated as,

( x - 7 ) : ( 7 - 6 ) :: 5 : 4

(x - 7 ) / 1 = 5 / 4

4x - 28 = 5

4x = 33

x = 33 / 4 = ` 8.25

Question 4:

How many kgs of wheat costing Rs. 8 per kg must be mixed with 36 kg of rice costing Rs. 5.40 per kg so that 20% gain may be obtained by selling the mixture cost Rs. 7.30 per kg?

SP of 1 kg mixture = 7.20 with 20% gain
CP of 1 kg mixture = 100 / 1120 * 7.20 = Rs. 6

Wheat 1 : Wheat 2 = 60 : 200 = 3 : 10

Let x kg of Wheat 1 be mixed with 36 kg of Wheat 2,

Then 3 / 10 = x / 36

10 x = 108

x = 10.8 kg

Question 5:

The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio, the liquids in both the vessels be mixed to obtain a new mixture in vessel C containing half milk and half water?

Let the CP of milk be Re. 1.
Milk in 1 litre mixture of A = 4 / ( 4 + 3 ) = 4 / 7 = Re. 4 / 7
Milk in 1 litre mixture of B = 2 / 5 = Re . 2 / 5

Milk in 1 litre mixture of C = 1 / 2 = Re . 1 / 2

Required Ratio = 1/10 : 1/14 = 14/140 : 10/140 = 7 : 5

Time to Think:

Two vessels A and B contain spirit and water mixed in the ratio 5 : 2 and 7 : 6 respectively. Find the ratio in which these mixtures be mixed to get a new mixture in vessel C containing spirit and water in the ratio 8 : 5?