We will take the Weighted Average rule
discussed in the previous section and rewrite the formula such that the
quantity terms come on one side and the price terms come on the other side. If
we do this we get the rule 
This is called the RULE OF ALLIGATION. This
rule connects quantities and prices in mixtures. This can also be written as 
In a descriptive manner, the Rule of
Alligation can be written as 
= 
This rule is a very powerful rule and is
useful in solving problems on weighted averages and mixtures. This rule is also
useful in a number of problems which can be treated as mixtures and applied to
parameters other than price also. .
If there is P litres of pure liquid initially
in a vessel and in each operation, Q litres is taken out and replaced by water,
then at the end of n such operations, the concentration (k) of Q the liquid in
the solution is given by 
= k
This gives the concentration (k) of the
liquid as a PROPORTION of the total volume of the solution. .
If the concentration has to be expressed as a
percentage, then it will be equal to 100k. .
If the volume of the liquid is to be found
out at the end of n operations, it is given by kP, i.e., the concentration k
multiplied by the total volume P of the solution. .
Example.1
Alok purchased 16 items. The average cost of
those 16 items is Rs.59. He then returned four items with an average price of Rs.60
and three other items costing Rs.39, Rs.49 and Rs.40. Find the average cost of
the remaining items. .
Solution
Total
cost of 16 items – Rs. (16 x 59). .
Total cost of the final 9 items
= Rs.[(16 x 59) - (4 x 60) - (39 + 49 + 40)] .
= Rs.576 .
New average = 576/9 = Rs.64 .
Example.2
A dealer sells a mixture of two varieties A
and B of tea at no per kg making 25% profit. Variety A costs Rs.22 per kg. If
the two varieties were mixed in the ratio 1 : 1, find the cost of variety B (in
Rs.per kg). .
Solution
Let the
cost price of the mixture be Rs.x per kg .
x + 
x = 30
5x/4 = 30
x = 24
Whenever two varieties (measured in kg) are
mixed in the ratio 1 : 1, cost of the mixture is given by the average of the
costs of the two varieties (in Rs per kg)
= 
= 24
Cost of variety B = 2(24) – Cost of variety A
= Rs.26 per kg. .
Example.3
In a group of 19 boys, one boy named Bhavan
ate 18 chocolates more than the average number of chocolates eaten by all the
boys. If the remaining 18 children ate an average of 5 chocolates, find the
number of chocolates eaten by Bhavan. .
Solution
Let the average
number of chocolates eaten by the boys bo x. .
Number of chocolates eaten by Bhavan = x +
18. .
Total number of chocolates eaten by the
others = (5)
(18) = 90
&⇒ 19x = x + 18 + 90
x = 6
So Bhavan ate 6 + 18 = 24 chocolates
Example.4
The average weight of 20 students of a class
is 25 kg. If one student named Amar leaves the class, the average weight of the
class decreases by 0.2 kg. Find the weight of Amar. (in kg.) .
Solution
Weight of Amar = Weight of 20 students of the
class -Weight of other 19 students of the class – (20) (25) -(19) (24.8) =
28.80 kg .
