**Direct Formula / Rule 2 : **

**Theorem: **
The quantity of salt at Rs x per kg that a man
must mix with n kg of salt at Rs y per kg, so that he may, on
selling the mixture at Rs z per kg, gain p% on the outlay is given by

= [
^{100z - y(100 + p)}/_{x(100 + p) - 100z}
] X n

**Note : ** if we suppose that the quantity of salt at Rs x be m,
then we have.

**Example : **

How many kg of salt at 42 P per kg must a man mix
with 25 kg of salt at 24 P per kg, so that he may, on
selling the mixture at 40 per kg, gain 25% on the outlay?

**Detail Method : **
Let the required amount of salt be * x* kg

As per Question,

Because Selling price of the mixture = 40 per kg given

so Cost price,of the mixture = 40 X ^{100}/_{125} X (*x* + 25)

or, 42*x* + 24 x 25 = 32*x* + 32 x 25

or, 10*x* = 25 x 8

so *x* = 20 kg.

**Ailigation Method : **

Cost price,of the mixture = 40 + ^{100}/_{125 } P = 32 p per kg..

By the rule of fraction :

Ratio = 4 : 5

Thus for every 5 kg of salt at 24 P, 4 kg of salt at 42 P is used.

the required no. of kg = 25 x ^{4}/_{5} = 20

**Quicker Method : **Here you can use direct formula :

= [
^{100z - y(100 + p)}/_{x(100 + p) - 100z}
] X n

= [
^{100 x 40 - 24(100 + 25)}/_{42 x (100 + 25) - 100 x 40}
] X 25

= [
^{4000 - 3000}/_{5250 - 4000}
] X 25

= [
^{1000}/_{1250}
] X 25 = 20 kg.