## Class 9 Mathematics Note

# UNITARY METHOD

**Unitary method **

The **unitary method** is a technique in mathematics for solving a problem finding the value of single unit, and then finding the necessary value by multiplying the single unit value.

**Variation **

This shows the relation between the two quantities

**Types of variation**

**Direct variation:**

If the increase in one quantity results in increase in other quantities and decrease in one quantity results in decrease in other quantities, then the relation between the quantities is said to be in direct variation.

Egif the cost of 2 chair is Rs 400 then, the cost of 4 chair is Rs 800

**Indirect variation:**

If the increase in one quantity results in decreasein other quantities and decrease in one quantity results in increasein other quantities, then the relation between the quantities is said to be in indirect variation. Time and work

Eg20 men can do a piece of the work in 40 days

1 man can do a piece of the work in 80 days

**Some important formula **

a. A can do a piece of the work in x days

A can do $\frac{1}{{\rm{x}}}{\rm{\: }}$of the work in 1 days${\rm{\: }}$

b. No of days =$\frac{{{\rm{total\: work\: done}}}}{{{\rm{work\: done\: in\: }}1{\rm{\: days\: }}}}$

c. Time and work (direct variation)

More work →more time

Less work → less time

d. men andtime (indirect variation)

More men↑ less time ↓

Less men ↓more time ↑

e. If A_{1} men can do W_{1 }work in D_{1}days and A_{2 }can do W_{2 }work in D_{2 }days

$\frac{{{{\rm{A}}_1}{{\rm{D}}_1}}}{{{{\rm{W}}_1}{\rm{\: }}}} = \frac{{{{\rm{A}}_2}{{\rm{D}}_2}}}{{{{\rm{W}}_2}{\rm{\: }}}}{\rm{\: }}$

In case of hours

$\frac{{{{\rm{A}}_1}{{\rm{D}}_{1{\rm{\: }}}}{{\rm{H}}_1}}}{{{{\rm{W}}_1}{\rm{\: }}}} = \frac{{{{\rm{A}}_2}{{\rm{D}}_2}{{\rm{H}}_2}}}{{{{\rm{W}}_2}{\rm{\: }}}}$

f. If A can do a work in ‘x’ days and B can do the same work in ‘y’ days, then the number of days required to complete the work if A and B work together is$\frac{{{\rm{xy}}}}{{{\rm{x\: }} + {\rm{\: y}}}}$

**Ratio and proportion **

The term in the form of p/q is called ratio. A proportion is two ratios are equal. It can be written in two ways: two equal fractions, or, using a colon, a: b = c: d.

**Problem of the allegation **

In our daily life, we face many problems related to allegation. If the two quantities m and n are mixed and the cost of these object are a and b and the average cost of the mixture = ${\rm{Rs}}\frac{{{\rm{ma}} + {\rm{nb}}}}{{{\rm{m}} + {\rm{n}}}}$

**Examples **

1.A takes twice as much as B and thrice as much as C to finish a work; working together they can finish the work in 10 days. Find the time each will take to finish the work separately.

Soln

The working capacity of B = 2 times the working capacity of A

B = 2 A

The working capacity of C= 2 times the working capacity of A

C = 3 A

Now, A + B+ C can finish 1 work 10 days

A + 2A+ 3A can finish 1 work 10 days

6 A can finish 1 work in 10 days

A can do 1 work in 60 days

B can do 1 work in $\frac{1}{2}{\rm{\: }}$time taken by A = $\frac{1}{2}$*60 = 30 days${\rm{\: }}$

C can do 1 work in$\frac{1}{3}{\rm{\: time\: taken\: by\: A\: }} = {\rm{\: }}\frac{1}{3}{\rm{*}}60{\rm{\: }} = {\rm{\: \: }}20{\rm{\: days}}$

2. 3 duck eggs can be exchanged with 4 hen eggs .Similarly with 4 eggs of swan, 7 eggs of duck can be exchanged .If the cost of 2 hen eggs is Rs 7.5, then what is the cost of swan’s egg.

Soln

Cost of 7 eggs of duck = Cost of 4 eggs of swan

Cost of 1 egg of duck = Cost of 4/7 eggs of swan

Cost of 3 duck eggs = Cost of $\frac{4}{7}{\rm{*}}3$ = $\frac{{12}}{3} = 4{\rm{\: }}$hen eggs

Cost of 2 hen eggs = Rs 7.5

Cost of 1 hen egg = Rs 3.75

Cost of $\frac{{12}}{3}{\rm{\: }}$hen eggs = Rs 3.75 *$\frac{{12}}{3}$ = 4 *3.75 = Rs 15

i.e Cost of 3 duck eggs = Rs 15

Cost of 1 duck egg = Rs 5

Cost of 4/7 eggs of swan = Rs 5

Cost of 1 egg of swan = Rs 35/4 = 8.75

3. Two types of rice costing Rs 40 and Rs 30 per kg are mixed in the ratio of 3:4. Find the cost of rice mixture per kg.

Soln

a =40

b = 30

m : n= 3 : 4

The cost of per kg = $\frac{{{\rm{ma}} + {\rm{nb}}}}{{{\rm{m}} + {\rm{n}}}}$= $\frac{{{\rm{\: }}3{\rm{*\: }}40{\rm{\: }} + 4{\rm{*\: }}30}}{{3 + 4{\rm{\: }}}}$=$\frac{{120 + 120}}{7}$ =$\frac{{240}}{7}$ = 34.28