 ## Unitary Method

Subject: Compulsory Maths

#### Overview

Unitary Method is a technique in mathematics for solving a problem finding the value of the single unit and then finding the necessary value by multiplying the single unit value. Time and amount of work done are in direct proportion.
##### Unitary Method

Unitary Method

It is a technique in mathematics for solving a problem finding the value of the single unit and then finding the necessary value by multiplying the single unit value. If the quantities are in direct proportion, unit value is obtained by division. Similarly, if the quantities are in inverse proportion, unit value is obtained by multiplication. For examples,

Suppose the cost of 5 pens is Rs. 50.

Then, the cost of 1 pen is Rs. $\frac{50}{5}$ = Rs. 10

Here, 1 pen is the unit of quantity and Rs. 10 is the unit cost. The number of pens and their cost is in direct proportion.

Again, suppose 4 men can finish a work in 6 days.

Then, 1 man can finish the work in 4 × 6 days = Rs. 24 days.  Here, the number of men and their working days are in inverse proportion.

Time and Work

In less time we do less amount of and in more time we do the greater amount of work. So, Time and amount of work done are in direct proportion. In this case, the amount of work done in unit time is obtained by division. For example,

In 8 days, Hari can do 1 work.

In 1 day, Hari can do $\frac{1}{8}$ part of work.

$\frac{1}{8}$ parts of work are done in 1 day.

Similarly, in 2 days, Hari can do$\frac{2}{8}$ =$\frac{1}{4}$ parts of the work

In 3 days, Hari can do$\frac{3}{8}$ parts of the work.

In 4 days, Hari can do$\frac{4}{8}$ =$\frac{1}{2}$ parts of work and so on.

##### Things to remember
•  If the quantities are in direct proportion, unit value is obtained by division.
• If the quantities are in inverse proportion, unit value is obtained by multiplication.
•  The amount of work done in unit time is obtained by division.
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### Videos for Unitary Method

Solution:

The cost of 12 m of clothes = Rs 600
The cost od 1 m of clothes = Rs $\frac{600}{12}$
= Rs 50
The cost of 8 m of clothes = 8 × Rs 50
= Rs 400
So, the required cost of 8 m of clothes is Rs 400.

Solution:

Here,
the length of the floor (l) = 14 m
the breadth of the floor (b) = 8.5m
∴ Area of the floor = l × b
= 14m × 8.5m
= 119 m2
Now,
the cost of carpeting  119 m2 is 119 × Rs 75
= Rs  8925
So, the required cost of carpeting the floor is Rs 8925.

Solution:

Here,
Rs 2 is the tax for the income of Rs 100
Re 1 is the tax for the income of Rs $\frac{100}{2}$
= Rs 50
Rs 525 is the tax for the income of Rs 525 × 50
= Rs 26250
So, the required income of the man is Rs 26250

Solution:

Here,
the cost of $\frac{3}{4}$ parts of a land = Rs 15,000
The cost of 1 (whole) land = Rs $\frac{15000}{\frac{3}{4}}$
= Rs $\frac{15000 × 4}{3}$
= Rs 20,000
The cost of $\frac{2}{5}$ parts of land = Rs 20,000 × $\frac{2}{5}$
= Rs 8,000
so, the required cost is Rs 8,000.

Solution:

Here,
In 20 days, 7 workers build a wall
In 1 days, 20 × 7 workers build the wall
In 14 days, $\frac{20 × 7}{14}$ workers build the wall = 10 workers build the wall
∴ The required number of workers to be added = 10 - 7
= 3 workers

Solution:

In 10 days, (X + y) can finish 1 work
In 1 days, (X = y) can finish $\frac{1}{10}$ parts of the work
Also,
In 15 days, X can finish 1 work
In 1  days , X can finish $\frac{1}{15}$ parts of the work.
Now,
In 1 days (X + Y - X) can finish ($\frac{1}{10}$ - $\frac{1}{15}$) parts of the work
In 1 days Y can finish ($\frac{3 - 2}{30}$) parts of the work = $\frac{1}{30}$ parts of the work.
∴ Y can finish $\frac{1}{30}$ parts of the work in 1 day.
Y can finish 1 work in $\frac{1}{\frac{1}{30}}$ days = 1 × $\frac{30}{1}$ = 30 days.
So, Y alone can finish the work in 30 days

Solution:

Here,
In 8 days, Hari can do 1 work
In 1 day, Hari can do $\frac{1}{8}$ parts of the work.
In 6 days, Hari can do $\frac{6}{8}$ parts of the work = $\frac{3}{4}$ parts of the work.
Now,
Remaining parts of work = (1 - $\frac{3}{4}$)
= $\frac{1}{4}$ parts  of the work
So, Gopal finished $\frac{1}{4}$ part of the work

Solution:

In 6 days, A can do 1 work
In 1 day, A can do $\frac{1}{6}$ parts of the work
Also,
In 12 days, B can do 1 work
In 1 day, B can do $\frac{1}{12}$ parts of the work
Now,
In 1 day, (A + b) can do ($\frac{1}{6}$ + $\frac{1}{12}$) partss of the work
= ($\frac{2 + 1}{12}$) parts of the work = $\frac{3}{12}$ = $\frac{1}{4}$ parts of the work
∴ (A + B) can do $\frac{1}{4}$ parts of the work = 1
(A + B) can do 1 work in $\frac{1}{\frac{1}{4}}$ day = 1 × $\frac{1}{4}$ = 4 days
So, A and B finish the work in 4 days working together.

Solution:

Here,
after joining 10 more people, total number of the people in the group = 0 + 10 = 50
40 people had provisions for 60 days
1 people had provisions for 40 × 60 days
50 people had provisions for $\frac{40 × 60}{50}$ days = 48 days
So, the provisions would long last for 48 days.

Solution:

Here, Cost of 5 pens = Rs. 100

or, Cost of 1 pen = $\frac{Rs.100}{5}$ = Rs. 20

Then, The cost of 2 pens = 2 × Rs. 20 = Rs. 40 ans.

Solution:

6 men finish a piece of work in 5 days.

Then, 1 man finishes the same piece of work in 6 × 5 days = 30 days.

Now, 3 men finish the same piece of work in $\frac{30}{3}$ = 10 days ans.

Solution:

Here, The cost of 12 litres of milk = Rs. 432

or, The cost of 1 litres of milk = Rs. $\frac{432}{12}$ = Rs. 36

The cost of 5 litres of milk = 5 × Rs. 36 = Rs. 180

So the required cost of 5 litres of milk is Rs. 180.