Percentage


A percentage is a fraction whose denominator is 100 and the numerator of the fraction is called the rate percent.
Percent is denoted by the sign '%'.
The term percent means for every hundred.

If we have to find x % of a number, then x % of number =
x
times of that number
100
For example: 20% of 700 =
20
× 700 = 20 × 7 = 140
100

Some fast result :-

10% of a number =
Number
10
11.11% of a number =
Number
9
12.50% of a number =
Number
8
14.28% of a number =
Number
7
6.66% of a number =
Number
6
20% of a number =
Number
5
25% of a number =
Number
4
33.33% of a number =
Number
3
50% of a number =
Number
2
Example: Find 16.66% of 252.
Solution: 16.66% of 252 =
1
× 252 = 42
6
Example: Find 14.28% of 504.
Solution: 14.28% of 504 =
1
× 504 = 72
7
Example: Find 33.33% of 369.
Solution: 33.33% of 369 =
1
× 369 = 123
3

Conversion of Percent into Number :-

To convert any percent into number we should divided by 100.

Example : Express 20% in fraction.

Solution: 20% =
20
=
1
100 5
Example : Express 75 % in fraction.
Solution: 75% =
75
=
3
100 4
Example : - Express 64% in decimal.
Solution: 64% =
64
= 0.64
100

Conversion of Fraction into Percentage :-

To convert fraction into per cent we should multiply by 100 .

Example : Convert
3
into percent.
5
Solution:
3
× 100 = 60%
5
Example : Express 31/4 in percent.
Solution: 31/4 × 100 =
13
× 100 = 13 × 25 = 325
4

Expressing one quantity as a percent with respect to other .

The quantity to be expressed in percent
× 100 %
other quantity

Note : For this formula, both the quantity must be in same metric unit.
Example: 50 kg is what per cent of 250 kg?

Solution:
The quantity to be expressed in percent
× 100 %
other quantity
=
50
× 100 % = 20%
250

Example: 20 g is what per cent of 2 kg?
Solution: First quantity = 20 g and second quantity = 2 kg = 2000 g

The quantity to be expressed in percent
× 100 %
other quantity
=
20
× 100 % = 1%
2000

Increasing percent and Decreasing percent :-

Percentage increase =
Increase
× 100
original value
Increase = New value - Original value

Example: Rent of the house is increased from ₹ 8000 to ₹ 10000. Express the increase in price as a percentage of the original rent.
Solution: Given, Initial rent = ₹ 8000 , New rent = ₹ 10000
Increase = New rent - Original rent
= 10000 - 8000 = ₹ 2000

Percentage increase =
Increase
× 100
original value

Percentage increase =
2000
× 100 = 25%
8000

Percentage decrease =
decrease
× 100
original value
Decrease = Original value - New value

Example: The cost of a bike last year was ₹ 50000. Its cost this year is ₹ 45000. Find the percent decrease in its cost.
Solution: Given, last year cost = ₹ 50000
This year cost = ₹ 45000
Decrease = Original cost - New cost
Decrease = 50000 - 45000 = 5000

Percentage decrease =
decrease
× 100
original value

Percentage decrease =
5000
× 100 = 10%
50000

Use of Percent :-

Trick: Successive change :-

Net Percentage Change = A + B +
A × B
%
100

where, A = first change and B = second change

Example: If the salary of a person increased by 10% of income. And again increased by 10% in next year. Find net percent change in his total salary.
Solution: Let A = 10% and B = 10%

Net Percentage Change = A + B +
A × B
%
100

Net percent change in salary = 10 + 10 +
( 10 × 10 )
= 20 + 1 = 21%
100

Example: The price of an article is first increased by 20% and later on the price were decreased by 25% due to reduction in sales. Find the net percentage change in final price of article.
Solution:
Let A = 20% and B = -25%

Net Percentage Change = A + B +
A × B
%
100

Net percent change in final price = 20 - 25 +
{ 20 × ( - 25 ) }
= -5 - 5 = -10%
100
10 % decrease in the price of an article.

Trick: To solve "Expenditure Constant" questions

Reduction in consumption =
x
× 100
100 + x

Increase in consumption =
x
× 100
100 - x
where, x = increase or decrease
Use '-'sign for decrease

Example: Price of sugar is increased by 10%. Find by how much % the consumer should the consumption of sugar to keep the expenditure constant.
Solution: Here, x = 10%
Using the above given formula , we have

Reduction in consumption =
10
× 100 =
10
× 100
100 + 10 110

Reduction in consumption =
100
= 9.09%
11


Example: Price of rice is decreased by 20 %. Find by how much per cent the consumption should be increased to keep the expenditure constant.
Solution: Here x = 20%
Using the above given formula , we have

Increase in consumption =
20
× 100 =
20
× 100 = 25%
100 - 20 80

Trick: If the population of a town is P and it increases ( or decreases ) at the rate of r% per year, then

Population after n year = P 1 +
r
n
100

Population n year ago = P 1 -
r
n
100


Use '-' sign for decrement
Example: The population of a town is 472300. If it increases at the rate of 10 % per year, them what will be its population after 2 yr.
Solution:- given, P = 473200 , r = 10% , n = 2 yr

Population after 2 year = P 1 +
r
n
100

Population after 2 year = 472300 1 +
10
2
100

Population after 2 year = 472300 1 +
1
2
10

Population after 2 year = 472300
11
2
10

= 4723 × 11 × 11 = 571483

Trick: If the present population of a city is P and there is a increment or decrement of r1%, r2% and r3% in first, second and third year respectively, then

Population of city after 3 yr = P 1 +
r1
1 +
r2
1 +
r3
100 100 100

Use'-' sign for decrement
Example: Population of a city in 2010 was 200000. If in 2011 there is an increment of 10%, in 2012 there is a decrement of 20% and in 2013 there is increment of 30%, then find the population of city at the end of year 2013.
Solution:- given, P = 200000, r1 = 10%, r2 = 20%, r3 = 30%
Using above given formula ,

Population of city at the end of year 2013 = P 1 +
10
1 -
20
1 +
30
100 100 100

Population of city at the end of year 2013 = 200000 ×
11
×
4
×
13
10 5 10

= 228800