Problems involving percent are called percentage problems. There are three types of percentage problems. They are:

1. Finding a percent of a given number,
2. Finding what percent one number is of another,
3. Finding the number when the percent of the number is given.

Concept of Percentage
By a certain percent, we mean that many hundredths. Thus [latex]x[/latex] percent means [latex]x[/latex] hundredths, written as [latex]x[/latex]%
To express [latex]x[/latex]% as a fraction: We have , [latex]x[/latex]% = [latex]\frac{x}{100}[/latex]
Examples:

1. 20% = [latex]\frac{20}{100}[/latex] = [latex]\frac{1}{5}[/latex];
2. 48% = [latex]\frac{48}{100}[/latex] = [latex]\frac{12}{25}[/latex].

To express [latex]\frac{a}{b}[/latex] as a percent: We have, [latex]\frac{a}{b}[/latex] = [latex](\frac{a}{b})[/latex] x 100%
Examples:

1. [latex]\frac{1}{4}[/latex] = [[latex]\frac{1}{4}[/latex] x 100] = 25%
2. 0.6 = [latex]\frac{6}{10}[/latex] = [latex]\frac{3}{5}[/latex] = [[latex]\frac{3}{4}[/latex] x 100]%= 60%

If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is:

=[([latex]\frac{R}{(100 + R)}[/latex]) x 100]%

If the price of the commodity decreases by R%, then to maintain the same expenditure by increasing the consumption is:

=[([latex]\frac{R}{(100 - R)}[/latex]) x 100]%

Examples 1 In the new budget, the price of kerosene oil rose by 25%. By how much percent must a person reduce his consumption so that his expenditure on it does not increase?
Solution:

Reduction in consumption = [[latex]\frac{R}{(100 + R)}[/latex] x 100]% = ([latex]\frac{25}{125}[/latex] x 100)% = 20%.

Examples 2 The price of wheat falls by 16%. By what percentage a person can increase the consumption in order to maintain the same budget?
Solution:

Increase in Consumtion = [([latex]\frac{R}{(100 - R)}[/latex]) x 100]% = ([latex]\frac{16}{84}[/latex] x 100)% = [latex]\frac{400}{21}[/latex]% = 19.04% = 19%.

Let the population of the town be P now and suppose it increases at the rate of R% per annum, then:

1. Population after [latex]n[/latex] years = [latex]p[1 + (\frac{R}{100})]^{n}[/latex]
2. Population [latex]n[/latex] years ago = [latex]\frac{p}{[1 + (\frac{R}{100})]^{n}}[/latex]

Examples 1 The population of a town is 1,76,400. If it increase at the rate of 5% per annum, what will be its polulation 2 years hence? What was it 2 years ago?
Solution:

Population after 2 years = 176400 x [latex](1 + \frac{5}{100})^{2}[/latex] = (176400 x [latex]\frac{21}{20}[/latex] x [latex]\frac{21}{40}[/latex]) = 194481
Population 2 years ago = [latex]\frac{176400}{(1 + \frac{5}{100})^{2}}[/latex] = (176400 x [latex]\frac{20}{21}[/latex] x [latex]\frac{20}{21}[/latex]) = 160000.

Examples 2 The population of a town increases by 5% anually. If its population in 2001 was 1,38,915, what it was in 1998?
Solution:

Population in 1998 = [latex]\frac{138915}{(1 + \frac{5}{100})^{3}}[/latex] = (138915 x [latex]\frac{20}{21}[/latex] x [latex]\frac{20}{21}[/latex] x [latex]\frac{20}{21}[/latex]) = 120000.

Let the present value of a machine be [latex]P[/latex]. Suppose it depreciates at the rate [latex]R[/latex]% per annum. Then:

1. Value of the machine after [latex]n[/latex] years = [latex]p[1 - (\frac{R}{100})]^{n}[/latex]
2. Value of the machine [latex]n[/latex] years ago = [latex]\frac{p}{[1 - (\frac{R}{100})]^{n}}[/latex]

Examples 1 The value of a machin depreciates at the rate of 10% per annum. If its present value is Rs. 1,62,000, what will be its worth after 2 years? What was the value of he machine 2 years ago?
Solution:

Value of the machine after 2 years
= Rs. [162000 x [latex](1 - \frac{10}{100})^{2}[/latex]] = Rs. (162000 x [latex]\frac{9}{10}[/latex] x [latex]\frac{9}{10}[/latex]) = Rs. 131220.
Value of the machine 2 years ago
= Rs. [latex][\frac{162000}{(1 - \frac{10}{100})^{2}}][/latex] = Rs. (162000 x [latex]\frac{10}{9}[/latex] x [latex]\frac{10}{9}[/latex]) = Rs. 200000.

Examples 2 Depreciation applicable to an equipment is 20%. The value of the equipment 3 years from now will be less by:
Solution:

Let the preent value be Rs. 100.
Value after 3 years = Rs. [100 x [latex](1 - \frac{20}{100})^{3}[/latex]] = Rs. (100 x [latex]\frac{4}{5}[/latex] x [latex]\frac{4}{5}[/latex] x [latex]\frac{4}{5}[/latex]) = Rs. 51.20
Therefore, Reduction in value = (100 - 51.20)% = 48.8%.

1. To express [latex]x[/latex]% as a fraction: [latex]x[/latex]% = [latex]\frac{x}{100}[/latex]
2. To express [latex]\frac{a}{b}[/latex] as a percent: [latex]\frac{a}{b}[/latex] = ([latex]\frac{a}{b}[/latex] x 100)%
3. If the price rate increases by R%, then decrease in consumption so as not to increase the expenditure is [[latex]\frac{R}{(100 + R)}[/latex] x 100]%
4. If the price rate decreases by R%, then increase in consumption so as not to decrease the expenditure is [[latex]\frac{R}{(100 - R)}[/latex] x 100]%
5. Let the population ⇒ P, then

(a). Increase in rate R% per annum, then

(i). Population after n years = [latex]{P(1 + \frac{R}{100})}^n[/latex]
(ii). Population n years ago = [latex]{\frac{p}{(1+ \frac{R}{100})^n}}[/latex]

(b). Decrease in rate R% per annum, then

(i). Population after n years = [latex]{P(1 - \frac{R}{100})}^n[/latex]
(ii). Population n years ago = [latex]{\frac{p}{(1 - \frac{R}{100})^n}}[/latex]

1. Express each of the following as a fraction: a). 0.8% b). 28% c). 76% d). 0.006% ?
Solution:

a). Given 0.8%

0.8% = [latex]\frac{0.8}{100}[/latex] = [latex]\frac{8}{1000}[/latex] = [latex]\frac{1}{125}[/latex]

b). Given 28%

28% = [latex]\frac{28}{100}[/latex] = [latex]\frac{7}{25}[/latex]

c). Given 76%

76% = [latex]\frac{76}{100}[/latex] = [latex]\frac{19}{25}[/latex]

d). Given 0.06%

0.06% = [latex]\frac{0.06}{100}[/latex] = [latex]\frac{6}{10000}[/latex] = [latex]\frac{3}{5000}[/latex]

Therefore,
0.8% = [latex]\frac{1}{125}[/latex]
28% = [latex]\frac{7}{25}[/latex]
76% = [latex]\frac{19}{25}[/latex]
0.06% = [latex]\frac{3}{5000}[/latex]

2. Determine the value of 14% of 350 + 48% of 250?
Solution:

Given that
14% of 350 + 48% of 250
=[latex]\frac{14}{100}[/latex] x 350 + [latex]\frac{48}{100}[/latex] x 250
=49 + 120
=169
Therefore, 14% of 350 + 48% of 250 = 169

3. Find the missing number 10% of ? = 250.
Solution:

Given that 10% of ? = 250
Assume the unknown value as [latex]x[/latex]
⇒[latex]\frac{10}{100}[/latex] x [latex]x[/latex] = 250
⇒[latex]x[/latex] = 250 x 10
⇒[latex]x[/latex] = 2500
Therefore, missing value is [latex]x[/latex] = 2500

4. Difference between two numbers is 122. If 6.8% of one number is 12.4% of the other number, Find the two numbers?
Solution:

Let the two numbers be [latex]x[/latex] and [latex]y[/latex]
Given that
6.8% of [latex]x[/latex] = 12.4% of [latex]y[/latex]
⇒68[latex]x[/latex] = 124 [latex]y[/latex]
⇒[latex]x[/latex] = [latex]\frac{124}{68}y[/latex]
⇒[latex]x[/latex] = [latex]\frac{31}{17}y[/latex]
also given [latex]x[/latex] - [latex]y[/latex] = 122 -------(i)
Now substitute value of [latex]x[/latex] in equation (i)i.e.
⇒[latex]\frac{31}{17}y[/latex] - [latex]y[/latex] =122
⇒[latex]\frac{(31-17)y}{17}[/latex] = 122
⇒[latex]14y[/latex] = 122 x 17
⇒[latex]y[/latex] = [latex]\frac{122 * 17}{14}[/latex]
⇒[latex]y[/latex] = 148.14
Now substitute the value of y in equation (i)
[latex]x[/latex] - [latex]y[/latex] = 122
⇒[latex]x[/latex] = 122 + 148.4
⇒[latex]x[/latex] = 270.4
Therefore the two numbers are 270.4 and 148.14

5. The population of a town is 2,00,000. If it increases at the rate of 10% per annum, what will be its population 2 years hence? what was it 2 years ago?
Solution:

Consider, Population after n years = [latex]{P(1 + \frac{R}{100})}^n[/latex]
i.e. Population after 2 years = [latex]{200000(1 + \frac{10}{100})}^2[/latex]
⇒P = 200000 x [latex]\frac{121}{100}[/latex]
⇒P = 242000
Now Consider, Population n years ago = [latex]{\frac{p}{(1+ \frac{R}{100})^n}}[/latex]
i.e. Population 2 years ago = [latex]{\frac{200000}{(1+ \frac{10}{100})^2}}[/latex]
⇒P = 200000 x [latex]\frac{100}{121}[/latex]
⇒P = 165289.25
Therefore, Population after 2 years =242000 and
Population 2 years ago = 165289.25

6. Express [latex]5\frac{3}{2}[/latex] as rate percent
Solution:

Given [latex]5\frac{3}{2}[/latex]
⇒[latex]\frac{13}{2}[/latex]
⇒([latex]\frac{13}{2}[/latex] x 100)%
⇒(13 X 50)%
⇒650 %
Therefore, [latex]5\frac{3}{2}[/latex] = 650 %