  # Percentage Problems

#### Chapter 11 5 Steps - 3 Clicks

# Percentage Problems

### Introduction

Percentage Problems: A Percentage is a dimensionless ratio or number expressed as a fraction of 100. It is often denoted by the character (%).

Example: $$10$$% = $$\frac{10}{100}$$ = $$\frac{1}{10}$$

### Methods

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 $$x$$ percent means $$x$$ hundredths, written as $$x$$%

To express $$x$$% as a fraction: We have , $$x$$% = $$\frac{x}{100}$$

Examples:

1. 20% = $$\frac{20}{100}$$ = $$\frac{1}{5}$$;

2. 48% = $$\frac{48}{100}$$ = $$\frac{12}{25}$$.

To express $$\frac{a}{b}$$ as a percent: We have, $$\frac{a}{b}$$ = $$(\frac{a}{b})$$ x 100%

Examples:

1. $$\frac{1}{4}$$ = [$$\frac{1}{4}$$ x 100] = 25%

2. 0.6 = $$\frac{6}{10}$$ = $$\frac{3}{5}$$ = [$$\frac{3}{4}$$ 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:

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

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

=[($$\frac{R}{(100 – R)}$$) 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 = [$$\frac{R}{(100 + R)}$$ x 100]% = ($$\frac{25}{125}$$ 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 = [($$\frac{R}{(100 – R)}$$) x 100]% = ($$\frac{16}{84}$$ x 100)% = $$\frac{400}{21}$$% = 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 $$n$$ years = $$p[1 + (\frac{R}{100})]^{n}$$

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

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 $$(1 + \frac{5}{100})^{2}$$ = (176400 x $$\frac{21}{20}$$ x $$\frac{21}{40}$$) = 194481

Population 2 years ago = $$\frac{176400}{(1 + \frac{5}{100})^{2}}$$ = (176400 x $$\frac{20}{21}$$ x $$\frac{20}{21}$$) = 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 = $$\frac{138915}{(1 + \frac{5}{100})^{3}}$$ = (138915 x $$\frac{20}{21}$$ x $$\frac{20}{21}$$ x $$\frac{20}{21}$$) = 120000.

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

1. Value of the machine after $$n$$ years = $$p[1 – (\frac{R}{100})]^{n}$$

2. Value of the machine $$n$$ years ago = $$\frac{p}{[1 – (\frac{R}{100})]^{n}}$$

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 $$(1 – \frac{10}{100})^{2}$$] = Rs. (162000 x $$\frac{9}{10}$$ x $$\frac{9}{10}$$) = Rs. 131220.

Value of the machine 2 years ago

= Rs. $$[\frac{162000}{(1 – \frac{10}{100})^{2}}]$$ = Rs. (162000 x $$\frac{10}{9}$$ x $$\frac{10}{9}$$) = 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 $$(1 – \frac{20}{100})^{3}$$] = Rs. (100 x $$\frac{4}{5}$$ x $$\frac{4}{5}$$ x $$\frac{4}{5}$$) = Rs. 51.20

Therefore, Reduction in value = (100 – 51.20)% = 48.8%.

### Formulae

1. To express $$x$$% as a fraction: $$x$$% = $$\frac{x}{100}$$

2. To express $$\frac{a}{b}$$ as a percent: $$\frac{a}{b}$$ = ($$\frac{a}{b}$$ x 100)%

3. If the price rate increases by R%, then decrease in consumption so as not to increase the expenditure is [$$\frac{R}{(100 + R)}$$ x 100]%

4. If the price rate decreases by R%, then increase in consumption so as not to decrease the expenditure is [$$\frac{R}{(100 – R)}$$ x 100]%

5. Let the population ⇒ P, then

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

(i). Population after n years = $${P(1 + \frac{R}{100})}^n$$

(ii). Population n years ago = $${\frac{p}{(1+ \frac{R}{100})^n}}$$

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

(i). Population after n years = $${P(1 – \frac{R}{100})}^n$$

(ii). Population n years ago = $${\frac{p}{(1 – \frac{R}{100})^n}}$$

### Samples

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% = $$\frac{0.8}{100}$$ = $$\frac{8}{1000}$$ = $$\frac{1}{125}$$

b). Given 28%

28% = $$\frac{28}{100}$$ = $$\frac{7}{25}$$

c). Given 76%

76% = $$\frac{76}{100}$$ = $$\frac{19}{25}$$

d). Given 0.06%

0.06% = $$\frac{0.06}{100}$$ = $$\frac{6}{10000}$$ = $$\frac{3}{5000}$$

Therefore,

0.8% = $$\frac{1}{125}$$

28% = $$\frac{7}{25}$$

76% = $$\frac{19}{25}$$

0.06% = $$\frac{3}{5000}$$

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

Solution:

Given that

14% of 350 + 48% of 250

=$$\frac{14}{100}$$ x 350 + $$\frac{48}{100}$$ 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 $$x$$

⇒$$\frac{10}{100}$$ x $$x$$ = 250

⇒$$x$$ = 250 x 10

⇒$$x$$ = 2500

Therefore, missing value is $$x$$ = 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 $$x$$ and $$y$$

Given that

6.8% of $$x$$ = 12.4% of $$y$$

⇒68$$x$$ = 124 $$y$$

⇒$$x$$ = $$\frac{124}{68}y$$

⇒$$x$$ = $$\frac{31}{17}y$$

also given $$x$$ – $$y$$ = 122 ——-(i)

Now substitute value of $$x$$ in equation (i)i.e.

⇒$$\frac{31}{17}y$$ – $$y$$ =122

⇒$$\frac{(31-17)y}{17}$$ = 122

⇒$$14y$$ = 122 x 17

⇒$$y$$ = $$\frac{122 * 17}{14}$$

⇒$$y$$ = 148.14

Now substitute the value of y in equation (i)

$$x$$ – $$y$$ = 122

⇒$$x$$ = 122 + 148.4

⇒$$x$$ = 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 = $${P(1 + \frac{R}{100})}^n$$

i.e. Population after 2 years = $${200000(1 + \frac{10}{100})}^2$$

⇒P = 200000 x $$\frac{121}{100}$$

⇒P = 242000

Now Consider, Population n years ago = $${\frac{p}{(1+ \frac{R}{100})^n}}$$

i.e. Population 2 years ago = $${\frac{200000}{(1+ \frac{10}{100})^2}}$$

⇒P = 200000 x $$\frac{100}{121}$$

⇒P = 165289.25

Therefore, Population after 2 years =242000 and

Population 2 years ago = 165289.25

6. Express $$5\frac{3}{2}$$ as rate percent

Solution:

Given $$5\frac{3}{2}$$

⇒$$\frac{13}{2}$$

⇒($$\frac{13}{2}$$ x 100)%

⇒(13 X 50)%

⇒650 %

Therefore, $$5\frac{3}{2}$$ = 650 %