 # Percentage Problems

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# 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 %