Percentage: Basic Concept And  Examples

1. The Basic Concept Of Percentage

MaNiSH manevandra 01 Jan 1970

The percentage can be divided into 'per-cent-age' which means ‘per every hundred’. It is denoted by the symbol %

Following calculations should be kept in mind:

Comparison between two values of X and Y :

If we compare x to y then we assume y is always equal to 100% When any question asked, what percent of x is y, then y will be written in the denominator.

Q. If x is 80% of y, what percent of x is y?

1st method :
\( y = {100×100 \over 80} x \)
\( y = 125\% {of} x \)

2nd method :
Let y is 100 then x = 80
\(∴ Required \ \% = {100 \over 80}×100 = 125\% \)

Example1 – K is what % of N ?
\[ {K \over N}×100  = {K \over N}\% \]

If A is R % more than B then B is less than A by – $$ {\brack {R \over 100+ R}×100}\% $$

If A is R% less than B, then B is more than A by – $$ {\brack {R \over 100-R}×100}\% $$

Example. If Ram’s income is 10% more than that of Shyam’s income then, how much Percentage Shyam’s income is less than that of Ram’s income?
\( {{10 \over 100+10} × 100} \)  \( = {{10 \over 110}×100} \) \(= 9 {1 \over 11}\% \)

2. Percentage: Formula And Examples Related To Population

MaNiSH manevandra 01 Jan 1970

In Case of population Increase
Eg. Let the population of a town be p now and suppose increase at the rate of R % per annum then,

Population after n years \( = P (1+{R \over 100})^n \)

Population n years ago  \( = {P \over (1+{R \over 100})^n}\)

Example Present population of a town 1.21 crore and it is increasing at 10 % per annum. Then find out the population after two years as well as before 2 years?
Sol. Given, Present population = 1.21 crore,
  Two years after
 \( = P{(1+{R \over 100})}^n \)
  \( = 1.21(1+{10 \over 100})^2 \)
  \( = 1.21 \times 1.21 = 1.4641 \ crore \)

Two years before
 \( = {P \over (1+{R \over 100})^n} \)
\( = {1.21 \over (1+{10 \over 100})^2} \)
\( = { 1.21 \over 1.21} =1 \ crore \)


In Case of population decrease
Let the population of a town be p now and suppose decrease at the rate of R % per annum then
Population after n year \(= P(1-{R \over 100})^n\)
Population n years ago \( = {P \over (1-{R \over 100})^n }\)

Example: If the present population of a town is 72600 and it is decreased by 10 % per annum. What will be its population 2 years hence?
Sol.
As we Know Population after n year \( = P ({1+R \over 100})^n \)
Here given,
p = 72600,
R = 10%,
N = 2  years
Population after 2 years
\( = 72600({1-10 \over 100})^2 \)
\( = 72600 \times{ 90 \over 100} \times {90 \over 100} \)
\( = 58806 \ ( Answer ) \)

Note : this formula is also used for the depreciation value of machine. 
Net % change \(= {\pm X \pm Y \pm {XY \over 100}} \)
‘ + ’ means increment
‘ – ’ means decrement

Example. if the length of a rectangle is increased by 30 % and breadth is decreased by 20 % find the net % change in the area of that rectangle.
Sol. We know that,  Net % change  \(  = x \pm y \pm {(x \times y) \over 100} \)
\( Let \ x = 30\% \) \( ….( we \ put \ x = +30) \)
\( and \ y = 20\% \) \( …( we \ put \ y =-20 ) \)
Now net % change
 \( =  +30-20+{((+30) \times (-20)) \over 100}\)
 \(= + 10 – 6 = 4 % \)
\( Increase \% = 4 \% \ ( Answer) \)