Percentage

Basic concepts:-

1)      X% = x/100

e.g. 20% = 20/100 = 1/5

25% = 25/100 = 1/4

50% = 50/100 = 1/2

75% = 75/100 =3/4

80% = 80/100 = 4/5

When we want to convert a % value into fraction then

divide the given by 100

2)      x/5 = 100 x /5 = 20% of x

e.g.1/5= 100/ 5= 20%

1/4 = 100/4 = 25%

3/4 = 3*100/4 =75%

3)      Use ‘ + ‘ sign when the word ‘more’ comes  ,  ‘ – ‘

sign when the word ‘less’ comes and ‘ x ‘ when the word

‘ of ’ comes in word problem

When we want to convert a fraction into % value then

multiply the given fraction by 100

Exercise:-

1)      The price of an article is reduced by 20%. If the new

price is 825. Find the original price of the article.

Solution:- Assume that the original price of the article is

x Rs.

New price= x- 25% of X = 75% of x

75% of x = 825……………….given that the new price is 825

(75/100) * x=825

X=825*100/75 = 1100

2)      The price of an article is increased by 25%. If the new

price is 1200. Find the original price of the article.

Solution:- Assume that the original price is x Rs.

New price= x + 25% of x = 125% of x

125% of x = 1200………………….given that the new price is 1200

(125/100)*x = 1200

X=1200*100/125 = 960

Formula 1 :-

Suppose that A be a value / price / amount / quantity

which changes twice. First time increases by x% and

then by y%. Then the net % change in amount will be

( 100% of A + x% of A ) + y% of ( 100% of A + x% of A )

(100 + x)% of A + y% of [(100 + x)% of A ]

[( 100+x )% of A ]( 1+ y%)

[(100 + x)/100 ]*[1 + y/100] * A

1/100 [(100 + x)*(100 + y)] * A/100

1/100[10,000 + 100x + 100y + xy] * A/100

[100 + x + y + xy/100] * A/100

A + (x + y + xy/100)% of A

Net % increment = (x + y + xy/100)%

Remark:-

·       

if both x an y increases then formula will be

(x + y + xy/100)%

·       

if x increases and y decreases then formula will be

(x - y - xy/100)%

·       

If x and y both decreases then formula will be

( - x - y + xy/100)%

3)      The price of an article is first increased by 25% and

then reduced by 20%. If the new price is 1540.

Find the original price of the article.

Solution:- Here x=25% and y=20%

Net % change = (x + y + xy/100)%

Net % change = (25 – 20 – 500/100)%

= (5 – 5)% = 0%

There will not be any change in the amount.

Original price will remain same i.e. 1540

4)      Suppose a person gives 25% of his amount to his

elder son and 40% of the remainder to his younger

son if he is left with 150 Rs.then how  much money

he was having earlier?

Solution:- Assume that he has x Rs

Net % change in x=(25+40+25*40/100)%

Net % change in x=(65+1000/100%

Net % change in x=75%

He is left with=100-75=25% of x

But he is left with 150 Rs

150=25% of x

150*100/25=x

600=x

Earlier he was having 600 Rs.

 

5)      Mr. Robert spends 25% of his income on food

and 20% of the remaining on house rent. If at the

end he is left with 30,000 then what is his salary?

Solution:- Net % percent change will be

=(25+20+500/100)%

Net % change=50%

This means he spends his 50% salary

èhe is left with 50% salary which is 30,000

ð 

His salary is 60,000 Rs.

Formula 2 :-

Case 1:- The population of a town is suppose 5000

and it increase by 10% annually then the population

after 2 years will be

5000 ( 1 + 10/100 )^2 = 5000 ( 11/10 )^2

= 6050

Generalization: - Assume that the population of a

town is P and it increase R% annually then the

population after n years will be

·        

P ( 1 + R/100 )^2

Case 2:- The population of a town is 5000 and it

decreases by 10% annually due to some disease

then population after 2 years will be

5000 ( 1 – 10/100 )^2 = 5000 ( 9/10 )^2 = 4050

Generalization:- Assume that the population of a

town is P and it decreases by 10% annually then

the population after n years will be

·        

P ( 1 – R/100 )^2

Case 3:- The population of a town is suppose 5000

and it increase by 10% annually then the population

before 2 years will be

5000 / ( 1 + 10/100 )^2 = 5000 / (11/10)^2

5000 *100 /121 = 4132

Generalization:- Assume that the population of a

town is P and it increase R% annually then the

population before n years will be

·        

P / ( 1 + R/100 )^2

Case 4:- The population of a town is suppose 5000 and

it decreases by 10% annually then the population before

2 years will be

5000 / ( 1 – 10/100 )^2 = 5000 / ( 9/10)^2

5000*100/81 = 6173

Generalization:- Assume that the population of a

town is P and it decrease R% annually then the

population before n years will be

·        

P / ( 1 - R/100 )^2

Exercise

6)      The population of a town is 84000 and it increases by

12% annually what will it be in 3 years?

Solution:-  here P=84000, R=5 & n=3. We have

P ( 1 + R/100 )^n = 84000 ( 1+ 5/100 )^3

= 84000 ( 21/20 )^3 =97241