A fraction with its denominator as `100'
is called a percentage. Percentage means per
hundred.
So it is a fraction of the form
6
1 0 0
3 7
1 0 0
1 5 1
1 0 0
, and and these fractions
can be expressed as 6%, 37% and 151%
respectively.
In such a fraction, the numerator is called
rate percent.
To express x% as a fraction or a decimal,
divide x by 100.
If the price of an item increases by r%,
then the reduction in consumption, so that the
expenditure remains the same is
r
r
x


 

 
1 0 0
1 0 0%
If the price of the commodity decreases
by r%, the increase in consumption, so that
the expenditure remains the same is
r
r
x
1 0 0
1 0 0 %


 

 
If the value is first increased by x% and
then by y%, the final increase is
x y
   xy
 

 
100
%
If there is a decrease instead of increase, a
negative sign is attached to the corresponding
rate percent.
If the value of a number is first increased
by x% and later it is decreased by x% then net
change is always a decrease which is equal to
x2
100

 

 
%
If pass marks in an examination is x% and
if a student secures y marks and fails by z
marks, then the maximum mark
=
1 0 0 (y  z)
x
A candidate scores x% in an examination
fails by `a' marks while another candidate
who scores y% gets `b' marks more than
the minimum required for a pass, then the
maximum mark =
100 (a b)
y  x
If the length of a rectangle is increased
by x% and the breadth is decreased by y%,
then the area is increased or decreased by
(x  y  xy )%
1 0 0
according to the (+) ve or
(-) ve sign obtained.
If the present population is P which increases
R% annually, then
(i) the population after n years
= P 100 R n
100
 
 

 
(ii) the population n years ago
=
n
100 R
100 P 





If the present value of a machine is P
which depreciates at R% per annum, then
(i) the value of the machine after n years
= P 100 R n
100
 
 

 
(ii) the value of the machine n years ago
=
P
R
n 100
100 

 

 