This document provides information on calculating percentages of numbers and dealing with percentage problems. It discusses: 1) How to calculate a percentage of a number by replacing the % with a decimal and multiplying. For example, 225% of 24 is 2.25 x 24 = 54. 2) A trick for breaking percentages into 10% increments, like calculating 60% of 42 as (50% of 42) + (10% of 42). 3) Common mistakes made when dealing with percentages that are increased or decreased by a certain amount, like confusing "increased by 200%" versus "increased to 200%". 4) How to calculate the percentage one number is more or less than another, and examples of

1. Quantitative Aptitude
Section :- Arithmatic
Chapter 1 Percentages

2. How to find the Percent of
Number
 Step 1:- Replace the % sign with decimal sign (.) ,
move the decimal Two places to the left .
Example :- 75%  0.75 , 63%  0.63
 Step 2:- Multiply the obtained decimal form with
Number .
 Example :- 225% of 24  2.25x24 = 54
75% of 65  0.75x65 = 48.75

3. Simple Trick to Solve Percentage of
Number
 Example :-
1} 15% of 80  (10% of 80)+(5% of 80)
 8+4=12
2} 60% of 42  (50% of 42)+(10% of 42)
 21+4.2=25.2
3} 95% of 720  (100% of 720)-(5% of 720)
720-36=684

4. General mistakes done during
calculation of Percentages
Increased/decreased by “ X% ”
 P is increased by 200%
P’= 3P { (100+200)% of P }
 P is Decreased by 75%
P’= 0.25P { (100-75)% of P }
 P is increased by 175% of Q
P’= P+1.75Q
Increased/decreased to “ X% ”
 P is increased to 200%
P’= 2P { (200)% of P }
 P is Decreased to 75%
P’= 0.75P { (75)% of P }
 P is increased to 175% of Q
P’= 1.75Q

5. Comparing Two Percentages
 If A is r% more/less than B
 Then B is
100𝑟
100±𝑟
% less/more than A
 Example :-
1} A is 10% more than B
B is
100(10)
100+10
=
1000
110
= 9.09% less than A
2} A is 10% less than B
B is
100(10)
100−10
=
1000
90
= 11.11% more than A

6. Successive Percentage changes
 The concept of successive percentage change deals with two or more
percentage changes applied to quantity consecutively. In this case,
the final change is not the simple addition of the two percentage
changes .
 If value of an object/number P is successively changed by x%, y% and
then z%, then final value P’
 Percentage change for this kind of situation is (P’-P)100%