helpful for aptitude preparation for various exams like ssc , bank and cat

1. PERCENTAGES
• PERCENT -----MEANS OUT OF 100
• WHEN WE COMPARE SOME• WHEN WE COMPARE SOME
VALUE WITH 100
• PERCENT ==== PER + CENT
• CENT MEANS 100

5. • z% is z percent which means z
z% = z⁄100
• p⁄ as percent: (p⁄ x 100)%• p⁄q as percent: (p⁄q x 100)%

7. COMMODITY, CONSUMPTION AND EXPENDITURE
• Expenditure( in Rs) = = = =
( commodity) x (consumption)
• RS = (RS /KG) x( KG)
• Total rupees = rate of sugar x
quantity of sugar

8. When expenditure remains constant
• If the price of a commodity increases by R%, then the
reduction in consumption so as not to increase the
expenditure is:
• [R⁄(100 + R)x 100]%• [ ⁄(100 + R)x 100]%
• If the price of a commodity decreases by R%, then
the increase in consumption so as not to decrease
the expenditure is:
• [R⁄(100 - R)x 100]%

9. • Q 3- If the cost of tea is expanded by 20%, by what
amount of percent must the utilization of tea be
lessened so as not to build the consumption?
• A - 50/3%
• B - 100/3%
• C - 20/3%
• D - 200/3%• D - 200/3%
• Answer - A
• Explanation
• Reduction % in consumption = {R/((100+R) )*100}%
• (20/120*100)% = 50/3%

10. Percentages - Solved Examples
• Q 1 - What is fraction equivalent of 32%.
• A - 6/30
• B - 8/25
• C - 7/50
• D - 11:10
• Answer - B
• Explanation
• 32% = 32/100 = 8/25.

11. • Q 2 - 65% of a number is 21 less than 4/5 th of
that number. Find the number.
• A - 140
• B - 130
• C - 120
• D - 110
• Answer - A• Answer - A
• Explanation
• Let the number be x. Then, (4/5 x)-(65% of x)= 21
=> 4x/5-65x/100=21 =>
• 4x/5-13x/20=21 =>
• 16x-13x=420 =>3x=420 ?x=140. Required number
= 140.

12. • STANDARD PERCENTAGE VALUES
TO MEMORIZE
FRACTION VALUE
X 100
===
PERCENTAGE
VALUEVALUE
1/3 =
============
33(1/3)%
1/2 ---------------
--------------
50 %

15. basics of percent change
• (percent of increase or decrease).
• You are given the initial and final quantities
• you have to calculate the percentage change
(percent change).(percent change).
• DIFFERENCE/ORIGINAL (= decimal) x 100
• = percentage value

16. Concept of base value
• % increase ===== Difference /base lower x 100
• % decrease ===== Difference / base higher x 100
• BASE ==== TOTAL QUANTITY OR WHOLE QUANTITY• BASE ==== TOTAL QUANTITY OR WHOLE QUANTITY
OR DENOMINATOR VALUE
• EG : ----- 1/4 ===== 25% BASE = 4
• 1/10======10% BASE =10
• HENCE HIGHER BASE GIVES LOWER % VALUE IF
NUMERATOR IS SAME

17. Q]Salary of A is 20% less than salary of B, how much is
salary of B more than salary of A
• Method 1 ------------let sal of B =100 sal of A = 20%
less = 100 -20 = 80 ====
B sal more than A -------------------THAN- A –
BASE is ABASE is A
DIFF = 100 – 80 = 20
Hence we need to compare with A
% CHANGE = 20 / 80 = 25%

18. TECHNIQUE FOR % DIFF
• WE ASSUME THE BASE AS EASY
VALUE LIKE 100 OR SOME OTHER
VALUE SO THAT CALCULATIONVALUE SO THAT CALCULATION
BECOMES EASY AND
COMPARISON IS FAST
• ANSWER IS GOT FAST

20. Successive % increase and decrease

21. Q] If the length and breadth of rectangle
increase and decrease by 10% respectively
find the % change in area
• Use the formula of the previous slide with + 10%
and – 10% and solve
• Ans ==== 10 - 10 + (10)(-10)/100
= = -1 % reduction of 1% in the area of rectangle

22. Successive % change in 1 quantity
• Let P be the population and it changes by 10
% increase , 10 % decrease and 20 % increase
in 3 years continously then find the population
after 3 yearsafter 3 years
• ------ FINAL POPULATION =
• P( 1 + 10/100)(1-10/100)(1+20/100)
• HENCE INITIAL POPULATION IS MULTIPLIED BY
MULTIPLYING FACTOR

23. MULTIPLYING FACTOR---USED TO GET
THE FINAL VALUE DIRECTLY
% INCREASE MF
10% 1.1 = 1 +(10/100)
20% 1.2 = 1+ (20/100)
30% 1.330% 1.3
% DECREASE MF
20% 0.8 = 1 – (20/100)
60% 0.4 = 1 - (60/100)
INITIAL VALUE = 1

24. • FINAL VALUE =
(INITIAL VALUE) X( MF)
EG ----- INITIAL VALUE = 30EG ----- INITIAL VALUE = 30
20 % INCREASE == 30 + ( 20% OF 30)
30 ( 1+ 0.2) ----------------30 ( Multiplying factor)
FINAL VALUE ===== (30)x(MF)
MF = 1.2

25. • Population
• The population of a city is P and let it increases at
the rate of R% per annum:
• Population after t years: P(1 + R⁄100)t
• Population t years ago: P⁄(1 +
R
⁄100)
t
• Depreciation ------means reduction• Depreciation ------means reduction
• Let V be the present value of machine. Suppose it
depreciates at the rate of R% per annum:
• Machine's value after t years:P(1 - R⁄100)t
• Machine's value t years ago: P⁄(1 -
R
⁄100)
t