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1 | P a g e
STUDY NOTES FOR ALL COMPETATIVE EXAM
Write us : [email protected] www.mahendras.org www.mahendraguru.com myshop.mahendras.org
2 | P a g e
STUDY NOTES FOR ALL COMPETATIVE EXAM
Write us : [email protected] www.mahendras.org www.mahendraguru.com myshop.mahendras.org
CONCEPT OF PERCENTAGE
Percentages are special types of fractions in
which the denominator is always hundred.
e.g.
7
100
It is written as 10% or its simplified form is
1
10
PERCENTAGE INCREASE /
DECREASE
If the price of a commodity increases by a%,
then the reduction in consumption so as not
to increase the expenditure is:
100 %100
a
a
If the price of a commodity decreases by a%,
then the increase in consumption so as not to
decrease the expenditure is:
100 %100
a
a
Results on Population:
Populations after n year = 1
100
nr
p
Populations before n years = 1
100
n
p
r
If the value of a number is first increased by
a% and later decreased by a%. Then the net
effect is always decreased which is equal to
a% of a and is written is
2
%100
a
SOME IMPORTANT FRACTION
VALUES
1 100%
150%
2
1 133 % 33.33%
3 3
125%
4
120%
5
or
1 214 %
7 7
112.5%
8
1 111 %
9 9
110%
10
1 19 %
11 11
1 18 %
12 3
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1 97 %
13 13
1 17 %
14 7
1 26 %
15 3
1 16 %
16 4
1 155 %
17 17
1 55 %
18 9
1 55 %
19 19
15%
20
1 14 %
24 6
14%
25
1 12 %
40 2
3 137 %
8 2
5 162 %
8 2
4 157 %
7 7
SOME QUESTIONS BASED ON
PERCENTAGE
EX: - If ram earns 25% more than Shyam
then what % less does Shyam earn than
ram?
125%
4
It means Shyam income is 4 and Ram
income is 5
So ram is % less than Shyam
=
1100 20%
5
EX: A's income is 10% more than B's
income. How much per cent is B's income
less than A's income?
Sol: Let B's Income = Rs. 100
A's income =
100 110110
100
Required percentage =
10 1100 9 %
110 11
EX: If ram earns 𝟏𝟔𝟐
𝟑% more than
Shyam what % less does Shyam earn than
Ram?
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2 116 %
3 6
Shyam = 6 and Ram = 7
Shyam is % less than Ram
=
1 2100 14 %
7 7
EX: Sixty – five percent of a number is 21
less than three – fourth of that number.
What is the number?
SOL:
Three fourth = ¾ = 75%
75% - 65% = 21
10% = 21
100% = 210
EX: If 𝟔𝟔𝟐
𝟑% of a no. is added with itself
then result becomes 3900. Find the
original number?
SOL:
2 266 %
3 3
It means 3 is the original number
And 2 is added to the number then it
becomes 5 and
5 unit = 3900
1 unit = 780
3 unit = 2340
EX: If 96 is added in a number then
number becomes 157𝟏
𝟕% of itself. Find the
numbers?
SOL:
1 1 114157 % 100 57 1
7 7 7 7
It means 7 is the original number and after
adding 96 it becomes 11 that means
4 unit = 96
1 unit = 16
7 unit = 112
EX: If the sides of a square is increased by
40% find the percentage change in its
area ?
100
xyx y
40 4040 40
100
80 16
96
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Ex: If the radius of a circle is increased by
25% then find the percentage change in
its area ?
125%
4
It mean we can say radius of circle is 4
And new radius is 5
So area of original circle = 16
So new area of original circle = 25
Percentage change in area =
25 16 9100 100 56.25%
16 16
Ex: If the base radius of a right circular
cylinder is increased by 40% and its
height is reduce by 37.5% then find the
percentage change its volume?
2 340% ,37.5%
5 8
It means original radius is 5 and original
height is 8
Volume of cylinder = 25 8 200
New radius is 7 and new height is 5
New volume of cylinder = 49 5 245
Percentage change in area
245 200100 22.5%
200
Ex : If the length of the rectangle is
increased by 37𝟏
𝟐% and its breadth is
decreased by 16𝟐
𝟑%. Find the
approximate % change in its area ?
Sol:
1 3 2 137 % ,16 %
2 8 3 6
It means original length of rectangle is 8 and
original breadth of rectangle is 6
Area of rectangle = 48
New length of rectangle is 11
New breadth of rectangle is 5
Area of rectangle is 55
So percentage change in area
=
55 48100 14.58%
48
EX: A man saves a certain parts of his
income every month, so that he can
purchase a car in one year . By what
percent he must increase in his saving, so
that he can purchase the same car in nine
month?
SOL:
Let the amount of the car is 36
So he saves 3 rupees per month if he has to
purchase to it in 12 month
If he has to purchase it in 9 month
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So he saves 4 rupees per month
His saving increased by
4 3 1100 33 %
3 3
EX: The price of sugar is increased by
16𝟐
𝟑% and the consumption of a family is
decreased by 20% . Find the % change in
expenditure?
SOL:
2 116 %
3 6
120%
5
It means the original price of sugar is 6 and
consumption of family is 5
Total expenditure is 30
And new price of sugar is 7 and new
consumptions is 4
Total expenditure is 28
Percentage change in expenditure is
30 28 200 2100 6 %
30 30 3
Ex: The sale of a cinema ticket is
increased by 57𝟏
𝟕% and the price of ticket
is increased by 16𝟐
𝟑%. Find the change in
his revenue?
Sol:
1 457 %
7 7
2 116 %
3 6
It means no. of sale of cinema ticket is 7 and
original price of ticket is 6
Therefore total revenue is 42
New sale of ticket is 11
New price of ticket is 7
New revenue is 77
Percentage change in its revenue is
77 42 35100 100 83.33%
42 42
Ex: A man can type 20 line in 10 minute
but he leaves 8% margin on each line. In
how many time he will type 23 page with
40 lines on each page on which he leaves
25% more margin of before
Sol:
We know that
20 line in 10 minute
It means 2 lines per minute
If he leaves 8% margin that means
2 × 92 line/minute
But he has to type 23 page with 40 lines
leaves 25% more margin of before
Which means
23 × 40 × 90 = 2 × 92 × x
x = 450 min
Ex : If the income tax is increased by 19%
then net income is reduced by 6%. Find
the rate of income tax?
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Sol:
19 6
100 100
6
19
tax NI
tax
NI
Total income is 6 + 19 = 25
Rate of income tax is
6100 24%
25
Ex : Deepak spends 40% of his salary on
house rent, on remaining 10% spends on
travels , on remaining 16𝟐
𝟑 % spends on
food and remaining is saved. If he saved
Rs. 6750 what amount he spent on food ?
Sol:
Let the income be 100
House rent = 40
Remaining = 60
Travel = 60 × 10% = 6
Remaining = 54
Food = 54 × 16.66% = 9
Remaining = 45
45 = 6750
1 = 150
Amount spent on food
150 × 9 = 1350
Ex: A man spends 75% of his income. His
income is increased by 20% and he
increases his expenditure by 10%. His
saving are increased by :
Sol:
75% = ¾
Let the income be = 4
Expenditure = 3
Saving = 1
New income = 4 × 120% = 4.8
New expenditure = 3 ×110% = 3.3
New saving = 1.5
Percentage change in saving
1.5 1100 50%
1
A reduction of 25% in the price of sugar
enables a person to purchase 4 kg more
for Rs. 800. Find the original and current
price per kg ?
Sol :
25% = ¼
Original price of sugar = 4
New price of sugar = 3
We know that quantity is inversely
proportional to price
Original quantity of sugar = 3
New quantity of sugar = 4
1 unit = 4
3 unit = 12 (original quantity)
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4 unit = 16 (new quantity)
Original price per kg =
Amount / original quantity
= 800/12 = 66.67
Current price per kg
Amount / current quantity
800/16 = 50
Ex: In an examination 35% is passing
marks one candidate obtain 135 marks
and failed by 40 marks then find the
maximum marks ?
Sol: 35% = 135 + 40
100% =
175100 500
35
Ex: Rohit obtained 480 marks out of 600
and Mohit obtained 560 marks out of 800.
Whose performance is better and how
much %?
Sol:
Rohit got =
480100 80%
600
Mohit got =
560100 70%
800
Rohit performance is better.
% better =
10 2100 14 %
70 7
Ex: In an election between two
candidates, 65 votes were declared invalid
one candidate get 52% votes and win by
98 votes .Find the total number of votes?
Winner 52% Looser 48%
4% = 98
100% = 2450
Valid vote = 2450
Total vote = 2450 +65 = 2515
Ex: The length and breadth of rectangle
is increased by 30% and 20%
respectively, so find the net % change in
its area?
Sol:
Net % change = x + y + X × Y
100
= 30 + 20 + 30×20
100
= 56%
Ex: A batsman scored 120 runs which
included 3 boundaries and 8 sixes. What
percent of his total score did he make by
running between the wickets?
Number of runs made by running
= 110 - (3 × 4 + 8 × 6)
= 120 - (60)
= 60
Now, we need to calculate 60 is what percent
of 120.
=> 60/120 × 100 = 50 %
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Ex: Rahul's Mathematics test had 75
problems, 10 arithmetic, 30 algebra, 35
geometry problems. Although he
answered 70% of arithmetic, 40% of
arithmetic and 60% of geometry
problems correctly, still he got less than
60% problems right. How many more
questions he would have to answer more
to get passed
Sol:
Number of questions attempted correctly =
(70% of 10 + 40% of 30 + 60% of 35)
= 7 + 12 + 21 = 40.
Questions to be answered correctly for 60%
= 60% of total questions
= 60 % of 75 = 45.
He would have to answer 45 - 40 = 5
Ex: 10% of inhabitants of a village having
died of cholera, a panic set in, during
which 25% of the remaining inhabitants
let the village. The population is then
reduced to 4050. Find the original
inhabitants
Sol:
Let the total number is x,
then,
(100 - 25)% of (100 - 10)% x = 4050
75% of 90% of x = 4050
75/100 × 90/100 × x = 4050
x = (4050 × 50)/27 = 6000
Two numbers are less than third number
by 30% and 37% respectively. How much
percent is the second number less than by
the first?
Let the third number is x.
Then first number = (100-30)% of x
= 70% of x = 7x/10
Second number is (63x/100)
Difference = 7x/10 - 63x/100 = 7x/10
So required percentage is, difference is what
percent of first number
=> (7x/100 × 10/7x × 100 )% = 10%
Ex: In an election between two
candidates, one got 55% of the total valid
votes, 20% of the votes were invalid. If the
total number of votes was 7500, the
number of valid votes that the other
candidate got, was:
Total number of votes = 7500
Given that 20% of Percentage votes were
invalid
=> Valid votes = 80%
Total valid votes = 7500 × (80/100)
1st candidate got 55% of the total valid
votes.
Hence the 2nd candidate should have got
45% of the total valid votes
=> Valid votes that 2nd candidate got = total
valid votes x (45/100)
7500 × (80/100) × (45/100) = 2700
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Ex: 405 sweets were distributed equally
among children in such a way that the
number of sweets received by each child
is 20% of the total number of children.
How many sweets did each child receive?
Let the total number of children be x.
Then, x × (20% of x) = 405
=> x × 20x/100 = 405
1/5 x2 = 405 => x = 45
Number of sweets received by each child =
20% of 45 = 9.
Ex: Entry fee in an exhibition was Rs. 1.
Later, this was reduced by 25% which
increased the sale by 20%. The
percentage increase in the number of
visitors is :
Let the total original sale be Rs. 100. Then,
original number of visitors = 100.
New number of visitors = 120/0.75 = 160.
Increase % = 60 %.
Ex: In a City, 35% of the population is
composed of migrants, 20% of whom are
from rural areas. Of the local population,
48% is female while this figure for rural
and urban migrants is 30% and 40%
respectively. If the total population of the
city is 728400, what is its female
population?
Total Population = 728400
Migrants = 35 % of 728400 = 254940
Local population
= (728400 - 254940) = 473460.
Rural migrants = 20% of 254940 = 50988
Urban migrants
= (254940 - 50988) = 203952
Female population
= 48% of 473460 + 30% of 50988 + 40% of
203952
= 324138
Ex: In an examination, 34% of the
students failed in Mathematics and 42%
failed in English. If 20% of the students
failed in both the subjects, then the
percentage of students who passed in both
the subjects was :
n (A) = 34, n (B) = 42, n (A ∩ B) = 20.
So, n (A U B) = n (A) + n(B) - n(A ∩B) =
34 + 42 - 20 = 56.
Percentage failed in either or both the
subjects = 56.
Hence, percentage passed
= (100 - 66)% = 44%.
Ex: The ratio of earnings of A and B is
4:5. If the earnings of A increase by 20%
and the earnings of B decrease by 20%,
the new ratio of their earnings becomes
6:5. What are A's earnings?
Sol:
Cannot be determined
Ex: 30% of the men are more than 25
years old and 80% of the men are less
than or equal to 50 years old. 20% of all
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men play football. If 20% of the men
above the age of 50 play football, what
percentage of the football players are less
than or equal to 50 years?
Let total number of men = 100
Then,
20 men play football.
80 men are less than or equal to 50 years old.
Remaining 20 men are above 50 years old.
Number of football players above 50 years
old
2020 4
100
Number of football players less than or equal
to 50 years old
= 20 − 4 = 16
Ex: A number is mistakenly divided by
10 instead of being multiplied by 10.What
is the percentage error in the result?
Sol:
Let the number is 10.
Actual result = 10 × 10 = 100
Wrong result = 10
10 = 1
% change = 100−1
100 × 100 = 99%
Ex: The numerator of a fraction is
increased by 400% and denominator is
increased by 500%, so the resultant
fraction is 15/22 . Find the original
fraction?
Sol:
500% 15
600% 22
ofx
ofy
5 15
6 22
x
y
Original Fraction =
15 6
22 5
x
y
Original Fraction = 9/11
Ex: A, B and C are three persons. A
earned 40% more than B and B earned
20% less than C, so what % more earned
A than C.
Sol:
A B C
112 80 100
A more earned than C = 112 – 100 = 12
% more earned =
12 10012%
100
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Ex: Due to reduction of 20% in the price
of apples enables a person to purchase 16
apples more for Rs.320, so find the
reduced rate of 10 apples?
Sol:
Reduced rate = 320×20
100 = 64
Reduced rate of 1 apple = 64
16
= Rs. 4
Reduced rate of 10 apples = 4 × 10
= Rs.40
Ex: A mixture of 40 litres of milk and
water contains 10% water. How much
water should be added, so water may be
20% in the new mixture?
Sol:
40×10
100= 4 lit.
80% = 36
1% = 36
80
20% = 36
80× 20
= 9 Lit.
req. water = 9 – 4
= 5 lit.
Ex: Fresh fruit contains 68% water and
dry fruit contains 20% water. How much
dry fruit can be obtained from 100 kg. of
fresh fruit ?
Sol:
fruit content in 100 kg. of fresh fruits
= 100×32
100 = 32 kg
Since , dry fruits contain 80% fruit
content,
80% = 32
1% = 32
80
100% = 32
80× 100
= 40 kg.
Ex: In an examination, 35% of the
candidates failed in Mathematics and
38% failed in English. 28% passed in both
subjects, then how much percent failed in
both the subjects?
Sol: Failed in Maths = 35%
Passed in Maths = 100% − 35% = 65%
Failed in English = 38%
Passed in Maths = 100% − 38% = 62%
Failed in both subjects
= 100% – (37%+28%+34%)
= 1%
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Ex: Deepak salary is first increased by
25% and then decreased by 20%. The
resultant salary of Deepak is same as the
resultant salary of Naimish which was
first decreased by 25% and increased by
20%. Find the ratio of the salary of
Naimish and Deepak?
Sol:
125 80 75 120
100 100 100 100
20 18
D N
D N
N : D = 10 : 9
EX: A get 30% of the maximum mark and
get failed by 25 mark whereas B get 40%
of total marks in the same exam and get
25% of the passing marks more than the
passing marks. Find the passing marks of
the exam?
B’s marks =
10040 32%
125
32% - 30 % = 25
2% = 25
32% = 25 × 16
= 400
Ex: In a class 25% students were absent
for an exam. 30% failed in exam and
there average marks are 20 less then the
passing marks, 10% of the student passed
with 5 marks of grace and the remaining
student pass with the average marks of 60.
If 33 is the passing marks. Find the
average score of the class?
Sol:
25% are absent it means they got 0 marks
Average marks of the class =
25 0 30 13 10 33 35 602820
100
= 28.20
The sum of salary of A,B and C is
72000.They spend 80% ,85% and 75% of
their salary. If the ratio of their savings is
8 : 9 : 20. What is the salary of A ?
Sol:
20% = 8 100% = 40
15% = 9 100% = 60
25% = 20 100% = 80
Ratio of their income is
40 : 60 : 80
2 : 3 : 4
Salary of A =
272000 16000
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In an examination in which maximum
marks were 500, A got 10% less than B .
B got 25% more than C. C got 20% less
than D. If A got 360 marks, What
percentage of maximum marks was
obtained by D ?
Sol:
According to questions
D = 100
C = 80
B = 100
A = 90
So we can say that
90 = 360
100 = 400
Maximum marks obtained by D = 400
Percentage of marks obtained by D
400100 80%
500
A train has a capacity of 500 seats of
which 10% are in AC and rest in sleeper
class. On the particular day when it is
booked up to the capacity of 85% if AC
class was booked 96% of its capacity, then
how many sleeper class seats were vacant
on that particular day.
Sol:
Capacity of train = 500
Seats in AC = 50
Seats in Sleeper = 450
On particular day
Total passengers in train
500 ×85% = 425
AC seats = 50 × 96% = 48
It means sleeper seats are
425 – 48 = 377
So sleeper seats were vacant on that
particular day
= 450 – 377 = 73
A solution of salt and water contains 20%
salt by weight. If 20 litre of water is
evaporated from the solution, the solution
contains 25% salt. What is the original
quantity ( in litre ) of the solution ?
Sol:
Salt : water = 20 : 80
Salt : water = 1 : 4
After evaporation
Salt : water = 25 : 75
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Salt : water = 1 : 3
1 unit = 20 litre
Original quantity 5 = 5 × 20 = 100
Fresh grapes contains 80% of water ,
while dry grapes contains 10% of water.
If the weight of dry grapes is 500 kg. What
is the total weight when it is Fresh ?
When grapes Fresh
Water = 80%
Pulp = 20%
When fruit is dry
Water = 10%
Pulp = 90%
500 90% 20%
2250
x
x
x = 30
Fresh fruit contains 68% water and dry
fruit contains 20% water. How many kg
of dry fruits can be made from 75 kg of
fresh fruits ?
When fruit is fresh
Water = 68%
Pulp = 32
When fruit is dry
Water = 20
Pulp = 80
75 32% 80%x
x = 30
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