SS Bharti Maths Notes PDF for SSC CGL

SS Bharti Maths Notes PDF for SSC CGL

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SS Bharti Maths Notes PDF for SSC CGL:-Here we are sharing today SS Bharti Maths Notes PDF for SSC CGL. This pdf will increase your knowledge and help you to crack your target. In this  S S Bharti Sir Maths Class Notes PDF Download  we are providing you SS Bharti Sir Maths Book Notes PDF In Hindi. We are sharing a compilation of various individual section in the Mega SS Bharti Sir Maths Class Notes PDF Download, which is very important for all exams like-SSC CGL, CHSL, MTS, IBPS, RRB, UPSC, IAS,CTET,REET,HTET, Insurance, Bank exams and other govrnment exams.

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SS Bharti Maths Notes PDF for SSC CGL

Basic math formulae

A formula is a mathematical expression or definite rule that is derived from the relation between two or more quantities and the derived final product is expressed in symbols. The formulas of mathematics included numbers known as constants, letters that represent the unknown values and are known as variables, mathematical symbols known as signs, and exponential powers in some cases.

Arithmetic is the oldest method of calculation known till now. The word arithmetic is derived from the Greek words ‘arithmos’ which literally means numbers. Brahmagupta the Indian mathematician is known as the ‘father of arithmetic‘. And, the Fundamental theory of number theory was proposed by Carl Friedrich Gauss in 1801.

The basic operations involved in arithmetic are addition, subtraction, multiplication, and division.

Arithmetic formula

Arithmetic mean (average) = Sum of values/Number of values.

Algebra is an elementary subject of mathematics that deals with the study of the evaluation of numbers and symbols. The algebraic operations are carried out to determine the unknown values which are expressed by letters. Algebraic equations are the expressions formed by the combination of variables, constants, factors, and coefficients of variables.

Basic algebra formula

a2 – b2 = (a – b)(a + b)
(a + b)2 = a2 + 2ab + b2
a2+ b2 = (a + b)2 – 2ab
(a – b)2 = a2 – 2ab + b2
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(a – b – c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca
(a + b)3 = a3 + 3a2b + 3ab2 + b2
(a – b)3 = a3 – 3a2b + 3ab2 – b3
a3 – b3 = (a – b)(a2 + ab + b2)
a3 + b3 = (a + b)(a2 – ab + b2)
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
(a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4
a4– b4 = (a – b)(a + b)(a2 + b2)
(am)(an) = am + n
(ab)m = ambm
(am)n = amn

Geometry is a part of mathematics that is concerned with the study of shapes, sizes, parameters, measurement, properties, and dimensions. There are generally three types of geometry. They are Euclidean geometry, Spherical geometry, and Hyperbolic geometry.

Basic geometry formula

Perimeter of Rectangle = 2(l + b)
The area of Rectangle = l × b
Where ‘l’ is Length and ‘b’ is Breadth

The area of Square = a2
The perimeter of Square = 4a
Where ‘a’ is the length of the sides of a Square

Area of Triangle= 1/2 × b × h

Where ‘b’ is the base of the triangle and ‘h’ is the height of the triangle

Area of Trapezoid = 1/2 × (b1 + b2) × h

Where b1 and b2 are the bases of the Trapezoid

And, h = height of the Trapezoid

Area of Circle = π × r2
Circumference of Circle = 2πr
Where ‘r’ is the radius of a Circle

Surface Area of Cube = 6a2

Where ‘a’ is the length of the sides of the Cube

The curved surface area of Cylinder = 2πrh
The total surface area of Cylinder = 2πr(r + h)
The volume of Cylinder = V = πr2h
Where ‘r’ is the radius of the base of the Cylinder

And, ‘h’ is the height of Cylinder

The curved surface area of a cone = πrl
Total surface area of cone = πr(r + l) = πr[r + √(h2 + r2)]
Volume of a Cone = V = 1/3× πr2h
Here, ‘r’ is the radius of the base of Cone and h = Height of the Cone

Surface Area of a Sphere = S = 4πr2
Volume of a Sphere = V = 4/3 × πr3
Where, r = Radius of the Sphere

Probability is the mathematical term used to determine the chance of occurring a particular event. Probability can simply be defined as the possibility of the occurrence of an event. It is expressed on a linear scale from 0 to 1. There are three types of theoretical probability, experimental probability, and subjective probability.

Basic probability formula

P(A) = n(A)/n(S)


P(A) is the probability of an event.

n(A) is the number of favorable outcomes

n(S) is the total number of events


A fraction is a number expressed with integers in which a numerator is divided by the denominator. A fraction is basically the quotient of a division.

Basic fractions formula

(a + b/c) = (a × c) + b/c
(a/b + d/b) = (a + d)/b
(a/b + c/d) = (a × d + b × c/b × d)
a/b × c/d = ac/bd
(a/b)/(c/d) = a/b × d/c

A percentage is a numerical value or ratio expressed as a fraction of 100. It is generally symbolized by the sign %.

Basic percentage formula

Percentage = (Amount in the category/Total value) × 100

Sample problems

Question 1: Determine the probability to get an ace from a card taken from a deck.



Total number of favorable outcomes n(S) = 52

Number of face cards in the deck = 12

Number of favorable outcomes n(A) = 12


P(A) = n(A)/n(S)

=> 12/52

=> 3/13

Hence, the probability to get a face card from a card deck is 3/13.


Question 2: Simplify 3/(x – 1) + 1/(x(x – 1) = 2/x


=> 3x + 1/x(x – 1) = 2(x – 1)/x(x – 1)

=> 3x + 1 = 2(x – 1)

=> x = -3


Question 3: If x + 1/x = 3. Find the value of x2 + 1/x2.


=> (x + 1/x)2 = (3)2

=> x2 + 2 × x × 1/x + (1/x)2 = 9

=> x2 + 1/x2 + 2 = 9

=> x2 + 1/x2 = 7


Question 4: If the radius of a circle is 21cm. Find the area of the given circle.



The radius of the circle is 21cm.

We have,

Area of the circle (A) = πr2

=> 22/7 × 21 × 21

=> 1386cm2

Hence, the area of the given circle is 1386cm2


Question 5: Find the area of a triangle having a base of 100cm and a height of 20cm.



The base of the triangle is 100cm.

The height of the triangle is 20cm.

We have,

Area(A) = 1/2 × b × h

=> 1/2 × 10 × 20

=> 1000cm2


Question 6: Punam has 4/5 parts of the field among which she uses 2/5 parts for farming. How much part of the farm is left for other purposes?



The total fraction of land 4/5.

The total fraction used for farming 2/5.


=> 4/5 – 2/5

=> 4 – 2/5

=> 2/5

Hence, 2/5 part of the field left.

Question 7: What will be the 20% of 240kg?


=> 20/100 × 240

=> 48kg

Hence, the 20% of 240kg will be 48kg.


Percentage Questions For Competitive Exam: Fraction values

Percentage is denoted by the symbol %. Percentage means per every 100 i.e. if we say 25% then it can be written as 25/100 and when you will simplify it then it becomes 1/4. In order to solve the percentage questions for competitive exam in less time candidates should learn some important fraction values of percentage. Any fraction can be converted into percentage by multiplying to 100. In the table given below candidates can check the fraction value of percentage.

Percentage                                                                Fraction Form
100%                                                                                    1
50%                                                                                     1/2
33+1/3% or 33.33                                                               1/3
25%                                                                                      1/4
16+2/3% or 16.66%                                                             1/6
14+2/7% or 14.28%                                                              1/7
12.5%                                                                                     1/8
100/9% or 11.11%                                                                 1/9
10%                                                                                       1/1
100/11 or 9.09%                                                                   1/11
81/3 %                                                                                  1/12

Percentage Questions For Competitive Exam: Formulas

Before solving percentage questions for competitive exam candidates first checkout the percentage formulas given below

Change in quantity: There are two types of changes occur when any number is increased or decreased from its original value and they are actual change as well as percentage change.

Actual Change– Rahul bought 1kg mangoes for rupees 150 in the month of April and next month the price gets increased and now it becomes 200. So here the price of the mangoes gets increased from 150 to 200 so the actual change would be

Actual change: 200-150= 50

Percentage change: Now lets suppose if we have to calculate the price change in percentage then it would be calculated by the given below formula

Actual change/original value*100

50/200*100= 25%

To calculate x% of y – (x/100) * y = (x*y)/100

x% of y = y% of x

Increase A by x%= A(1 + x/100)

Decrease A by x% = A(1 – x/100)

If Y is x% more /less than Z, then Z is 100x/(100 + x) % less/more than Y

Successive Percentage Change – If there are successive percentage increases of x % and y%, the effective percentage increase is: {(x + y + (xy/100)}%

Percentage Questions For Competitive Exam

The major concern which remains in the candidate’s mind is whether the percentage questions are different for every competitive exam. Well the topic is common but candidates need to be exam oriented as the level of question varies exam wise. Some of the percentage questions for competitive exam is given below and for more questions candidates can give free quizzes which are available on adda247 app.

Q1. Nandita scored 80% marks in five subjects together Hindi, Science, Mathematics, English and Sanskrit, where the maximum marks of each subject were 105. How many marks did Nandita scored in Science, if she scored 89 marks in hindi 92 marks in Sanskrit, 98 marks in Mathematics and 81 marks in English?

(a) 60

(b) 75

(c) 65

(d) 70

(e) 80

SS Bharti Maths Notes PDF for SSC CGL

Q2. A number is divided into two parts in such a way that 80% of 1st part is 3 more than 60% of 2nd part and 80% of 2nd part is 6 more than 90% of the 1st part. Then the number is:

(a) 125

(b) 130

(c) 135

(d) 145

(e) 155

SS Bharti Maths Notes PDF for SSC CGL

Q3. The expenses on rice, fish and oil of a family are in the ratio 12 : 17 : 3 respectively. The price of these articles is increased by 20%, 30% and 50% respectively. The total expenses increased by:

(a) 113/8 %

(b) 57/8 %

(c) 449/8 %

(d) 225/8 %

(e) None of these

SS Bharti Maths Notes PDF for SSC CGL

Q4. The cost of a piece of diamond varies with the square of its weight. A diamond of Rs. 5184 value is cut into 3 pieces whose weights are in the ratio 1 : 2 : 3. Find the loss involved in the cutting.

(a) Rs. 3068

(b) Rs. 3088

(c) Rs. 3175

(d) Rs. 3168

(e) None of these


Q5.  The total strength of capital Education is 5000. The number of males and females increases by 8% and 16% respectively and strength becomes 5600. What was the number of males at Capital Education?

(a) 2500

(b) 2000

(c)  3000

(d) 4000

(e) 4500

SS Bharti Maths Notes PDF for SSC CGL

Q6. The rates of simple interest in two banks A and B are in the ratio 5 : 4. A person wants to deposit his total savings in these two banks in such a way that he receives equal half yearly interest from both. He should deposit the savings in banks A and B in the ratio:

(a) 2 : 5

(b) 4 : 5

(c) 5 : 2

(d) 5 : 4

(e) 3 : 5

Q7. One year ago the ratio between Laxman’s and Gopal’s salary was 3 : 4. The ratios of their individual salaries between last year’s and this year’s salaries are 4 : 5 and 2 : 3 respectively. At present the total of their salary is Rs. 4160. The salary of Laxman, now is:

(a) Rs. 1040

(b) Rs. 1600

(c) Rs. 2560

(d) Rs. 3120

(e) Rs. 4210

SS Bharti Maths Notes PDF for SSC CGL

Percentage Questions For Competitive Exam

Q8. In a village, each of the 60% of families has a cow; each of the 30% of families has a buffalo and each of the 15% of families has both a cow and a buffalo. In all there are 96 families in the village. How many families do not have a cow or a buffalo?

(a) 20

(b) 24

(c) 26

(d) 17

(e) 28

Q9. A businessman imported Laptop, worth Rs. 210000, Mobile phones worth Rs. 100000 and Television sets worth Rs. 150000.  He had to pay 10% duty on Laptops, 8% on phones and 5% on Television sets as special case. How much total duty (in rupees) he had to pay on all items as per above details?

(a) 36500

(b) 37000

(c) 37250

(d) 37500

(e) 42500

SS Bharti Maths Notes PDF for SSC CGL

Q10. 60% students of a college study math, 55% students study commerce and 15% students study both. If ratio of boys to girl who study math only is 3 : 2 and ratio of boys to girls who study commerce only is 2 : 3, then total number of boys who study only math and only commerce together is what percent of total number of girls who study only math and only commerce together? (in approximate)

(a) 102%

(b) 109%

(c) 117%

(d) 123%

(e) 112%

Q11. Mr. Raghav owned a flat for Rs. 13.2 lacs. He spends 20% of his monthly salary on its paintings. 25% on lighting and furniture, 15% on foods and other supplements. If Mr. Raghav’s annual salary is 10.8 lacs then find the total spending made by Mr. Raghav.

(a) 137.4 lacs

(b) 1.374 lacs

(c) 13.74 lacs

(d) 11.74 lacs

(e) 17.34 lacs

SS Bharti Maths Notes PDF for SSC CGL

Q12. The entrance ticket of an exhibition was Rs. 5. Later it was decreased by 20% and thus the sale amount is increased by 44%. What was the percentage increase in the number of visitors?

(a) 64%

(b) 24%

(c) 80%

(d) 20%

(e) 40%

Q13. A tree’s height increases annually by 1/8th of its height. By how much will its height increase after 2.5 years, if it stands today at 10 m height?

(a)3 m

(b)3.44 m

(c)3.6 m

(d)3.88 m

(e)4 m

SS Bharti Maths Notes PDF for SSC CGL

Q14. A number X is  less than another number Y. Number Y is three fourth of 24. Find the sum of the two numbers X and Y.

(a) 30

(b) 33

(c) 36

(d) 24

(e) 42

Q15. Udita purchased a hand bag in Rs. 2,520. Later she felt that she had given 5% extra of the actual price of hand bag. Find the actual price of hand bag.

(a) Rs. 2,646

(b) Rs. 2,500

(c) Rs. 2400

(d) Rs. 2,450

(e) None of these

SS Bharti Maths Notes PDF for SSC CGL

Some Important Maths Questions:-

1. The value of log 10000 is-
(A) 4 (B) 8 (C) 5 (D) 1 (Ans : a)
Hint: log 10000 = log10 104=4

2. A boy takes 20 minutes to reach the school at an average speed of 12 km/h. If he has to reach the school in 15 minutes, then the average speed should be…..
(A) 14 km/h (b) 15 km/h (c) 16 km/h (d) 18 km/h (Ans: c)
Hint: Distance of school =12×20/60=4 km
Now his speed=distance/time=4/15/60
=16 km/h

3. If the side of a square is doubled, then the area-
(A) doubles (B) 4 times (C) 8 times (D) 16 times (Ans : b)
Hint: According to the question, A1=a2
So obviously it will be four times.

4. 15% of 220 =?
(A) 33 (B) 22 (C) 24 (D) 26 (Ans : a)
Hint: 220×15/100=33

5. 8888 + 888 + 88 + 8 = ?
(A) 9784 (B) 9792 (C) 9072 (D) None of these (Ans : d)

6. Out of 450 apples, 30% are rotten. How many apples are fine?
(A) 135 (B) 140 (C) 125 (D) 315 (Ans : a)
Hint: Number of rotten apples = 30% of 450
= 450×30/100=135

7. A star is about 8.1×1013 km away from the earth. Suppose light travels at a speed of 3.0×105 km/s. Find the time taken by the light from the star to reach the earth.
(A) 2.7×108 (B) 2.7×1011 (C) 7.5×104 (D) 7.5×103 (Ans : a)

Hint: Time taken from star to type to reach Earth 1
= distance/speed = 8.1×1013/3.0×105 = 2.7×108 seconds

8. An article worth Rs. By selling it for 220, Neeta made a profit of 10%. Accordingly for how much should he sell it so that the profit becomes 30%?
(a) Rs. 220 (b) Rs. 230 (c) Rs. 260 (d) Rs. 280 (Ans : c)
Hint: Cost price of article =200 × 100/110 = Rs. 200
Selling price of article to make 30% profit
= 130% of 200 = Rs. 260

9. 100 × 10 – 100 + 2000 » 100 is equal to?
(a) 20 (b) 920 (c) 980 (d) 1000 (Ans : b)
Hint: 100 × 10 – 100 + 2000 » 100
= 1000 – 100 + 20 = 900 + 20 = 920

10. A man gave 1/4th of his property to his daughter, to his sons and 1/5 to charity. Accordingly, how much did he give in total?
(a) 1/20 (b) 19/20 (c) 1/10 (d) 9/10 (Ans : b)
Hint: Total part given by the person

maths gk questions and answers

1. Who is the father of mathematics?
Answer: Archimedes

2. Who discovered Zero (0)?
Answer: Aryabhata, 458 AD
Explanation: Aryabhatta invented zero but he did not give any symbol for zero, Brahmagupta was the first to give the symbol of zero and the rules for counting with zero.

3. What is the average of the first 50 natural numbers?
Answer: 25.5

4. When is Pi Day celebrated across the world?
Answer: 14 March

5. The value of pi?
Answer: 3.14159

6. The value of cos is 360°?
Answer 1

7. What is an angle greater than 180 degrees but less than 360 degrees called?
Answer: Reversible angle

8. Who discovered the laws of lever and pulley?
Answer: Archimedes

9. Scientist who was born on Pi Day?
Answer: Albert Einstein

10. Who discovered the Pythagorean theorem?
Answer: Pythagoras of Samosas

11. Who discovered the symbol infinity “∞”?
Answer: John Wallis

12. Father of Algebra?
Answer: Muhammad ibn Musa al-Khwarizmi (Persian mathematician)

13. Who discovered the Fibonacci sequence?
Answer: Leonardo Pisano Bigolo

14. The first person to use the Greek letter pi (π) to denote a constant?
Answer: William Jones in 1706

15. Who discovered logarithm and decimal point?
Answer: John Napier

16. Who invented the equal sign (=)?
Answer: Robert Ricarde

17. Who invented the slide rule?
Answer: William Otrade

18. Who invented the moon?
Answer: Joseph Huddart

19. Who discovered the center of gravity?
Answer: Archimedes

20. Where was the abacus invented?
Answer: China


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