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NCERT Solutions for Class 10 Maths Chapter 14 Probability

NCERT Solutions for Maths Chapter 14 Probability Class 10 - Free PDF Download

Class 10 Maths NCERT Solutions for Chapter 14 helps students understand the concept of set theory, like unions and intersections, which can be applied to represent events and analyse their probabilities. Probability provides a way of quantifying uncertainty and helps in making informed decisions based on the likelihood of various events. Understanding and applying probability concepts is essential for solving real-world problems effectively.

The free PDF of Class 10 Maths Chapter 14 Solutions PDF download is available on Vedantu, giving students a better understanding of the problems. It covers solutions to every exercise in this chapter and is updated according to the latest CBSE syllabus . 

Glance on Maths Chapter 14 Class 10 - Probability

Chapter 14 of Class 10 Maths deals with the foundation by defining probability as the measure of how likely an event is to occur. It introduces the concept of outcomes, events, and sample spaces.

You'll learn how to identify favourable outcomes (those that satisfy a specific condition) and calculate the probability of an event by dividing the number of favourable outcomes by the total number of possible outcomes.

Probability in class 10 explains Fundamental theorems like the addition and multiplication rules of probability are introduced. These rules help you calculate the probability of combined events.

This article contains chapter notes, important questions, exemplar solutions, and exercise links for Chapter 14 - Probability, which you can download as PDFs.

There is one exercise (25 fully solved questions) in class 10th maths chapter 14 Probability.

Access Exercise wise NCERT Solutions for Chapter 14 Maths Class 10

Exercises under NCERT Solutions for Class 10 Maths Chapter 14 Probability

Exercise 14.1: This exercise involves solving word problems by paying close attention to key terms like event, sample space, and favourable outcome definitions. Break down complex problems into simpler events and apply the relevant rules.

Access NCERT Solutions for Class 10 Maths Chapter 14 – Probability

Exercise 14.1.

1. Complete the following statements:

i. Probability of an event E + Probability of the event ‘not E’ = _____. 

If the probability of an event be $p$, then the probability of the ‘not event’ will be, $1-p$ . Thus, the sum will be, $p+1-p=1$.

ii. The probability of an event that cannot happen is _____. Such an event is called _____.

The probability of an event that cannot happen is always $0$. iii. The probability of an event that is certain to happen is _____. Such an event is called _____.

The probability of an event that is certain to happen is $1$ . Such an event is called, sure event.

iv. The sum of the probabilities of all the elementary events of an experiment is _____.

The sum of the probabilities of all the elementary events of an experiment is $1$.

v. The probability of an event is greater than or equal to and less than or 

equal to _____.

The probability of an event is greater than or equal to $0$  and less than or equal to $1$ .

2. Which of the following experiments have equally likely outcomes? Explain.

i. A driver attempts to start a car. The car starts or does not start.

Equally likely outcomes defined as the outcome when each outcome is likely to occur as the others. So, the outcomes are not equally likely outcome.

ii. A player attempts to shoot a basketball. She/he shoots or misses the shot. 

The outcomes are not equally likely outcome.

(iii) A trial is made to answer a true-false question. The answer is right or wrong. 

The outcomes are equally likely outcome.

(iv) A baby is born. It is a boy or a girl. 

3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

We already know the fact that a coin has only two sides, head and tail. So, when we toss a coin, it will either give us the result head or tail. There is no chance of the coin landing on his edge. And on the other hand, the chances of getting head and tail are also just the same. So, it can be concluded that the tossing of a coin is a fair way to decide the utcome, as it can not be biased and both teams will have the same chance of winning.

4. Which of the following cannot be the probability of an event?

(A) $\frac{2}{3}$

The probability of an event have to always be in the range of $[0,1]$ .

Now, let us the check the given values.

We can see, $\frac{2}{3}=0.67$ . This is in the given range. It can be a probability of an event.

We can see, $-1.5$ , which is a negative number and not inside the given range. It can not be a probability of an event.

We can see, $15%=\frac{15}{100}=0.15$ . This is in the given range of $[0,1]$ . It can be a probability of an event.

(D) $0.7$ 

We can see, $0.7$ , which is in the given range. It can be a probability of an event.

5. If $P(E)=0.05$ , what is the probability of an event ‘not $E$’? 

The sum of the probabilities of all events in always $1$ .

Thus, if $P(E)=0.05$ , the probability of the event ‘not E’ is, $1-0.05=0.95$.

6. A bag contains lemon flavored candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out

i.an orange flavored candy?

There is no orange candy available in the bag, so, the probability of taking out an orange flavored candy is $0$.

(ii) a lemon flavored candy? 

All the candies in the bag are lemon flavored candies only. Thus, any candy Malini takes out will be a lemon flavored candy. 

So, the probability of taking out a lemon flavored candy is $1$.

7. It is given that in a group of 3 students, the probability of $2$  students not having the same birthday is $0.992$ . What is the probability that the $2$  students have the same birthday?

It is provided to us that, probability of 2 students not having the same birthday is, $0.992$ .

So,$P(2\text{ students }\text{having }\text{the}\text{ same}\text{ birthday})+P(\text{2}\text{ students }\text{not }\text{having }\text{the}\text{ same}\text{ birthday})=1$

$\Rightarrow P(\text{2}\text{ students}\text{ having}\text{ the }\text{same}\text{ birthday})+0.992=1$

Simplifying further,

$P(2 \text{students having the same birthday})=0.008$.

8. A bag contains $3$  red balls and $5$  black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red ?

The bag is having $3$ red balls and $5$ black balls. 

Now, the probability of getting a red ball will be, $\frac{number\,of\,red\,balls}{total\,number\,of\,balls}$ .

Putting the values, we get, $\frac{3}{3+5}=\frac{3}{8}$ .

(ii) not red? 

And, the probability of getting a red ball will be, $1-\frac{number\,of\,red\,balls}{total\,number\,of\,balls}$.

Again, putting the values, $1-\frac{3}{8}=\frac{5}{8}$.

9. A box contains $5$ red marbles, $8$ white marbles and green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red ?

The box is containing, $5$ red marbles, $8$ white marbles and green marbles.

The probability of getting a red marble will be, $\frac{number\,of\,red\,marbles}{total\,number\,of\,marbles}$

Putting the values, $\frac{5}{5+8+4}=\frac{5}{17}$.

(ii) white ? 

Again, the probability of getting a white marble will be, $\frac{number\,of\,white\,marbles}{total\,number\,of\,marbles}$

Putting the values, $\frac{8}{5+8+4}=\frac{8}{17}$.

(iii) not green? 

And, the probability of getting a green marble will be, $\frac{number\,of\,green\,marbles}{total\,number\,of\,marbles}$.

Putting the values, $\frac{4}{5+8+4}=\frac{4}{17}$ .

So, the probability of the marble not being green will be, $1-\frac{4}{17}=\frac{13}{17}$.

10. A piggy bank contains hundred $50$ p coins, fifty ` $1$  coins, twenty ` $2$  coins and ten ` $5$  coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a $50$  p coin ?

We are provided with the fact that, the piggy bank contains, hundred $50$ p coins, fifty $1$ rs coins, twenty $2$ rs coins and ten $5$ rs coins.

So, the total number of coins, $100+50+20+10=180$ .

Thus, the probability of drawing a $50$ p coin, $\frac{100}{180}=\frac{5}{9}$.

ii. will not be a ` $5$  coin?

Similarly, the probability of drawing a $5$ rs coin, $\frac{10}{180}=\frac{1}{18}$ .

Thus, the probability of not getting a $5$ rs coin, $1-\frac{1}{18}=\frac{17}{18}$ .

Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing $5$  male fish and $8$  female fish (see Fig. 15.4). What is the probability that the fish taken out is a male fish?

The total number of fishes in the tank, $5+8=13$ .

Thus, the probability of getting a male fish, $\frac{no\,of\,male\,fishes}{total\,no\,of\,fishes}=\frac{5}{13}$.

12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers $1,2,3,4,5,6,7,8$  (see Fig. 15.5), and these are equally likely outcomes. What is the probability that it will point at

We can see, there are $8$  numbers on spinner then the total number of favorable outcome is $8$ .

Thus, the probability of getting the number $8$ is, $\frac{no\,of\,digit\,8\,on\,the\,spinner}{total\,number\,of\,digits}=\frac{1}{8}$.

(ii) $\text{an odd number?}$

There are 4 odd digits, $1,3,5,7$ .

Thus, the probability of getting an odd number is,  

$\frac{no\,of\,odd\,digits\,}{total\,number\,of\,digits}=\frac{4}{8}=\frac{1}{2}$ .

(iii) $\text{a number greater than 2?}$ 

There are 6 numbers greater than 2, say $3,4,5,6,7,8$ .

Thus, the probability of getting a number greater than 2, $\frac{no\,of\,digits\,greater\,than\,2\,on\,the\,spinner}{total\,number\,of\,digits}=\frac{6}{8}=\frac{3}{4}$.

(iv) $\text{a number less than 9}$$?$ 

As we can see, every number in the spinner is less than $9$ , thus, we get, 

The probability of getting a number less than $9$ , will be, $1$.

13. A die is thrown once. Find the probability of getting (i) a prime number; 

There is $6$ results can be obtained from a dice. 

There are 3 prime numbers, $2,3,5$ among those results.

Thus, the probability of getting a prime number,$=\frac{\text{no }\text{of }\text{prime }\text{numbers }\text{in }\text{a}\text{ dice}}{\text{total}\text{ numbers}\text{ on }\text{dice}}=\frac{3}{6}=\frac{1}{2}$

(ii) a number lying between $2$ and $6$ 

There are 3 numbers between 2 and 6, 3,4,5.

Thus, the probability of getting a number between 2 and 6,

$=\frac{\text{no }\text{of }\text{numbers }\text{between }\text{2}\text{ and}\text{ 6 }\text{in}\text{ a}\text{ dice}}{\text{total}\text{ numbers }\text{on}\text{ dice}}=\frac{3}{6}=\frac{1}{2}$

(iii) an odd number. 

There are 3 odd numbers among the results, 1,3,5.

Thus, the probability of getting a odd number,

$=\frac{\text{no }\text{of}\text{ odd }\text{numbers}\text{ between}\text{ in }\text{a }\text{dice}}{\text{total}\text{ numbers }\text{on}\text{ dice}}=\frac{3}{6}=\frac{1}{2}$

14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting

(i) a king of red color

We know there are 52 numbers in the deck.

There are 2 kings of red color in the deck.

Thus, the probability,

$=\frac{total\,number\,of\,red\,kings}{total\,number\,of\,cards}=\frac{2}{52}=\frac{1}{26}$

(ii) a face cards 

There are 12 face cards in the deck.

$=\frac{total\,number\,of\,face\,cards}{total\,number\,of\,cards}=\frac{12}{52}=\frac{3}{13}$

(iii) a red face cards 

There are 6 red face cards in the deck.

$=\frac{total\,number\,of\,red\,face\,cards}{total\,number\,of\,cards}=\frac{6}{52}=\frac{3}{26}$

(iv) the jack of hearts

There are 1 jack of hearts card in the deck.

$=\frac{total\,number\,of\,red\,face\,cards}{total\,number\,of\,cards}=\frac{1}{52}$

(v) a spade

There are 13 spade cards in the deck.

$=\frac{total\,number\,of\,spade\,cards}{total\,number\,of\,cards}=\frac{13}{52}=\frac{1}{4}$

(vi) the queen of diamonds 

There are 1 queen of diamonds card in the deck.

15. Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.

(i) What is the probability that the card is the queen?

There are total 5 cards given in our deck.

Thus, the probability of getting a queen card among the 5 cards,

\[=\frac{number\,of\,queen}{number\,of\,total\,cards}=\frac{1}{5}\]

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen? 

Now, the queen is put aside, so there will be 4 cards left.

a. Thus, the probability of getting a queen card among the 5 cards,

\[=\frac{number\,of\,ace}{number\,of\,total\,cards}=\frac{1}{4}\] 

b. There are no queen cards left in the deck.

Thus, the probability of getting a queen card among the 4 cards,

\[=\frac{number\,of\,queen}{number\,of\,total\,cards}=\frac{0}{4}=0\]

16. 12 defective pens are accidentally mixed with 132  good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

There are total (132+12)=144 number of pens in the lot.

And also there are 132 good pens in the given collection.

Thus, the probability of getting a good pen,

$=\frac{number\,of\,good\,pens}{number\,of\,total\,pens}=\frac{132}{144}=\frac{11}{12}$

17. A lot of 20  bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

There are 4 defective bulbs among 20 bulbs.

Thus, the probability of getting a defective bulb,

$=\frac{number\,of\,defective\,bulb}{number\,of\,total\,bulb}=\frac{4}{20}=\frac{1}{5}$

ii. Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?

After the first draw, there are 19 bulbs left in the lot. Again, as the bulb was a non-defective bulb, the total non-defective bulbs are 15.

Therefore, the probability of not getting a defective bulb this time,

$=\frac{15}{19}$ .

18. A box contains 90 discs which are numbered from $1$  to $90$ . If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number. 

There are 81 two digit numbers between 1 to 90.

Thus, the probability of getting a two digit number in the draw,

$=\frac{the\,total\,number\,of\,two\,digit\,numbers}{total\,numbers}=\frac{81}{90}=\frac{9}{10}$

(ii) a perfect square number 

The number of perfect square numbers between 1 to 90.

Thus, the probability of getting a perfect number in the draw,

$=\frac{the\,total\,number\,of\,perfect\,number}{total\,numbers}=\frac{9}{90}=\frac{1}{10}$

(iii) a number divisible by $5$. 

The number of numbers divisible by 5, 18.

Thus, the probability of getting a number divisible by 5 in the draw,

$=\frac{the\,total\,number\,of\,number\,divisible\,by\,5}{total\,numbers}=\frac{18}{90}=\frac{1}{5}$

19. A child has a die whose six faces show the letters as given below: A, A, B, C, D, E. The die is thrown once. What is the probability of getting

(i) A? 

There are two A’s in the six faces, so, the probability of getting an A,

$=\frac{total\,number\,of\,A's}{total\,number\,of\,sides}=\frac{2}{6}=\frac{1}{3}$

(ii)  D? 

There are one D in the six faces, so, the probability of getting an D,

$=\dfrac{total\,number\,of\,D's}{total\,number\,of\,sides}=\dfrac{1}{6}$

20. Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is the probability that it will land inside the circle with diameter 1m?

It is given that it is a rectangle with sides 3 m and 2 m.

Thus, the area of the rectangle,

$=$ length $\times$ breadth

$=3 \times 2=6 \mathrm{~m}^{2}$

The radius of the circle, half of diameter $=\frac{1}{2}\,m$ .

The area of the circle, $=\pi .{{\left( \frac{1}{2} \right)}^{2}}=\frac{\pi }{4}\,{{m}^{2}}$

Thus, the probability of the die landing inside the circle is,

$=\frac{area\,of\,the\,circle}{area\,of\,the\,rectangle}=\frac{\frac{\pi }{4}}{6}=\frac{\pi }{24}$

21. A lot consists of $144$  ball pens of which $20$ are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

(i) She will buy it ? 

There are total 144 ball pens in the lot and 20 of them are defective.

Thus, the total number of non-defective pens, $(144-20)=124$ .

Nuri will not buy the pen if it is defective, thus,

The probability of getting a good pen is,

$=\frac{total\,no\,of\,good\,pens}{total\,no\,of\,pens}=\frac{124}{144}=\frac{31}{36}$

(ii) She will not buy it ? 

Now, the probability of Nuri not buying the pen is,

$=1-\frac{31}{36}=\frac{36-31}{36}=\frac{5}{36}$

22. Refer to example 13: (i) Complete the following table:

If there are two dices thrown simultaneously, then we can get the following results,

$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$,

$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$

Thus, the total number of results is 36.

Probability of getting a sum of 2, $=\dfrac{1}{36}$ 

Probability of getting a sum of 3, $=\dfrac{2}{36}=\dfrac{1}{18}$ 

Probability of getting a sum of 4, $=\dfrac{3}{36}=\dfrac{1}{12}$ 

Probability of getting a sum of 5, $=\dfrac{4}{36}=\dfrac{1}{9}$ 

Probability of getting a sum of 6, $=\dfrac{5}{36}$ 

Probability of getting a sum of 7, $=\dfrac{6}{36}=\dfrac{1}{6}$ 

Probability of getting a sum of 8, $=\dfrac{5}{36}$ 

Probability of getting a sum of 9, $=\dfrac{4}{36}=\dfrac{1}{9}$ 

Probability of getting a sum of 10, \[=\dfrac{3}{36}=\dfrac{1}{12}\] 

Probability of getting a sum of 11, $=\dfrac{2}{36}=\dfrac{1}{18}$ 

Probability of getting a sum of 12, $=\dfrac{1}{36}$ 

Thus, we get the values of our table.

(ii) A student argues that there are $11$ possible outcomes$2,3,4,5,6,7,8,9,10,11\text{ and 12}$ .Therefore, each of them has a probability $\frac{1}{11}$ . Do you agree with this argument?

Justify your answer.

As we can see different values all over the table, we can conclude that, the given statement is wrong. The probability of each of them can never be $\dfrac{1}{11}$. 23. A game consists of tossing a one rupee coin $3$  times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Hanif will win if he gets 3 heads and 3 tails consecutively.

The probability of Hanif losing the game, = The probability of not getting 3 heads and 3 tails.

The possible outcomes of the tosses, $(HHH,HHT,HTH,HTT,THH,THT,TTH,TTT)$ 

The total number of outcomes is, 8.

Thus, the probability of not getting 3 heads and 3 tails, 

$=1-\frac{2}{8}=\dfrac{6}{8}=\dfrac{3}{4}$.

24. A die is thrown twice. What is the probability that

(i) $5$  will not come up either time? 

(Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment)

We can see that two dices are thrown altogether, thus the total number of outcomes =36.

Now, the total cases where atleast 5 occurs is, 11, i.e,$(1,5),(2,5),(3,5),(4,5),(5,5),(6,5),(5,1),(5,2),(5,3),(5,4),(5,6)$  

So, the probability of not getting 5 either time is,$=1-\frac{11}{36}=\frac{25}{36}$.

(ii) $5$  will come up at least once?

And, probability of getting 5 atleast once is,$\frac{11}{36}$.

25. Which of the following arguments are correct and which are not correct? Give reasons for your answer.

(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is $\frac{1}{3}$. 

In this problem, two coins are tossed simultaneously, thus we get 4 outcomes, i.e, $(HH,HT,TH,TT)$ .

So, the probability of getting both heads and tails, $=\frac{2}{4}=\frac{1}{2}$ .

Thus, the statement is wrong.

(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is $\frac{1}{2}$. 

By throwing a die, we get 6 possible outcomes.

The odd numbers are $1,3,5$ .

Thus the probability of getting a odd number, $=\frac{3}{6}=\frac{1}{2}$ .

So, the statement is true.

Overview of Deleted Syllabus for CBSE Class 10 Maths Probability

Vedantu's NCERT Solutions for Class 10 Maths Chapter 14 - Probability offers an exceptional resource for students seeking to grasp the complexities of probability theory. 

You will be able to solve problems involving coin tosses, dice rolls, card games, and other scenarios where chance plays a role.

With comprehensive and well-structured explanations, the platform empowers learners to tackle real-world challenges with confidence. 

Other Study Material for CBSE Class 10 Maths Chapter 14

Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 10 Maths Chapter 14 Probability

1. What is the curriculum of CBSE Board Class 10 Mathematics?

There are 15 units in Class 10 Mathematics. These include Arithmetic Progressions, Pair of Linear Equations in Two Variables, Real numbers, Polynomials, Quadratic Equations, Triangles, Some Applications of Trigonometry, Introduction to Trigonometry, Constructions, Circles, Coordinate Geometry, Surface Areas and Volumes, Areas Related to Circles, Probability and Statistics.

2. What are the applications of probability in day to day life?

Probability is a branch of mathematics that deals with the likelihood of events happening. It is used in a wide variety of applications in our daily lives, including:

Weather forecasting: Weather forecasters use probability to estimate the likelihood of rain, snow, or other types of precipitation.

Sports betting: Sports bettors use probability to assess the likelihood of a particular team or player winning a game.

Insurance: Insurance companies use probability to calculate the risk of an event happening and to determine the premiums that they charge their customers.

Medical diagnosis: Doctors use probability to assess the likelihood of a patient having a particular condition.

Stock market investing: Stock market investors use probability to assess the risk of an investment and to determine the potential returns.

Marketing: Marketers use probability to target their advertising campaigns to the most likely customers.

Decision making: In many situations, we need to make decisions based on uncertain information. Probability can help us to make more informed decisions by providing us with a framework for assessing the likelihood of different outcomes.

Here are some other examples of how probability is used in our daily lives:

When we decide what to wear, we are essentially using probability to assess the likelihood of different weather conditions.

When we choose a route to drive to work, we are using probability to assess the likelihood of traffic congestion.

When we decide what to eat, we are using probability to assess the likelihood of different foods being available at the grocery store.

When we decide what to watch on TV, we are using probability to assess the likelihood of different shows being interesting to us.

3. What are the subtopics of Ch 14 Maths Class 10?

Subtopics of Chapter 14 of the Class 10 Maths NCERT textbook:

Introduction

Experiment and Outcome

Sample Space

Elementary Events

Favourable Outcomes

Probability of an Event

Theoretical Probability

Experimental Probability

Addition Theorem of Probability

Complement of an Event

Multiplication Theorem of Probability

This chapter introduces the concept of probability, which is the likelihood of an event occurring. It discusses the different types of events, such as elementary events, favourable outcomes, and sure events. The chapter also covers the two methods of calculating probability: theoretical probability and experimental probability. The addition and multiplication theorems of probability are also discussed in this chapter.

4. What can we learn in the chapter Probability?

In the chapter Probability, we can learn about the following:

The definition of probability and how it is measured.

The different types of events, such as elementary events, favourable outcomes, and sure events.

The two methods of calculating probability: theoretical probability and experimental probability.

The addition and multiplication theorems of probability.

How to use probability to make predictions about the outcomes of random events.

Here are some specific things we can learn:

The probability of an event can be anywhere from 0 to 1. A probability of 0 means that the event is impossible, while a probability of 1 means that the event is certain.

The probability of an event can be calculated by dividing the number of favourable outcomes by the total number of possible outcomes.

If we repeat an experiment many times, the experimental probability of an event will approach the theoretical probability of the event.

The addition theorem of probability can be used to calculate the probability of two events occurring when the events are mutually exclusive.

The multiplication theorem of probability can be used to calculate the probability of two events occurring when the events are not mutually exclusive.

5. State the basic law behind probability?

The law of probability informs us how likely particular occurrences are to occur. According to the rule of large numbers, the more trials you have in an experiment, the closer you come to a precise probability. The multiplication rule is used to calculate the likelihood of two occurrences occurring at the same time. 

6. State an event where the probability is ½?

On tossing a random fair coin, there are two possible outcomes i.e. head and tail. 

When a coin is tossed, the probability of getting ahead is ½ and the probability of getting a tail is also ½. This is based on the probability of the specific event occurring.

7. What is the chapter probability about?

Probability is a field of mathematics that deals with numerical descriptions of what will happen in the future, or whether something is true or not. This is a high-scoring yet challenging Chapter in Class 10 Maths. As a result, you must be conversant with the tips and methods required to swiftly answer numerical issues. In this regard,  Vedantu  offers exact NCERT Solutions for Class 10 Maths Probability, which include all types of sums that may be encountered in examinations.

8. Why choose Vedantu for Chapter 14 Class 10 Maths?

In a highly competitive world, students are thriving for the best educational services to score as much as they can, Vedantu offers the best solutions. NCERT Solutions for Class 10 Maths Probability is one of the best study guides for students, to help them achieve their desired scores. These solution PDFs are available at free of cost on the Vedantu app and the Vedantu website.

9. What is the main focus of Chapter 14 - Probability Class 10 Maths?

The main focus of Chapter 14 - Probability Class 10 is to introduce students to the basic concepts of probability, including the theoretical approach, experiments, outcomes, events, and the probability of an event occurring.

10. How does NCERT Solutions for Class 10 Probability Chapter 14 help in understanding probability?

The NCERT Solutions for class 10 probability provides detailed explanations and step-by-step solutions to textbook problems, helping students understand the fundamental principles of probability and how to apply them to solve various problems.

11. What key concepts are covered in the NCERT Solutions for Probability Class 10 solutions Chapter 14?

Key concepts covered in class 10 probability include:

Probability of an event

Complementary events

Probability of simple and compound events

Use of probability in real-life situations

12. Are the probability class 10 solutions aligned with the latest syllabus?

Yes, the NCERT probability class 10 solutions are aligned with the latest CBSE syllabus for the academic year 2024-25, ensuring that all topics and concepts are covered as per the curriculum.

13. How do probability class 10 solutions assist in exam preparation?

The probability class 10 solutions include a variety of problems and their detailed solutions, which help students practice extensively. This thorough practice enhances problem-solving skills and prepares students effectively for exams.

14. Can these solutions help in clearing doubts about class 10 probability concepts?

Absolutely. In probability class 10 NCERT solutions, detailed and clear explanations provided in the solutions help in resolving any doubts or misconceptions students might have about probability concepts.

15. What types of questions are included in the probability class 10 NCERT solutions?

The probability class 10 NCERT solutions include a wide range of questions such as multiple-choice questions, short answer questions, and long answer questions. These questions cover theoretical concepts, practical problems, and real-life applications of probability.

16. How can students benefit from practising probability class 10 NCERT solutions regularly?

Regular practice of chapter 14 class 10 maths helps students build a strong foundation in probability, enhances their problem-solving abilities, and boosts their confidence in tackling various types of questions in exams.

17. Are there any real-life examples provided in the solutions to explain chapter 14 class 10 maths concepts?

Yes, the chapter 14 class 10 maths solutions include real-life examples to illustrate how probability is used in everyday situations, making the concepts more relatable and easier to understand.

18. Where can students access the class 10 maths ch 14 for Class 10 Maths Chapter 14 - Probability?

Students can access the class 10 maths ch 14 solutions on educational platforms like Vedantu, where they are available for free download in PDF format, providing easy and convenient access for study and revision.

NCERT Solutions for Class 10 Maths

Ncert solutions for class 10.

NCERT Solutions for Class 10 Maths Chapter 14 Probability

Get the NCERT Solutions for Class 10 Maths Chapter 14 Probability in Hindi and English Medium updated and revised for session 2024-25. As per the new syllabus and new NCERT Books published for academic year 2024-25.

10th Maths Chapter 14 Solutions for CBSE Board

• Class 10 Maths Exercise 14.1 in English
• Class 10 Maths Exercise 14.1 in Hindi

10th Maths Chapter 14 Solutions for State Boards

• Class 10 Maths Chapter 14 Exercise 14.1
• Class 10 Maths Chapter 14 Exercise 14.2
• Class 10 Maths Chapter 14 Exercise 14.3
• Class 10 Maths Chapter 14 Exercise 14.4
• Class 10th Maths Chapter 14 NCERT Book
• Class 10 Maths NCERT Solutions
• Class 10 all Subjects NCERT Solutions

Exercise 14.1 in English and Hindi medium for CBSE, UP Board, MP Board free to download updated for new academic session 2024-25. These NCERT Solutions are applicable for UP Board also as UP Board has implemented NCERT Textbooks as UP Board Textbooks for the academic session 2024-25. UP board Students can download UP Board solutions for Class 10 Maths Chapter 14 all exercises. These solutions are applicable for all the students who are following CBSE Board for academic session 2024-25. Not only NCERT Solutions but the NCERT Books also for all subjects along with Offline Apps of NCERT Textbook’s Solutions for class 10 all subjects. Videos related to Ex 14.1 of class 10 Maths are given below explaining each and every questions. Explanation is in simple language, so that students can understand easily.

Important Definitions related to Class 10 Maths Chapter 14

What do you understand by random experiment.

Random Experiment: A random experiment is one in which the exact outcome cannot be predicted.

What is meant by Trial in Probability?

Trial: Performing a random experiment is called a trial.

What is meant by Outcomes?

Outcomes: The result of a random experiment is called an outcome.

What is Sample space in terms of Probability?

Sample space: The collection of all possible outcomes of a random experiment is called a sample space.

What are Events in Probability?

Event: Any possible outcome or combination of outcomes of a random experiment is called an event.

Which events are called Equally likely events?

Equally likely events: Two or more events of a random experiment are said to be equally likely events if each one of them have an equal chance of occurrence.

What is meant by Probability of an event?

Probability of an event: The chance of occurrence of the event expressed quantitatively is known as the probability of an event and denoted by P(E).

NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.1 of Probability is given below for academic session 2024-25. NCERT Solutions for class 10 is updated according to CBSE Curriculum 2024-25. Join the discussion forum to ask your questions and answer the questions already asked by other users.

Important Questions on Class 10 Maths Chapter 14

Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game.

When we toss a coin, the possible outcomes are only two, head or tail, which are equally likely outcomes. Therefore, the result of an individual toss is completely unpredictable.

It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Probability that two students are not having same birthday P (not E) = 0.992 Probability that two students are having same birthday P (E) = 1 − P (not E) = 1 − 0.992 = 0.008

A die is thrown once. Find the probability of getting a prime number.

The possible outcomes when a dice is thrown = {1, 2, 3, 4, 5, 6} Number of possible outcomes of a dice = 6 Prime numbers on a dice are 2, 3, and 5. Total prime numbers on a dice = 3 Probability of getting a prime number = 3/6 =1/2

Five cards−−the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random. What is the probability that the card is the queen?

Total number of cards = 5 Total number of queens = 1 P(getting a queen) = (Number of favourable outcomes)/(Number of total possible outcomes) =1/5

A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

Total number of bulbs = 20 Total number of defective bulbs = 4 P(getting a defective bulb) = (Number of favourable outcomes)/(Number of total possible outcomes) = 4/20 = 1/5

A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears a two-digit number.

Total number of discs = 90 Total number of two-digit numbers between 1 and 90 = 81 P (getting a two-digit number) = 91/90 = 9/10

ONE MARK QUESTIONS A number is chosen at random from the numbers -3, -2,-1, 0, 1, 2, 3. What will be the probability that square of this number is less than or equal to 1? [CBSE 2017] THREE MARKS QUESTIONS Two different dice are thrown together. Find the probability that the numbers obtained (i) have a sum less than7 (ii) have a product less than 16 (iii) is a doublet of odd numbers. [CBSE 2017] FOUR MARKS QUESTIONS Peter throws two different dice together and finds the product of the two numbers obtained. Rina throws a die and squares the number obtained. Who has the better chance to get the number 25? [CBSE 2017]

The measure of certainty of events in numerical values, under certain conditions, is provided by the branch of mathematics called Theory of Probability . This theory has extensive use as one of the basic tools in statistics and wide range of applications in Science, engineering, biological science, medical, commerce, weather forecasting etc.

Historical facts!

1. The concept of probability was developed in a very strange manner. In 1654, a gambler by name Chevalier de Mere approached the well-known 17th century French philosopher and mathematician Blaise Pascal regarding certain dice problems. Pascal discussed them with another French mathematician Pierre de Fermat and they found solution to dice problems. This work was the beginning of probability theory. 2. Probability theory has its actual origin in the 16th century when an Italian physician and mathematician J. Cardan wrote the first book on the subject ‘The book on Games of Chance’. Since its inception, the study of Statistics and probability has attracted the attention of great mathematicians (James Bernoulli (1654 – 1705), A. de Moivre (1667 – 1754) and Pierre Simon Laplace (1749 – 1827)). 3. In 1812, Pierre Simon Laplace or Pierre de Laplace (1749 – 1827, France) proposed a mathematical system of inductive reasoning based on probability. He introduced many principles of probability, one among them is, “Probability is the ratio of the favoured events to the total possible events”. 4. Statistician Karl Pearson (1857 – 1936) had tossed the coin 24000 times and he got 12012 heads. Then calculated experimental probability 12012/2400 = 0.5005. 5. In the eighteenth century French De Buffon tossed a coin 4040 times and got 2048 heads. Then he calculated experimental probability 2048/4040 = 0.507.

How many examples are there in chapter 14 Probability of class 10 Maths?

There are two exercises in chapter 14, Probability (class 10 Maths). In the exercise 14.1, there are 25 questions. There are 13 examples in chapter 14 of class 10 math. All the examples and questions of this chapter are good.

What are real-life applications of Probability in Class 10 Maths Chapter 14?

Probability has something to do with a chance. It is the study of things that might happen or might not. We use it most of the time, usually without thinking of it. Some real-life applications of Probability:

• Weather Forecasting: Before planning for an outing or a picnic, we always check the weather forecast.
• Flipping a coin or Dice.
• Lottery Tickets: Winning or losing a lottery is one of the most interesting examples of Probability.
• Playing Cards: There is a probability of getting a desired card when we randomly pick one out of 52.

Is there any optional exercise in chapter 14 of class 10th Maths?

No, there is not any optional exercise in chapter 14 of class 10th Maths. The second exercise was earlier the optional exercise but now it is deleted from syllabus.

Is chapter 14 of class 10th math easy to understand?

Chapter 14 of class 10th Maths is not easy and not difficult. It lies in the middle of easy and difficult because some examples and questions of this chapter are easy, and some are difficult. However, the difficulty level of anything varies from student to student. So, Chapter 14 of class 10th Maths is easy or not depends on students also. Some students find it difficult, some find it easy, and some find it in the middle of easy and difficult.

« Chapter 13: Statistics

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NCERT Solutions Class 10 Maths Chapter 14 Statistics

NCERT Solutions for Class 10 Maths Chapter 14 Statistics deals with the classification of ungrouped as well as grouped frequency distributions. In the previous classes, the students must have learned how to represent the data pictorially in various graphical formats such as bar graphs, histograms, and frequency polygons. They must be familiar with topics like numerical representatives of ungrouped data and measures of central tendencies such as mean, median, and mode. In the NCERT Solutions Class 10 Maths Chapter 14, learning will switch to all the three measures such as mean, median, mode from ungrouped data to the grouped data.

The students will understand some new topics of statistics such as cumulative frequency, cumulative frequency distribution, cumulative frequency curves or ‘ogives,’ etc. A detailed pdf file of class 10 maths NCERT solutions Chapter 14 Statistics can be found below and also you can find some of these in the exercises given below.

• NCERT Solutions Class 10 Maths Chapter 14 Ex 14.1
• NCERT Solutions Class 10 Maths Chapter 14 Ex 14.2
• NCERT Solutions Class 10 Maths Chapter 14 Ex 14.3
• NCERT Solutions Class 10 Maths Chapter 14 Ex 14.4

NCERT Solutions for Class 10 Maths Chapter 14 PDF

NCERT Solutions Class 10 Maths Chapter 14 Statistics enables the students to understand the concept of statistics by outlining its key points. This chapter also helps in the revision of concepts such as mean, median, mode . Apart from this, the students will be able to explore some new topics with the help of this chapter, such as measures of central tendency in the ungrouped data. The following links mentioned below will help understand the sections in this chapter further in detail :

☛ Download Class 10 Maths NCERT Solutions Chapter 14 Statistics

NCERT Class 10 Maths Chapter 14   Download PDF

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

NCERT Solutions Class 10 Maths Chapter 14 Statistics forms the perfect study material for class 10 students to prepare for their board exams. With detailed explanations on the various methods of the step-deviation method, assumed mean method, direct method, etc., these solutions cover all the important points in a detailed step-by-step manner. A section-wise detailed analysis of NCERT Solutions class 10 chapter 14 Statistics can be seen below :

• Class 10 Maths Chapter 10 Ex 14.1 - 9 Questions
• Class 10 Maths Chapter 10 Ex 14.2 - 6 Questions
• Class 10 Maths Chapter 10 Ex 14.3 - 7 Questions
• Class 10 Maths Chapter 10 Ex 14.4 - 3 Questions

☛ Download Class 10 Maths Chapter 14 NCERT Book

Topics Covered: The Class 10 maths NCERT solutions Chapter 14 Statistics covers a variety of topics such as mean, median, and mode of grouped data, cumulative frequency , representation of cumulative frequency data in the graphical form, and obtaining the median of grouped data.

Total Questions: Class 10 Maths Chapter 14 Statistics consists of 25 questions. Out of these 25 problems, 15 are fairly easy to solve, 5 are of moderate level, while 5 are a bit complex in nature.

List of Formulas in NCERT Solutions Class 10 Maths Chapter 14

NCERT solutions class 10 maths Chapter 14 guides the students in studying the grouped data and finding their mean, median, and mode . Analyzing data is one of the most essential skills to have in today’s time hence this is the most important chapter of class 10 maths that has a lot of real-life applications. Statistics involves the use of formulas which are covered in detail in the chapter. Some of the important formulas that will help students study and analyze different data charts are as :

• Class mark = (Upper class limit + Lower class limit)/ 2
• Relationship between the three measures of central tendency : 3 Median = Mode + 2 Mean

Important Questions for Class 10 Maths NCERT Solutions Chapter 14

Video Solutions for Class 10 Maths NCERT Chapter 14

FAQs on NCERT Solutions Class 10 Maths Chapter 14

What is the importance of ncert solutions class 10 maths chapter 14 statistics.

NCERT Solutions Class 10 Maths Chapter 14 Statistics deals with the collection, analysis, presentation, and interpretation of data in different forms. This chapter enables the students to arrange data in a particular form in order to study the salient features of the same and present them in a way that can be easily understood by all. The modern world is highly inclined towards studying and analyzing data in order to make the right decisions, be it in any field, thereby these solutions become an important resource to study.

Do I Need to Practice all Questions Provided in NCERT Solutions Class 10 Maths Statistics?

There are a total of 25 questions in NCERT solutions class 10 maths Statistics. All the questions are curated to cover all the topics related to statistics, such as finding the mean, median, mode using different methods, cumulative frequency representation, etc. in detail. A consistent practice of all the examples and questions in the chapter will result in a better understanding and increased confidence in problem-solving. Thus, it is highly beneficial for the students to practice all the questions.

What are the Important Topics Covered in NCERT Solutions Class 10 Maths Chapter 14?

NCERT Solutions Class 10 Maths Chapter 14 will help you revise the numerical representation of ungrouped data, also known as the measures of central tendencies called mean, median, and mode. Along with that, these solutions also cover the concept of cumulative frequency, the cumulative frequency distribution, and steps involved in drawing ‘ogives’ or cumulative frequency curves.

How Many Questions are there in Class 10 Maths NCERT Solutions Chapter 14 Statistics?

NCERT Solutions Class 10 maths chapter 14 consists of a total of 25 questions distributed in 4 exercises. These questions challenge the students to think out of the box and solve complex problems based on the frequency distribution table, exclusive or continuous frequency, range, etc. Practicing all the questions in these NCERT solutions class 10 maths chapter 14 will result in an excellent preparation for board exams.

What are the Important Formulas in NCERT Solutions Class 10 Maths Chapter 14?

Some of the important formulas of this chapter is the mean of grouped data using the direct method, mean method, and step deviation method. Other formulas include median and mode for grouped data. These formulas can better be understood by solving questions in a stepwise manner. The derivation of these formulas is stated in the NCERT Solutions Class 10 Maths Chapter 14.

Why Should I Practice NCERT Solutions Class 10 Maths Statistics 14?

The NCERT Solutions Class 10 Maths Statistics 14 is a very important chapter not just for exams but for learning insights that have their applications in real-life situations as well. Through this chapter, students will learn how to arrange and analyze different kinds of data by arranging it in a definite or required order. This is one of the greatest skill-set in today’s time and can offer a life-long benefit to the students; hence they must practice NCERT solutions regularly.

NCERT Solutions for Class 10 Maths Chapter 14 Free PDF Download

Ncert solutions for class 10 maths chapter 14 statistics.

For every student, mathematics is the toughest subject and CBSE recommends NCERT books that aren’t decent. However these books square measure necessary for studies. Also, the scholars got to solve all the question and answer given within the exercise to fully learn the subject. Toppr provides the most effective NCERT solutions for class 10 maths chapter 14.

Apart from that our answer is extremely handy. They do not solely assist you to unravel the matter however conjointly teach you the way to unravel them. Our team of specialists makes the NCERT solutions for class 10 maths chapter Statistics for you. You’ll be able to download the answer by clicking the link below.

Also, our NCERT solutions of class 10 maths chapter 14 are printed for the scholars in keeping with the most recent CBSE board test guideline and course of study. These solutions can assist you to get sensible marks in your boards. You’ll be able to realize all the NCERT solutions for class 10 maths chapter 14 in a very single place. The NCERT solutions for class 10 maths chapter 14 is offered free from the value at Toppr to download. Besides we provide students several services like live doubt clearance sessions that save energy and times. Download supplementary subject solutions by clicking here. So, download Toppr for Android or iOS or you can signup for free.

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CBSE Class 10 Maths Chapter 14 NCERT Solutions

The solution is part of the NCERT Solutions for Class 10 Maths Chapter 14 in which they learn more advanced topics about Statistics that they have learned in class 9. In this chapter, there will be a shift to learn more about mean, median and mode of grouped data. Also, the students have to converse on the concept of cumulative frequency and its distribution, also to learn how to draw cumulative frequency curves.

The graphical illustration of cumulative frequency and mean, median, and mode of grouped data are the concept that the students will learn through NCERT Solutions for Class 10 Maths Chapter 14. Graphical representation gives the chapter a more striking and collaborating look. Also, this helps to clear the Chapter in a more broad way.

For 100% accurate result and answer and also for revision use NCERT Solutions for Class 10 Maths Chapter 14. They will not only help you to score higher grades but also help you to learn the concept fully. The solution is based on the guidelines of CBSE board and also follow the complete guidelines mentioned in the course of study. In the solutions, each and every topic is covered so that you can go through them in serial order.

Sub-topics covered under Chapter 14 NCERT Solutions

• Ex. 14.1 Introduction
• Ex. 14.2 Mean of Grouped Data
• Ex. 14.3 Mode of Grouped Data
• Ex. 14.4 Median of Grouped Data
• Ex. 14.5 Graphical Representation of Cumulative Frequency Distribution
• Ex. 14.6 Summary

You can download NCERT Solutions for Class 10 Maths Chapter 14 by clicking on the download button below

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NCERT Solutions for Class 10 Maths

• NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.1
• NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.2
• NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.3
• NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.4
• NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.1
• NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2
• NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3
• NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.4
• NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.5
• NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6

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NCERT Solutions for Class 10 Maths Ch 14 Statistics

• Exercise 14.1
• Exercise 14.2
• Exercise 14.3
• Exercise 14.4

How many exercises in Chapter 14 Statistics

What are the methods to calculate mean, mode = (...........) − 2 (mean), write the empirical relation between mean, mode and median., contact form.

CBSE NCERT Solutions

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Class 10 Mathematics Assignments

We have provided below free printable Class 10 Mathematics Assignments for Download in PDF. The Assignments have been designed based on the latest NCERT Book for Class 10 Mathematics . These Assignments for Grade 10 Mathematics cover all important topics which can come in your standard 10 tests and examinations. Free printable Assignments for CBSE Class 10 Mathematics , school and class assignments, and practice test papers have been designed by our highly experienced class 10 faculty. You can free download CBSE NCERT printable Assignments for Mathematics Class 10 with solutions and answers. All Assignments and test sheets have been prepared by expert teachers as per the latest Syllabus in Mathematics Class 10. Students can click on the links below and download all Pdf Assignments for Mathematics class 10 for free. All latest Kendriya Vidyalaya Class 10 Mathematics Assignments with Answers and test papers are given below.

Mathematics Class 10 Assignments Pdf Download

We have provided below the biggest collection of free CBSE NCERT KVS Assignments for Class 10 Mathematics . Students and teachers can download and save all free Mathematics assignments in Pdf for grade 10th. Our expert faculty have covered Class 10 important questions and answers for Mathematics as per the latest syllabus for the current academic year. All test papers and question banks for Class 10 Mathematics and CBSE Assignments for Mathematics Class 10 will be really helpful for standard 10th students to prepare for the class tests and school examinations. Class 10th students can easily free download in Pdf all printable practice worksheets given below.

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Mathematics assignments | pdf | 8 to 12, math assignment ch-13 class x | statistics.

Chapter: 13(Statistics) Formulas used to solve the Assignment on statistics

Range =  Highest value - Lowest value

Class size =  Upper limit - Lower limit

MODE of Data in Statistics

W here l = lower limit of the modal class,

 h = size of the class interval 

  f 1  = frequency of the modal class

f 0  = frequency of the class preceding the modal class,

f 2  = frequency of the class succeeding the modal class.

Median of Data in Statistics

Where l= lower limit of median class                           

N = Sum of all observations

f = frequency of the median class.       

Cf= cumulative frequency of class  preceding the median class

 h= height of  median class

Empirical Formula of Data in Statistics

      Mode= 3 Median  - 2 Mean

Empirical formula does not provide the same result in different situations. So, students should use it only when asked otherwise avoid it.

* If median class is the first class –interval then cumulative frequency (Cf) of the preceding class interval should be taken as zero.

* For finding Mean class interval need not to be continuous.

* For finding Mode and Median class interval should be continuous. 

* For making class interval continuous we should subtract 0.5 from the lower limits and add 0.5 to the upper limits of all class intervals. 

Assignment on Statistics Solve the following questions

Problems based on mean.

Do we use here step-deviation method ? Why or Why not   

No, we cannot use step deviation method here because height of the class intervals are not equal. So we can solve this question either by Direct Method or by Assumed mean method

PROBLEMS BASED ON MODE  

QUESTION 10. For the given data find mean & mode.              

Ans : Mean=26 & Mode=22.85

PROBLEMS BASED ON MEDIAN

QUESTION 15. Find median.         Ans: 40

Ans:  f 1 =35 &  f 2 = 25

MISCELLANEOUS PROBLEMS ON MEAN, MODE, MEDIAN  

QUESTION  21. 

If mode = 24.5 and mean is 29.75 then find median by using empirical formula.    [Ans: 28]

QUESTION 23. Calculate Mean, mode, median.     Ans : Mean = 51.63, Mode = 53.33, Median = 52

Marks no of students more than 0 80 more than 10 76 more than 20 71 more than 30 65 more than 40 57 more than 50 43 more than 60 28 more than 70 15 more than 80 10 more than 90 8 more than 100 0, higher order thinking skills (hots).

Answer:  f 1  = 28,  f 2  = 32 and  f 3  = 24

Solution Hint:   Use direct method to solve this question

For easy calculation use step deviation method and take assume mean (a) in front of f 1

QUESTIONS DELETED FROM CBSE SYLLABUS

QUESTION 25. Find median by using less than ogive & actual calculation.                               

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CBSE Assignments class 10 Mathematics

Lesson Plan Math Class X (Ch-8) | Trigonometry

• Assignment 10 15
• Assignment 11 12
• Assignment 12 15
• Assignment 8 8
• Assignment 9 5
• Lesson plan 10 15
• Lesson Plan 12 14
• Lesson Plan 8 10
• Maths 10 20
• Maths 11 21
• Maths 12 18

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NCERT Solutions for Class 10 Maths for 2025 Board Exams | Free PDF Download

Our professional faculty has curated these NCERT Solutions to aid students in their term-by-term exam preparation. Students looking for NCERT Solutions for Class 10 Maths can download complete chapter-by-chapter to help them solve questions more effectively. 

All the solutions of the NCERT books are without a doubt the best study material available for the students. These CBSE NCERT solutions for Class 10 Maths  will also help the students in developing a more in-depth grasp of subjects presented in the textbook. Practicing the questions from the textbook will assist pupils in assessing their degree of preparation and subject comprehension. This solved exercise has very helpful for your exam.

NCERT Solutions Class 10 Maths  All Chapters:

Practice Comprehensively with Oswaal360 CBSE Online Courses for Class 10 Board Exams 2025

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The NCERT Solutions for Class 10 Maths list contains all of the chapter-wise solutions to the NCERT Book for Class 10 Maths problems, given clearly and transparently while preserving the textbook's purpose. The NCERT Solutions for 10th Class  can be used as extra references and study tools by students. Practicing NCERT textbook exercise answers would undoubtedly aid students' test preparation.

These assist students in comprehending or solving the sorts of problems that a paper setter asks in the board examinations for  CBSE   Class 10 Maths . Furthermore, offering solutions to all areas relating to circles assists students in effectively preparing for the board examinations.

The subject matter experts make sure that the errors in the  NCERT  textbooks are resolved according to the  NCERT  rules and syllabus. 

The solutions to the  NCERT -based questions for all the chapters   of Class 10 Maths  are accessible at  Oswaal CBSE Class 10 Maths . These solutions will assist students in acing their board examinations by preparing for the examinations ahead of time. Students may clarify all of their conceptual questions or doubts with the assistance of the notes provided in Maths NCERT Solutions for Class 10 of all the chapters. Experts have created these NCERT Solutions based on the most recent update to the  CBSE  Syllabus to help students in Class 10 prepare well for their term-wise examinations.

Chapter Details and Exercises of Class 10 Maths NCERT Solutions:

The NCERT Solutions Chapter 1 Real Number for Class 10 Maths:

Students in grade 10 will investigate real and irrational numbers in Chapter 1. The chapter begins with Euclid's Division Lemma, which asserts that "given two positive numbers a and b, there are two distinct integers q and r fulfilling

a= bq + r, 0 ≤ r < b."

Based on this lemma, Euclid's Division technique is used to determine the HCF of two positive numbers. The Fundamental Theorem of Arithmetic, which is utilized to compute the LCM and HCF of two positive integers, is then defined. The idea of an irrational number, a rational number, and the decimal expansion of rational numbers are then described using a theorem.

Real Numbers:

A quantity that may be stated as a limitless decimal expansion in mathematics is a real number. Unlike natural numbers 1, 2, 3... derived from counting, real numbers are utilized to measure constantly altering quantities such as size and time. Real numbers are represented by R.

Real numbers are classified into two types, i.e., rational and irrational numbers . 

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Rational number:

A number may be expressed as the quotient p/q of two integers, with q ≠ 0. In addition to fractions, the set of rational numbers includes all integers. Each can be described as a quotient with the integer as the numerator and one as the denominator. Rational numbers are represented by Q.

Rational numbers are divided into two subparts, i.e., non-integer rational numbers and rational integer numbers.

Integer rational numbers are those rational numbers made up of integers, whereas non-integer rational numbers do not contain integers.

This class of rational numbers, i.e., integers, is further classified into two groups, i.e., negative and whole numbers. Thus, whole numbers are the numbers that are the combination of natural numbers along with 0.

Irrational numbers

Those types of real numbers, which are not rational, are referred to as irrational numbers. Irrational numbers, in other words, cannot be stated as the ratio of two integers. When the length ratio of two-line segments is an irrational number, the line segments are said to be incommensurable, which means that they have no "measure" in common. That is, no length ("the measure"), no matter how small, can be employed to convey the lengths of both given segments as integer multiples of itself. 

Chapter 1 Real Numbers:

The Fundamental Theorem of Arithmetic claims made after analysing previous work and showing and inspiring with examples. The representation of rational numbers in decimal terms in terms of terminating/non-terminating recurring decimals.

Important Steps to Take

To get the HCF of two positive numbers, say c and d, with c > d, apply the following steps:

Step 1:  Apply the division lemma of Euclid to c and d. So, we look for whole numbers, q and r, such that c = dq + r, 0≤ r< d.

Step 2:  If r = 0, the HCF of c and d equals d. We apply the division lemma to d and r if r = 0.

Step 3:  Repeat the process until the remaining equals zero. At this point, the divisor will be the desired HCF. This approach works because HCF (c, d) = HCF (d, r), where HCF (c, d) represents the HCF of c and d, respectively.

NCERT Solutions for 10th Grade Maths Chapter 1 Exercises:

Also available at  CBSE Class 10 Books  contains the following materials for NCERT Class 10 Chapter 1 Real Numbers :

• Notes for NCERT Real Numbers Class 10
• Important Real Numbers Class 10 Questions
• Class 10 NCERT Exemplar Solution of Chapter Real Numbers

Chapter 2 Polynomials –NCERT Solutions for Class 10 Maths:

The chapter Polynomials opens with a definition of the degree of a polynomial, a linear polynomial, a quadratic polynomial, and a cubic polynomial. This chapter has four exercises, one of which is optional. Exercise 2.1 covers questions about determining the number of zeroes in a graph. It necessitates a grasp of the Polynomial's Zeroes Geometrical Meaning. Exercise 2.2 is based on the Zeroes and Coefficients of a Polynomial's relationship. In some of the problems, students must discover the zeros of a quadratic polynomial and the quadratic polynomial. The idea of the division algorithm is described in Exercise 2.3, and students will find questions connected to it. The optional Exercise, 2.4, contains questions from all of Chapter 2's topics.

Polynomials

Polynomials are sums of k⋅xⁿ terms, where k can be any number and n can be any positive integer. 3x+2x-5, for example, is a polynomial. A polynomial is an expression in mathematics that consists of indeterminates (also known as variables) and coefficients and includes only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables. x 2 -4x + 7 is an example of a polynomial with a single indeterminate x. In three variables, consider x 3 + 2xyz + 2yz + 1. Polynomials may be found in many fields of mathematics and science.

Zeros of Polynomial

A polynomial's zeros are all the x-values that make the polynomial equal zero. They are fascinating to us for various reasons, one of which is that they reveal the x-intercepts of the polynomial graph. These are closely related to the polynomial factors.

Topics that are covered in Chapter 2: Polynomials for Class 10 Maths

The topics of the relationship between zeroes, coefficients of quadratic polynomials, and zeroes of a polynomial are covered in Chapter 2, i.e., Polynomials for Class 10 Maths.

Steps to Take

We begin by arranging the dividend and divisor words in decreasing order of their degrees. Remember that putting the terms in this sequence is referred to as "writing polynomials in standard form."

Step 1:  To get the first term of the quotient, divide the dividend's highest degree term by the divisor's highest degree term. Then proceed with the dividing procedure.

Step 2:  To get the second term of the quotient, divide the highest degree term of the new dividend by the divisor's highest degree term. Carry out the dividing procedure once again.

Step 3:  The remainder degree is now smaller than the divisor's degree. As a result, we cannot continue to divide.

We can see that Dividend = Divisor × Quotient + Remainder once more. We're using an approach similar to Euclid's division algorithm, which you learned about in Chapter 1.

There are two polynomials p(x) and g(x) with g(x) ≠ 0, we may evaluate polynomials h(x) and m(x) such that p(x) = g(x) × h(x) + m(x), where m(x) = 0 or degree of m(x) < degree of g (x).

This is known as the Polynomial Division Algorithm .

NCERT Solutions for Class 10 Maths Chapter 2 Exercises:

Our CBSE Class 10 books also has the following resources for NCERT Class 10 Chapter 2 Polynomials :

• Notes for NCERT Polynomials Class 10
• Important Polynomials Class 10 Questions
• Class 10 NCERT Exemplar Solutions for Chapter Polynomials

Class 10 Maths NCERT Solutions of Chapter 3: Pair of Linear Equations in Two Variables

The concept or idea of a "Pair of Linear Equations in Two Variables" is explained in this chapter. This chapter has seven problems, each of which describes a different approach to solving the pair of linear equations. Exercise 3.1 describes how to algebraically and visually depict a scenario. Exercise 3.2 illustrates how to solve the pair of linear equations using the graphical method. The Algebraic, Elimination, Cross-Multiplication Method, and Substitution Methods are described in Exercises 3.3, 3.4, 3.5, and 3.6. Exercise 3.7 is an optional exercise with a variety of questions. Students must practice these activities to grasp the process of solving linear equations.

Linear equation in two variables

It is defined as an equation which is in the form dx+ny+l, where d, n, and l are absolute values, and d and n are not equal to zero. It is known as the linear equation in two variables.

Topics that are covered in Chapter 3: Pair of Linear Equations in two variables for Class 10 Maths:

Several solutions have algebraic conditions: a pair of two-variable linear equations, their graphical solutions, and their consistency and inconsistency. A pair of linear equations' algebraic solutions in two variables using substitution and elimination. Simple situational and equation problems can be reduced to linear equations. 

Formulas of Importance

The general form of the pair of linear equations in two variables, i.e., x and y are f 1 x + g 1 y + h 1 = 0 and f 2 x + g 2 y + h 2 = 0, where f 1 , g 1 , h 1 , f 2 , g 2 , h 2 are all real integers and f 12 + g 12 ≠ 0, f 22 + g 22 ≠ 0

Class 10 Maths Chapter 3 NCERT Solutions Exercises

Our CBSE Class 10 books also has the following resources for NCERT Class 10 Chapter 3 Pair of Linear Equations in Two Variables:

• Notes for NCERT Class 10 Pair of Linear Equations in Two Variables
• Class 10 Maths Important Questions, Pair of Linear Equations in Two Variables
• Class 10 NCERT Exemplar Solutions Pair of Linear Equations in Two Variables

NCERT Solutions for Grade 10 Maths Chapter 4 Quadratic Equations:  

Students will learn the traditional manner of formulating a Quadratic Equation in this chapter. The chapter then explains how to solve a quadratic problem using the factorization approach and complete the square method. The chapter concludes with determining the nature of roots, which asserts that a quadratic equation ax² + bx + c = 0 has two separate real roots if b² – 4ac > 0 and two equal roots if b² – 4ac = 0.

If b² – 4ac < 0, there are no genuine roots.

Chapter 4 Quadratic Equations of Class 10 Maths:

The standard form of a quadratic equation is ax 2 + bx + c = 0 (a ≠ 0). Factorization and the quadratic formula solve quadratic equations (only real roots). The relationship between the discriminant and the nature of the roots. Situational questions based on quadratic equations that are relevant to everyday activities (problems on equations reducible to quadratic equations are excluded)

As a result, the roots of ax 2 + bx + c = 0 are both -b/2a.

As a result, in this example, the quadratic equation ax 2 + bx + c = 0 has two equal real roots.

There is no real integer whose square is b 2 – 4ac if b 2 – 4ac = 0. As a result, there are no true roots for the given quadratic equation in this example. 

Because b 2 – 4ac decides whether or not the quadratic equation ax 2 + bx + c = 0 has real roots, it is the discriminant of this quadratic equation.

As a result, the quadratic equation ax 2 + bx + c = 0 has

• If b 2 – 4ac > 0, there are two different real roots.
• If b2 – 4ac = 0, there are two equal real roots.

NCERT Solutions for Grade 10 Maths Chapter 4 Exercises

Also available in Our CBSE Class 10 books with the following materials for NCERT Class 10 Chapter 4 Quadratic Equations:

• Notes on Quadratic Equations for Class 10 NCERT
• Important Quadratic Equations Class 10 Questions
• Class 10 Quadratic Equations NCERT Exemplar Solutions

Chapter 5 Arithmetic Progressions NCERT Solutions for Class 10 Maths

This chapter introduces students to Arithmetic Progression, abbreviated as AP. There are four exercises in this chapter. Students will discover questions in Exercise 5.1 about describing a scenario in the form of AP, evaluating the first term and difference of an AP, and determining if a series is AP or not. Exercise 5.2 offers questions about calculating the nth term of an AP using the following formula:

a n = a + (n-1) d

The next Exercise, 5.3, offers questions on calculating the sum of the first n terms of an AP.

The last Exercise contains AP-based higher-level problems to help students improve their analytical and problem-solving abilities.

Class 10 Maths Chapter 5 Arithmetic Progressions:

Why do you want to learn about Arithmetic Progression? The nth term and the sum of the first n terms of A.P. are deduced and used to solve difficulties in everyday life.

(Applications based on an A.P.'s sum to n words are not permitted.)

If the terms of AP are a 1 , a 2 , a 3 , a 4 , a 5 , a 6 ... and d is the common difference between each term, then the sequence is an a+d, a+2d, a+3d, a+4d, a+5d,...., n th term..., where an is the first term. The nth term for mathematical progression is now;

a + (n-1) d = n th term

Arithmetic Progression's first n terms' sum;

S n = n/2 [2a + (n-1) d]

NCERT Solutions for Class 10 Maths Chapter 5 Exercises:

Our CBSE Class 10 books also has the following materials for NCERT Class 10 Chapter 5 Arithmetic Progressions:

• Class 10 NCERT Arithmetic Progressions Notes
• Important Arithmetic Progressions Class 10 Questions
• Class 10 NCERT Exemplar Solutions Progressions in Arithmetic

Chapter 6 Triangles : NCERT Solutions for Class 10 Maths

Students in 10th Grade CBSE Maths will study figures with the same shape but not necessarily the same size in Chapter 6. Triangles begin with the notion of a comparable and harmonious form. It goes on to describe the criteria for the similarity of two triangles as well as theorems about triangle similarity. Following that, the areas of identical triangles were defined using a theorem. The Pythagoras Theorem and its converse are discussed at the conclusion of this chapter.

Class 10 Maths Chapter 6 Triangles: Definitions, examples, and counter examples of related triangles

• (Demonstrate) If a line is drawn parallel to one side of a triangle and intersects the other two sides at separate locations, the other two sides are split in the same ratio.
• (Motivate) A line is parallel to the third side if it splits two sides of a triangle in the same ratio.
• (Motivate) If the corresponding angles in two triangles are identical, their corresponding sides are proportionate, and the triangles are comparable.
• (Motivate) If two triangles' corresponding sides are proportionate, their complementary angles are equal, and the two triangles are identical.
• (Motivate) If a triangle's one angle equals one angle of another triangle and the sides that include these angles are proportionate, the two triangles are comparable.
• (Inspire) When a perpendicular is drawn from the vertex of a right triangle's right angle to the hypotenuse, thus the triangles which are on either side of the perpendicular are identical to the complete triangle and each other.
• (Inspire) The ratio of the two identical triangle's areas is equal to the ratio of their corresponding side squares.
• (Prove) The square on the hypotenuse of a right triangle equals the sum of the squares on the other two sides.
• (Motivate) If the square on one side of a triangle equals the sum of the squares on the other two sides, the right angle is the angle opposite the first side.

Theorems of Importance

Theorem 6.1:  At one side of a triangle, a parallel line is drawn, which intersects the other two sides in distinct spots; the remaining two sides are split in the same ratio.

Theorem 6.2:  A line is parallel to the third side if it divides any of the two sides of a triangle in the same ratio.

Theorem 6.3:  If the corresponding angles in two triangles are identical, their corresponding sides are in the same ratio (or percentage), and the two triangles are comparable. 

Theorem 6.4:  If the sides of one triangle in two triangles are proportional to (i.e., in the same ratio as) the sides of the other triangle, then their corresponding angles are equal, and thus the two triangles are comparable.

Theorem 6.5:  If one angle of one triangle equals one angle of another triangle, and the sides containing these angles are proportionate, the two triangles are identical.

Theorem 6.6:  The ratio of the two comparable triangle areas is equal to the square of the ratio of their corresponding sides, according to Theorem 6.6.

Theorem 6.7 states that if a perpendicular is drawn from the vertex of a right triangle's right angle to the hypotenuse, triangles on both sides of the perpendicular are comparable to the complete triangle.

Theorem 6.8:  The square of the hypotenuse of a right triangle is equal to the sum of the other two sides square.

Theorem 6.9:  When the square of one side of a triangle equals the summation of the squares of the other two sides, the angle opposite the first side is right.

NCERT Solutions for Class 10 Maths Chapter 6 Exercises

Our CBSE Class 10 books also has the following materials for NCERT Class 10 Chapter 6 Triangles:

• Triangles Class 10 NCERT Notes
• Important Triangles Class 10 Questions
• Class 10 NCERT Exemplar Solutions Triangles

Chapter 7 Coordinate Geometry: NCERT Solutions for Class 10 Maths

Students will learn how to calculate or evaluate the distance between two locations whose coordinates are supplied and the area of a triangle formed by three specified points in this chapter. Students will also learn how to identify the coordinates of the point that splits a line segment connecting two specified points in a specific ratio. Students will study the Distance Formula, Section Formula, and Area of a Triangle in this chapter of Coordinate Geometry. 

Coordinate Geometry

The geometrical study by using coordinate points is known as coordinate geometry (or analytic geometry). It is possible to estimate the distance between two points, divide lines in an m:n ratio, identify the midpoint of a line, calculate the area of a triangle in the Cartesian plane, and so on using coordinate geometry.

Class 10 Maths Chapter 7 Coordinate Geometry:

Lines (In two dimensions)

Concepts of coordinate geometry and graphs of linear equations are reviewed.

Distance calculation formula

The formula for Sections (internal division)

Important formula

Distance Formula

NCERT Solutions for Class 10 Maths Chapter 7 Exercises

Our CBSE Class 10 Books also has the following materials for NCERT Class 10 Chapter 7 Coordinate Geometry:

• Notes for NCERT Coordinate Geometry Class 10
• Important Coordinate Geometry Class 10 Questions
• Class 10 NCERT Exemplar Solutions Coordinate Geometry Solutions

Chapter 8 Introduction to Trigonometry: NCERT Solutions for Class 10 Maths

Trigonometry will be introduced to pupils in this chapter. They will investigate trigonometric ratios of angles, ratios of a right triangle to its sharp angles. The chapter also defines trigonometric ratios for angles ranging from 00 to 900. Students will also understand how to compute trigonometric ratios for specific angles and create trigonometric identities employing these ratios.

Chapter 8 of Class 10 Maths: Trigonometry Overview

Trigonometric ratios of a right-angled triangle's acute angle Evidence of their existence (well defined). Trigonometric ratio values of 300, 450, and 600. Relationships among the ratios

IDENTITIES TRIGONOMETRIC

Application and proof of the identity sin 2 A + cos 2 A = 1. Only simple identities will be provided.

Important formulas

Trigonometry Maths Formulas for Class 10 address three key functions for a right-angle triangle: Sine, Cosine, and Tangent. Assume that a right-angled triangle ABC is right-angled at point B and has ∠θ.             

Trigonometry Table

Trigonometric Ratios of Complementary Angles

• sin (90° – A) = cos A
• cos (90° – A) = sin A,
• tan (90° – A) = cot A,
• cot (90° – A) = tan A,
• sec (90° – A) = cosec A,
• cosec (90° – A) = sec A
• sin 2 A + cos 2  A = 1,
• sec 2 A – tan 2  A = 1 for 0° ≤ A < 90°,
• cosec 2 A = 1 + cot 2  A for 0° < A ≤ 90°

NCERT Solutions pdf for Class 10 Maths Chapter 8 Exercises

Our CBSE Class 10 books also has the following materials for NCERT Class 10 Chapter 8 Trigonometry:

• Notes for NCERT Class 10 Introduction to Trigonometry
• Important Trigonometry Questions from Class 10
• Class 10 NCERT Exemplar Introduction to Trigonometry Solutions 

NCERT Solutions for Class 10 Maths Chapter 9 Trigonometry Applications

This chapter builds on the previous one by teaching students about trigonometry applications. It is used in geography, navigation, map production, and calculating an island's location in reference to longitudes and latitudes. Students will learn how to utilize trigonometry to find the heights and distances of various objects without having to measure them. They will learn about line of sight, angle of elevation, and angle of depression.

Chapter 9 of Class 10 Maths covers the following topics: Some Trigonometry Applications

HEIGHTS AND DISTANCES-Angle of Elevation and Angle of Depression

Simple height and distance issues. A problem should have no more than two right triangles. Elevation/depression angles should not exceed 30 degrees, 45 degrees, and 60 degrees.

Important Notes –

• The line of sight is the line drawn from an observer's eye to a point in the thing being observed.
• The angle created by the line of sight with the horizontal when the item being viewed is above the horizontal level, like when we elevate our head to look at it, is named the angle of elevation of the point observed.
• The angle of depression of a point on an item being observed is the angle created by the line of sight with the horizontal when the point is below the horizontal level, like when we lower our heads to look at the point being viewed.

You would need to be aware of the following: 

• The distance DE between the student's feet and the foot of the minar
• The angle of elevation, BAC, of the minar's top
• The student's AE height.

Assuming the three requirements mentioned above are met, how can we calculate the minar's height? 

CD = CB + BD in the diagram. In this case, BD = AE, which is the student's height.

To calculate BC, we shall utilize trigonometric ratios of BAC or A.

The side BC in ABC is the opposite side in regard to the known A because these ratios include AB and BC, our search narrows to tan A or cot A.

As a result, tan A = BC/AB or cot A = AB/BC, which, when solved, gives us BC.

The height of the minar may be calculated by adding AE to BC.

NCERT Solutions pdf for Class 10 Maths Chapter 9 Exercises

With our CBSE Class 10 books, you may also find the following materials for NCERT Class 10 Chapter 9 Some Applications of Trigonometry:

• Notes for NCERT Some Applications of Trigonometry Class 10
• Some Trigonometry Applications Class 10 Important Questions
• Class 10 NCERT Exemplar Solutions Some Trigonometry Applications

Chapter 10 Circles –NCERT Solutions for Class 10 Maths  

Students have previously learned about a circle and different circle-related terminology such as a chord, segment, arc, etc. Students will investigate the many circumstances when a circle and a line are presented in a plane in this chapter. As a result, they will get thoroughly acquainted with the concepts of Tangent to a Circle and Number of Tangents from a Point on a Circle.

Class 10 Math Topics Chapter 10 Circles:  

Point of contact tangent to a circle

• Establish that the Tangent at any point on a circle is perpendicular to the radius via the point of contact.
• (Prove) Tangents drawn from an exterior point to a circle have identical lengths.

Theorems of Importance –

• The Tangent at any point on a circle is perpendicular to the radius through the point of contact, according to Theorem 10.1.
• The lengths of tangents taken from an exterior point to a circle are identical, according to Theorem 10.2.

Number of Tangents from a Circle Point

Case 1:  There is no tangent to a circle that passes through a point located within the circle.

Case 2:  There is only one Tangent to a circle that passes through a point on the circle.

Case 3:  There are two tangents to a circle that intersects at a place outside the circle.

NCERT Solutions pdf for Class 10 Maths Chapter 10 Exercises

Our CBSE Class 10 books also has the following materials for NCERT Class 10 Chapter 10 Circles:

• Class 10 Important Questions for Circles
• Notes of NCERT Class 10 for Circles
• Circles NCERT Exemplar Solutions Class 10

Chapter 11 Constructions: NCERT Solutions Class 10 Maths

There are two exercises in this chapter. Whatever learners learned about building in previous sessions will also be helpful. Students will learn how to split a line segment in Exercise 11.1 and create tangents to a circle in Exercise 11.2. Methods and processes for building are discussed, and some examples are provided to help students understand.

Class 10 Math Topics Chapter 11 Constructions:

• Line segment division in a particular ratio (internally).
• Tangents to a circle from points outside of it.

Important Notes – 

11.1: To split a line segment in a specific ratio.

11.2: Build a triangle identical to a given triangle using the specified scale factor.

11.3: To build the tangents to a circle from a location outside it.

NCERT Solutions pdf for Class 10 Maths Chapter 11 Exercises

Our CBSE Class 10 books also has the following materials for NCERT Class 10 Chapter 11 Constructions:

• Notes for NCERT Constructions Class 10
• Important Constructions Class 10 Questions
• NCERT Exemplar Solutions Constructions Class 10

NCERT Solutions for Class 10 Maths Chapter 12 Circle Areas

This chapter begins with the notions of circle perimeter and area. The chapter goes on to describe how to find the area of a sector and segment of a circular region using this notion. Furthermore, students will gain confidence in calculating the areas of various plane figure combinations involving circles or their portions.

Topics Covered in Class 10 Maths Chapter 12 Circle:  

Motivate the area of a circle, the area of a circle's sections and segments. As mentioned earlier, area and perimeter/circumference problems are based on the plane. (When computing the area of a circle segment, issues should be limited to central angles of 60° and 90° only. Plane figures with triangles, basic quadrilaterals, and circles should be used.)

Circumference of the circle: A circle's perimeter is the linear distance travelled around the circle. The perimeter of a circle is known by a unique name, Circumference. The circumference multiplied by the diameter is given by the formula 2πr.

Area of circle: The circle area is πr 2 , where r is the circle's radius and π = 22/7 or 3.14 (maybe used interchangeably for problem-solving purposes). Thus, an area of the circle is nothing but is the ratio of a circle's circumference to its diameter.  

A Circle's Segment

A circular segment is a section of a circle "separated" from the portion of the circle by a secant or chord.  

A circle's sector

A circle sector is the portion of a circle bounded by an arc and two radii. The smaller region is referred to as the minor sector, while the larger area is the major sector.

A Sector's Angle

The angle of a sector is defined as the angle formed by the sector's two radii.

The Angle of a Sector

A sector's angle is defined as the angle created by the sector's two radii.

                                 L= (θ/360°) × 2πr

Where θ is the sector angle and r is the circle's radius.

Students may also utilize this NCERT Exemplar to complete even the most challenging problems and prepare for examinations. However, sample papers and previous year's question papers are quite helpful in understanding the sorts of questions a paper setter asked in the examination from this chapter 12 and the marking system. 

This chapter covers the Areas related to the Circles subjects listed below:  

• Areas related to Circles
• Segment and Sector of a circle
• Areas of a combination of a plane figure.

Students will learn to calculate various circle areas and combinations of plane figure areas, in Chapter 12.  

Students in the tenth grade are also given online study tools, including notes, example books, NCERT answers, and question papers, to help them prepare for the board examinations.

This chapter may take some time to complete, and students must pay close attention. The NCERT Exemplar for Class 10 Maths is provided here to assist students in solving chapter concerns and obtaining proper answers to complex questions.

Important Formulas –

• Circumference = 2πr
• Area of the circle = πr 2
• Area of the sector of angle θ = (θ/360) × π r 2
• Length of an arc of a sector of angle θ = (θ/360) × 2 π r, where r is the radius of the circle

NCERT Solutions pdf for Class 10 Maths Chapter 12 Exercises

Exercise 12.1 includes five solved questions. Four of them are short answer type questions , and the remaining one is the MCQ   type.

This exercise is primarily concerned with the perimeter and area of a circle. In this lesson, students learn how to determine the circumference and area using formulae and apply what they've learned to real-world problems. Solutions give an overview of the chapter's essential themes and assist students in becoming familiar with these concepts.

In this exercise 12.1 of Chapter 12 , students are given the solutions to many questions to perform well in their first term examinations.

The solutions to the  NCERT -based questions of  Chapter 12 for Class 10 Maths, Areas Related to Circles , is an essential study resource for Class 10 students. These assist students in comprehending or solving the sorts of problems that a paper setter asked in the first term examinations of  CBSE   Class 10 Maths . Furthermore, offering solutions to all areas relating to circles assists students in effectively preparing for the board examinations.

The subject matter experts make sure that the errors in the  NCERT  textbooks are resolved according to the  NCERT  rules and syllabus.

The solutions to the Exercise 12.1   NCERT -based questions of  Chapter 12 for Grade 10 Maths  are accessible at our Oswaal books. These solutions will assist students in acing their first term examinations by preparing for the examinations ahead of time.

Students may clarify all of their conceptual questions or doubts with the assistance of the notes of Maths NCERT Solutions for Class 10 of this chapter which we provide. Experts have created these NCERT Solutions based on the most recent update to the  CBSE  syllabus for 2024-25 to help students in Class 10 prepare well for their term-wise examinations

Our CBSE Class 10 books also has the following resources for NCERT Class 10 Chapter 12 Areas Related to Circles:

• Notes on Circles in Class 10 NCERT
• Notes on Circles for Class 10 NCERT Important questions
• NCERT Exemplar Solutions Class 10 Areas Related to Circles

Chapter 13 Surface Areas and Volumes - NCERT Solutions for Class 10 Maths

There are five exercises in Chapter 13. The first Exercise consists of questions on calculating the surface area of an object made by merging any two fundamental solids, namely a cuboid, cone, cylinder, sphere, and hemisphere. 13.2 questions in Exercise are focused on determining the volume of objects generated by joining any two of a cuboid, cone, cylinder, sphere, or hemispheres. Exercise 13.3 has questions on converting a solid from one shape to another. Exercise 13.4 involves calculating the volume, curved surface area, and total surface area of a cone's frustum.

Chapter 13 Surface Areas and Volumes in Class 10 Maths:

• Surface areas and volumes of any two of the following combinations: cubes, cuboids, spheres, hemispheres, and right circular cylinders/cones.
• Difficulties involving the transformation of one type of metallic solid into another and other mixed problems (Problems involving the combination of no more than two different solids are considered.)
• Total SA of new solid = CSA of one hemisphere + CSA of cylinder + CSA of another hemisphere
• Diameter of sphere = 2r
• The surface area of the sphere = 4 π r 2
• The volume of Sphere = 4/3 π r 3
• The curved surface area of the Cylinder = 2 πrh
• Area of two circular bases = 2 πr 2
• The total surface area of Cylinder = Circumference of Cylinder + Curved surface area of Cylinder = 2 πrh + 2 πr2
• Volume of Cylinder = π r 2 h
• Slant height of cone = l = √ (r2 + h2)
• Curved surface area of cone = πrl
• Total surface area of cone = πr (l + r)
• Volume of cone = ⅓ π r 2 h
• Perimeter of cuboid = 4(l + b +h)
• Length of the longest diagonal of a cuboid = √ (l 2 + b 2 + h 2 )
• Total surface area of cuboid = 2(l×b + b×h + l×h)
• Volume of Cuboid = l × b × h

NCERT Solutions pdf for Class 10 Maths Chapter 13 Exercises

Our CBSE Class 10 Books also has the following resources for NCERT Class 10 Chapter 13 Surface Areas and Volumes:

• Notes for NCERT Class 10 Surface Areas and Volumes
• Class 10 Important Questions on Surface Areas and Volumes
• Surface Areas and Volumes in NCERT Exemplar Solutions Class 10

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Students will learn how to convert ungrouped data to grouped data and how to calculate the Mean, Mode, and Median. In addition, the notion of cumulative frequency, cumulative frequency distribution, and how to design cumulative frequency curves will be discussed.

Class 10 Maths Chapter 14 Statistics:

Grouped data mean, median, and mode (bimodal situation to be avoided). Only the Direct Method and the Assumed Mean Method are used to calculate the mean.

The mean of the grouped data  can be found by 3 methods

 1. Direct Method:

4. The mode of grouped data: 

5. The median for a grouped data:

NCERT Solutions PDF for Class 10 Maths Chapter 14 Exercises

Our CBSE Class 10 books also has the following materials for NCERT Class 10 Chapter 14 Statistics:

• Notes for NCERT Statistics Class 10
• Important Statistics Class 10 Questions
• Statistics NCERT Exemplar Solutions Class 10

Class 10 Maths NCERT Solutions Chapter 15 Probability  

The final chapter is about Probability. The chapter begins with a theoretical look at probability. Following that, the chapter discusses the distinction between experimental and theoretical probability. There are several instances provided to illustrate it effectively. So, before proceeding with the practice questions, students must first complete the CBSE Maths examples. 

Chapter 15 Probability NCERT Class 10 Maths:  The classical definition of probability. Simple issues involve determining the likelihood of an event.

• The theoretical probability (also called classical probability) of an event E, written as P(E), is defined as
• The probability of an actual occurrence (or sure event) is 1.
• The probability of an impossible event is nil.
• The probability of an occurrence E is expressed as a number P(E) such that 0 ≤ P (E) ≤ 1
• An important event is one that has only one outcome. The total of all the probability of an experiment's elementary occurrences is 1.

NCERT Solutions pdf for Class 10 Maths Chapter 15 Exercises  

Our CBSE Class 10 books also has the following resources for NCERT Class 10 Chapter 15 Probability:

• Notes for NCERT Probability Class 10
• Important Probability Class 10 Questions
• Class 10 Probability NCERT Exemplar Solutions

NCERT Solutions for Class 10 Maths PDF in English Medium and Hindi Medium (for the academic year 2024-25 are used not only by CBSE but also by UP Board, Uttarakhand Board, and all other boards that use NCERT Textbooks.

Advantages of NCERT Solutions Class 10 Math:

The Class 10 NCERT Math Solutions in PDF format provided here offer various advantages, including:

• The answers provided here are simple to grasp.
• For ease of comprehension, solutions are presented in phases.
• Diagrams are provided to assist students in visualizing solutions.
• Each chapter's questions are answered.

Students are urged to prepare all of the chapters covered in the solution modules, as this will help them obtain a deeper understanding of the subjects. Students must comprehend all of the processes outlined in the solutions to make good preparations.

How Can CBSE Class 10 Maths Solutions from NCERT Help with Term Exams?

Math in CBSE Class 10 is an essential topic for students. We have offered full preparatory support to students here. Class 10 is the initial criterion for each student that will be reflected in subsequent accomplishment records. CBSE consistently recommends NCERT books for term-by-term test preparation.

Exam preparation is a time-consuming procedure that necessitates a thorough comprehension of individual chapters. This technique necessitates diligent study and a systematic approach to completing the solutions. NCERT 10 Class Maths Solutions play an essential part in preparing students for competitive admission tests. NCERT books are well-known for presenting topics in an easy-to-understand format. NCERT Class 10 Mathematics Books are written most understandably and straightforwardly, allowing complicated problems to be broken down effectively.

Continue to visit to obtain complete chapter-by-chapter NCERT Solutions for Class 10 Maths PDF free download for all courses. NCERT Solutions for Class 10 Science may also be found at class 10 . Students at class 10 benefit from a customized learning experience that helps them prepare for tests more successfully.

NCERT  Solutions assist students in solidifying their knowledge of the mathematics concept.

Diagrams are used to answer questions, making learning more dynamic and thorough.

The language used in  NCERT  Solutions is simple and easy to grasp.

Students benefit from a step-by-step approach to problem-solving.

Allows learners to work through challenging problems in their own time.

NCERT  Solutions for various courses and disciplines are also available for students to use. Well-experienced professors of class 10 books create these answers to offer clarity on essential ideas and problem-solving abilities.

Students can also obtain a firm grasp of critical topics by referring to additional study resources available at  CBSE class 10 books .

The NCERT Solutions for Class 10 Maths are essential in shaping a kid's future since the grades that student achieves on the board test ultimately form the student's potential. The examinations for Classes 10 CBSE board are the first public examinations that all students must attempt. The CBSE board exams assess the candidate's practical and theoretical knowledge. As a result, students must comprehend the principles, and connecting them to real-world applications will help them develop solid practical knowledge.

Our subject specialists develop the Class 10 NCERT Solutions to give a one-stop solution for all Maths queries. These Solutions for Class 10 NCERT provide an ideal platform for understanding all of the Maths ideas in the syllabus for students hoping to achieve in the CBSE Classes 10 Board Exam. The NCERT Solutions of solved examples and the exercise questions are explained with illustrated graphs and well-labelled diagrams that make learning effortless. It helps students create a strong conceptual point, which is crucial in the later phases of competing in competitive tests. The solutions to the  NCERT -based questions  for class 10 are easily accessible.

Solutions for Class 10 Maths NCERT include all the exercise problems from the NCERT Maths textbook. Our highly trained subject specialists have developed these Math answers to assist students in their test preparation. These answers will guide students in acquiring a higher grasp of the ideas taught in the Class 10 Maths curriculum. Students may assess their preparation and knowledge of statements by practicing NCERT Class 10 Maths textbook problems.

These assist students in comprehending or solving the sorts of problems that a paper setter asks in the first term examinations of  CBSE   Class 10 Maths . The subject matter experts make sure that the errors in the  NCERT  textbooks are resolved according to the  NCERT  rules and syllabus. The solutions to the  NCERT -based questions  for Grade 10 Maths  are accessible through class 10 books .  These solutions will assist students in acing their board examinations by preparing them ahead of time. Students may clarify all of their conceptual questions or doubts with the assistance of the notes of Maths  NCERT  Solutions for Class 10 of different chapters. Experts have created these NCERT Solutions based on the most recent update to the  CBSE  Syllabus to help students in Class 10 prepare well for their board examinations.

Students should complete Class 10 Maths NCERT Solutions after finishing the curriculum since it gives many problems from each unit to practice. Use these solutions to clarify any confusion, if anyone gets stuck when answering the issues, and supplementary references and study resources. All Math solutions are written unambiguously and prominently, considering the goal of textbooks. Students are urged to download and practice these solutions regularly to get a commendable rank in the Board exams. Students must comprehend all the processes outlined in the answers for successful preparation.

If you wish to strengthen your fundamental knowledge in several disciplines, the NCERT book is a fantastic place to start. It includes several diagrams, flowcharts, tables, and illustrations to supplement the various themes and concepts. It is a fantastic beginning point for people who wish to study a certain subject because it is designed to improve the fundamental understanding of pupils who read it. We think that after reading this, you will be grateful that we got your back, but this is not only for a subject.

The CBSE Class 10 math curriculum seeks to create learner knowledge and evaluation of basic concepts such as algebra, geometry, surface areas, volumes, statistics, and probability. The key to success in mathematics is practice. It is difficult to score well in this topic only by reading and memorizing the formulae; consequently, students are urged to practice them on a regular basis in order to obtain a good place in the board examinations. We provide class 10 Revision Notes for Mathematics so that students may quickly prepare all of the concepts included in the class 10 maths syllabus in a much better and more effective manner.

Maths Solutions for Class 10 include solutions to all of the questions and problems in the textbook. Real Numbers, Quadratic Equations, Triangles, Introduction to Trigonometry, Some Applications of Trigonometry, Area Related to Circles, Circles, Surface Areas and Volumes, and other topics are covered in Maths Solutions for Class 10. A team of subject specialists has developed Maths Solutions for Class 10, and the format follows the CBSE guidelines. It outlines each solution step by step so that students can comprehend it quickly, and marks are assigned to each step so that students may learn the answer writing skill from an exam standpoint.

We have supplied NCERT Class 10th Solutions to assist students in understanding the content, completing their assignments, and preparing for exams. With these answers, students will significantly improve their knowledge and problem-solving abilities. These solutions were developed under the supervision of professional educators and mentors while keeping the most recent NCERT curriculum in mind to provide you with up-to-date material. The questions and solutions are explained step-by-step in depth so that you can better comprehend them. Each subject presents solutions with brief details to give you the necessary context.

Frequently Asked Questions about NCERT Class 10 Maths Solutions

How can one understand the key ideas taught in the NCERT Solutions for Class 10 Maths?

Ans . Students who want to do well in their Class 10 examinations should get the NCERT Solutions for class 10 maths books . A group of teachers with extensive knowledge in the particular area curates the solutions with great care. Every minute topic is discussed engagingly to make studying easy for the pupils. The step-by-step answers are created using the marks weightage provided by the CBSE in mind.

Will the NCERT Solutions for Class 10 Maths make it easier for me to tackle the problems?

Ans. The Class 10 test is a watershed moment in the lives of learners. Mathematics is a topic that mainly consists of numerical, and comprehending ideas helps to obtain higher grades. Students should first comprehend the academic year's syllabus and then master the topics based on it for better performance. Students' logical thinking and analytical abilities will develop as they solve textbook problems using the answers PDF.

How can I obtain the NCERT Solutions for Class 10 Maths PDF?

Ans. Class 10 books  provide students with NCERT Solutions for Class 10 Maths PDF. Subject matter experts prepare the solutions with the students' learning capacity in mind. The primary goal of developing solutions is to assist students in learning complicated ideas without trouble. Students can also cross-check their answers while working through textbook problems to obtain a sense of other approaches to resolving topics quickly.

Are the NCERT Solutions for 10th Grade Maths beneficial to CBSE board students?

Ans. Some of the specific applications of the NCERT Maths Solutions for Class 10 -

• It establishes solid basics of essential topics and boosts confidence for term-based tests.
• The strategy for addressing complicated issues is simple to learn.
• It is an ideal study tool for students who want to complete their assignments on time and get higher grades in their term examinations.

How can I acquire a sense of the key concepts in the NCERT Solutions for Class 10 Maths?

Ans. Students should download and read the CBSE curriculum before the start of each academic year to have a thorough grasp of the subjects that will be assessed. Students must tackle NCERT Textbook issues by selecting the appropriate study resource that meets their needs. Students' problem-solving abilities improve when they practice the problems on a regular basis using the NCERT Solutions for Class 10 Maths.

Is it necessary to use the NCERT Solutions for Class 10 Maths?

Ans . NCERT Solutions for Class 10 Maths are designed in such a way that students may quickly comprehend the concepts and solve their difficulties. The answers include detailed explanations to help students understand the subject. Students can get PDF solutions and utilize them to strengthen their conceptual understanding.

Are the NCERT Solutions for Class 10 Maths the finest resource for CBSE students?

Ans . Students should choose the best study material for their needs from the many options available on the market. Class 10 books NCERT Solutions for Class 10 Maths are widely regarded as the finest reference tool for CBSE students. The solutions are given in chapter and practice forms, which students may download and utilize. Students can view the PDF links without regard to time limits.

Why do students require NCERT Solutions for Class 10 study material to qualify for the board exam?

Ans. Class 10 is an important time in a student's life since the grades they receive determine their future profession. It might be challenging to achieve good results in board examinations without adequate study materials. NCERT Solutions for Class 10 in a subject-by-subject manner. Students who are having problems answering textbook questions can use these solutions to ace the board test easily.

Explain the significance of class 10 books NCERT Solutions for Class 10.

Ans . The significance of class 10 books NCERT Solutions for Class 10 is as follows:

• NCERT Solutions are created in accordance with the CBSE syllabus.
• All of the textbook questions are answered so that students have no problem answering them.
• The experienced team of members creates solutions with the goal of offering high-quality study materials to pupils in grades 10.

How can one obtain free NCERT Solutions for Class 10 PDFs from OSWAAL?

Ans . As class 10 is a critical phase in students’ life, students should choose the appropriate study material that meets their demands. Keeping this in mind, the team of subject matter experts curate solutions that are clear and easy to grasp. The solutions include step-by-step explanations to enable students of all IQ levels to understand the ideas. The PDFs of the answers may be downloaded by visiting the our website. By inputting the necessary information, students will have access to both online and offline study resources for a particular subject.

Will the board exam questions be drawn from NCERT Solutions for Class 10 Maths?

Ans. The NCERT Solutions for Class 10 Maths provide the most questions in the board exam. Students can refer to the solutions offered by the professors at class 10 while answering textbook problems to gain a better knowledge of the subject. The answers are designed so that students may easily comprehend complicated concepts. Students can utilize online and offline solutions with no time limits.

How many exercises are there in the Grade 10 th Maths Chapter 12?

Ans . There are three exercises are present in the 10 th Grade Maths Chapter 12, i.e., Areas Related to Circles.

How many problems are present in each NCERT Exemplar Solutions for Class 10 Maths Chapter 12 exercise?

Ans. Three problems are included in Chapter 12 of the NCERT Solutions for Class 10 Maths . Exercise 12.1 consists of five questions, whereas Exercise 12.2 consists of 14 questions. 12.3 comprises 16 questions. Every problem is addressed with the utmost attention to give students accurate solutions following CBSE norms.

Explain the circumference of a circle in Chapter 12 of NCERT Exemplar Solutions for Class 10 Maths?

Ans.  A circle's perimeter is the linear distance travelled around the circle. The perimeter of a circle is known by a unique name, Circumference. The circumference multiplied by the diameter is given by the formula 2πr.

Where can I get online NCERT Exemplar Solutions for Class 10 Maths Chapter 12?

Ans.  Yes, the NCERT Exemplar Solutions for Class 10 Maths Chapter 12 are available on our website . Subject specialists evaluate the solutions to all problems to give students the most significant reference material. You can download chapter-by-chapter or exercise-by-exercise answers to understand better the problem-solving approaches used while solving textbook questions

Which chapter in class 10 math is the most difficult?

Ans. The most challenging chapter is trigonometry.

In Class 10, how many hours should I study?

Ans. A student should study between 7 and 10 hours every day. It is probable that if you dedicate more time, you may become weary. As a result, the maximum amount of time you should commit to your studies is 7 – 10 hours.

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NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

Extra Questions for Class 10 Maths with Solutions Chapter Wise

March 10, 2020 by Veerendra

Extra Questions for Class 10 Maths with Solutions

Here is the list of Extra Questions for Class 10 Maths with Solutions and Answers Chapter wise based on latest NCERT syllabus (http://ncert.nic.in/) prescribed by CBSE. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

• Chapter 1 Real Numbers Class 10 Extra Questions
• Chapter 2 Polynomials Class 10 Extra Questions
• Chapter 3 Pair of Linear Equations in Two Variables Class 10 Extra Questions
• Chapter 4 Quadratic Equations Class 10 Extra Questions
• Chapter 5 Arithmetic Progressions Class 10 Extra Questions
• Chapter 6 Triangles Class 10 Extra Questions
• Chapter 7 Coordinate Geometry Class 10 Extra Questions
• Chapter 8 Introduction to Trigonometry Class 10 Extra Questions
• Chapter 9 Applications of Trigonometry Class 10 Extra Questions
• Chapter 10 Circles Class 10 Extra Questions
• Chapter 11 Constructions Class 10 Extra Questions
• Chapter 12 Areas Related to Circles Class 10 Extra Questions
• Chapter 13 Surface Areas and Volumes Class 10 Extra Questions
• Chapter 14 Statistics Class 10 Extra Questions
• Chapter 15 Probability Class 10 Extra Questions

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Chapter 4 Class 10 Quadratic Equations

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Updated for Latest NCERT for 2023-2024 Boards.

Get NCERT Solutions for all exercise questions and examples of Chapter 4 Class 10 Quadratic Equations free at Teachoo. Answers to each and every question is provided video solutions. 

In this chapter, we will learn

• What is a Quadratic Equation
• What is the Standard Form of a Quadratic Equation
• Solution of a Quadratic Equation by Factorisation ( Splitting the Middle Term method)
• Solving a Quadratic Equation by Completing the Square
• Solving a Quadratic Equation using D Formula (x = -b ± √b 2 - 4ac / 2a)
• Checking if roots are real, equal or no real roots (By Checking the value of D = b 2 - 4ac)

This chapter is divided into two parts - Serial Order Wise, Concept Wise

In Serial Order Wise, the chapter is divided into exercise questions and examples.

In Concept Wise, the chapter is divided into concepts. First the concepts are explained, and then the questions of the topic are solved - from easy to difficult.

We suggest you do the Chapter from Concept Wise - it is the Teachoo (टीचू) way of learning.

Note: When you click on a link, the first question of the exercise will open. To open other question of the exercise, go to bottom of the page. There is a list with arrows. It has all the questions with Important Questions also marked.

Serial order wise

Concept wise.

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