NCERT Solutions for Exercise 13.2 Class 12 Maths Chapter 13 - Probability

# NCERT Solutions for Exercise 13.2 Class 12 Maths Chapter 13 - Probability

Edited By Ramraj Saini | Updated on Dec 04, 2023 11:45 AM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.2

NCERT Solutions for Exercise 13.2 Class 12 Maths Chapter 13 Probability are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Most of the questions in NCERT solutions for exercise 13.2 Class 12 Maths chapter 13 deal with multiplication theorem on probability, independent events, and its property. This NCERT book exercise is very important from the board's exam point of view as one question is generally asked from this exercise. There are 7 examples given before this exercise which you should try to solve before solving the exercises questions. You can take help from exercise 13.2 Class 12 Maths solutions if you are stuck while solving these NCERT problems. These Class 12 Maths chapter 13 exercise 13.2 solutions are created by the subject matter experts who know how best to answer in the board exams.

12th class Maths exercise 13.2 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.

Apply to Aakash iACST Scholarship Test 2024

Also, see

##### Aakash iACST Scholarship Test 2024

Get up to 90% scholarship on NEET, JEE & Foundation courses

##### JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

## CBSE NCERT Solutions for Class 12 Maths Chapter 13 probability-Exercise: 13.2

$P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5},$

Given : $A$ and $B$ are independent events.

So we have, $P(A\cap B)=P(A).P(B)$

$\Rightarrow \, \, \, \, P(A\cap B)=\frac{3}{5}\times \frac{1}{5}$

$\Rightarrow \, \, \, \, P(A\cap B)=\frac{3}{25}$

Two cards are drawn at random and without replacement from a pack of 52 playing cards.

There are 26 black cards in a pack of 52.

Let $P(A)$ be the probability that first cards is black.

Then, we have

$P(A)= \frac{26}{52}=\frac{1}{2}$

Let $P(B)$ be the probability that second cards is black.

Then, we have

$P(B)= \frac{25}{51}$

The probability that both the cards are black $=P(A).P(B)$

$=\frac{1}{2}\times \frac{25}{51}$

$=\frac{25}{102}$

Total oranges = 15

Good oranges = 12

Let $P(A)$ be the probability that first orange is good.

The, we have

$P(A)= \frac{12}{15}=\frac{4}{5}$

Let $P(B)$ be the probability that second orange is good.

$P(B)=\frac{11}{14}$

Let $P(C)$ be the probability that third orange is good.

$P(C)=\frac{10}{13}$

The probability that a box will be approved for sale $=P(A).P(B).P(C)$

$=\frac{4}{5}.\frac{11}{14}.\frac{10}{13}$

$=\frac{44}{91}$

A fair coin and an unbiased die are tossed,then total outputs are:

$= \left \{ (H1),(H2),(H3),(H4),(H5),(H6),(T1),(T2),(T3),(T4),(T5),(T6) \right \}$

$=12$

A is the event ‘head appears on the coin’ .

Total outcomes of A are : $= \left \{ (H1),(H2),(H3),(H4),(H5),(H6) \right \}$

$P(A)=\frac{6}{12}=\frac{1}{2}$

B is the event ‘3 on the die’.

Total outcomes of B are : $= \left \{ (T3),(H3)\right \}$

$P(B)=\frac{2}{12}=\frac{1}{6}$

$\therefore A\cap B = (H3)$

$P (A\cap B) = \frac{1}{12}$

Also, $P (A\cap B) = P(A).P(B)$

$P (A\cap B) = \frac{1}{2}\times \frac{1}{6}=\frac{1}{12}$

Hence, A and B are independent events.

Total outcomes $=\left \{ 1,2,3,4,5,6 \right \}=6$.

$\inline A$ is the event, ‘the number is even,’

Outcomes of A $=\left \{ 2,4,6 \right \}$

$n(A)=3.$

$P(A)=\frac{3}{6}=\frac{1}{2}$

$\inline B$ is the event, ‘the number is red’.

Outcomes of B $=\left \{ 1,2,3 \right \}$

$n(B)=3.$

$P(B)=\frac{3}{6}=\frac{1}{2}$

$\therefore (A\cap B)=\left \{ 2 \right \}$

$n(A\cap B)=1$

$P(A\cap B)=\frac{1}{6}$

Also,

$P(A\cap B)=P(A).P(B)$

$P(A\cap B)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}\neq \frac{1}{6}$

Thus, both the events A and B are not independent.

Given :

$P(E)=\frac{3}{5},P(F)=\frac{3}{10}$ and $P(E\cap F)=\frac{1}{5}.$

For events E and F to be independent , we need

$P(E\cap F)=P(E).P(F)$

$P(E\cap F)=\frac{3}{5}\times \frac{3}{10}=\frac{9}{50}\neq \frac{1}{5}$

Hence, E and F are not indepent events.

(i) mutually exclusive

Given,

$P(A)=\frac{1}{2},P(A\cup B)=\frac{3}{5}$

Also, A and B are mutually exclusive means $A\cap B=\phi$.

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$\frac{3}{5}=\frac{1}{2}+P(B)-0$

$P(B)=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}$

(ii) independent

Given,

$P(A)=\frac{1}{2},P(A\cup B)=\frac{3}{5}$

Also, A and B are independent events means

$P(A\cap B) = P(A).P(B)$. Also $P(B)=p.$

$P(A\cap B) = P(A).P(B)=\frac{p}{2}$

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$\frac{3}{5}=\frac{1}{2}+p-\frac{p}{2}$

$\frac{p}{2}=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}$

$p=\frac{2}{10}=\frac{1}{5}$

(i) $\inline P(A\cap B)$

$\inline P(A)=0.3$ and $\inline P(B)=0.4$

Given : A and B be independent events

So, we have

$P(A\cap B)=P(A).P(B)$

$P(A\cap B)=0.3\times 0.4=0.12$

(ii) $\inline P(A\cup B)$

$\inline P(A)=0.3$ and $\inline P(B)=0.4$

Given : A and B be independent events

So, we have

$P(A\cap B)=P(A).P(B)$

$P(A\cap B)=0.3\times 0.4=0.12$

We have, $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$P(A\cup B)=0.3+0.4-0.12=0.58$

(iii) $\inline P(A\mid B)$

$\inline P(A)=0.3$ and $\inline P(B)=0.4$

Given : A and B be independent events

So, we have $P(A\cap B)=0.12$

$P(A|B)=\frac{P(A\cap B)}{P(B)}$

$P(A|B)=\frac{0.12}{0.4}= 0.3$

(iv) $P(B\mid A)$

$\inline P(A)=0.3$ and $\inline P(B)=0.4$

Given : A and B be independent events

So, we have $P(A\cap B)=0.12$

$P(B|A)=\frac{P(A\cap B)}{P(A)}$

$P(B|A)=\frac{0.12}{0.3}= 0.4$

If $A$ and $B$ are two events such that $P(A)=\frac{1}{4},P(B)=\frac{1}{2}$ and $P(A\cap B)=\frac{1}{8},$

$P(not\; A\; and\; not\; B)= P(A'\cap B')$

$P(not\; A\; and\; not\; B)= P(A\cup B)'$ use, $(P(A'\cap B')= P(A\cup B)')$

$= 1-(P(A)+P(B)-P(A\cap B))$

$= 1-(\frac{1}{4}+\frac{1}{2}-\frac{1}{8})$

$= 1-(\frac{6}{8}-\frac{1}{8})$

$= 1-\frac{5}{8}$

$= \frac{3}{8}$

If $A$ and $B$ are two events such that $P(A)=\frac{1}{2},P(B)=\frac{7}{12}$ and $P(not \; A \; or\; not\; B)=\frac{1}{4}.$

$P(A'\cup B')=\frac{1}{4}$

$P(A\cap B)'=\frac{1}{4}$ $(A'\cup B'=(A\cap B)')$

$\Rightarrow \, \, 1-P(A\cap B)=\frac{1}{4}$

$\Rightarrow \, \, \, P(A\cap B)=1-\frac{1}{4}=\frac{3}{4}$

$Also \, \, \, P(A\cap B)=P(A).P(B)$

$P(A\cap B)=\frac{1}{2}\times \frac{7}{12}=\frac{7}{24}$

As we can see $\frac{3}{4}\neq \frac{7}{24}$

Hence, A and B are not independent.

(i) $\inline P(A \; and\; B)$

$\inline P(A)=0.3,P(B)=0.6,$

Given two independent events $\inline A$ and $\inline B$.

$P(A\cap B)=P(A).P(B)$

$P(A\cap B)=0.3\times 0.6=0.18$

Also , we know $P(A \, and \, B)=P(A\cap B)=0.18$

(ii) $\inline P(A \; and \; not\; B)$

$\inline P(A)=0.3,P(B)=0.6,$

Given two independent events $\inline A$ and $\inline B$.

$\inline P(A \; and \; not\; B)$$=P(A)-P(A\cap B)$

$=0.3-0.18=0.12$

(iii) $P(A\; or \; B)$

$\inline P(A)=0.3,P(B)=0.6,$

$P(A\cap B)=0.18$

$P(A\; or \; B)=P(A\cup B)$

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$=0.3+0.6-0.18$

$=0.9-0.18$

$=0.72$

(iv) $\inline P(neither\; A\; nor\; B)$

$\inline P(A)=0.3,P(B)=0.6,$

$P(A\cap B)=0.18$

$P(A\; or \; B)=P(A\cup B)$

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$=0.3+0.6-0.18$

$=0.9-0.18$

$=0.72$

$\inline P(neither\; A\; nor\; B)$ $=P(A'\cap B')$

$= P((A\cup B)')$

$=1-P(A\cup B)$

$=1-0.72$

$=0.28$

A die is tossed thrice.

Outcomes $=\left \{ 1,2,3,4,5,6 \right \}$

Odd numbers $=\left \{ 1,3,5 \right \}$

The probability of getting an odd number at first throw

$=\frac{3}{6}=\frac{1}{2}$

The probability of getting an even number

$=\frac{3}{6}=\frac{1}{2}$

Probability of getting even number three times

$=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}$

The probability of getting an odd number at least once = 1 - the probability of getting an odd number in none of throw

= 1 - probability of getting even number three times

$=1-\frac{1}{8}$

$=\frac{7}{8}$

(i) both balls are red.

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.

Total balls =18

Black balls = 10

Red balls = 8

The probability of getting a red ball in first draw

$=\frac{8}{18}=\frac{4}{9}$

The ball is repleced after drawing first ball.

The probability of getting a red ball in second draw

$=\frac{8}{18}=\frac{4}{9}$

the probability that both balls are red

$=\frac{4}{9}\times \frac{4}{9}=\frac{16}{81}$

(ii) first ball is black and second is red.

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.

Total balls =18

Black balls = 10

Red balls = 8

The probability of getting a black ball in the first draw

$=\frac{10}{18}=\frac{5}{9}$

The ball is replaced after drawing the first ball.

The probability of getting a red ball in the second draw

$=\frac{8}{18}=\frac{4}{9}$

the probability that the first ball is black and the second is red

$=\frac{5}{9}\times \frac{4}{9}=\frac{20}{81}$

(iii) one of them is black and other is red.

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.

Total balls =18

Black balls = 10

Red balls = 8

Let the first ball is black and the second ball is red.

The probability of getting a black ball in the first draw

$=\frac{10}{18}=\frac{5}{9}$

The ball is replaced after drawing the first ball.

The probability of getting a red ball in the second draw

$=\frac{8}{18}=\frac{4}{9}$

the probability that the first ball is black and the second is red

$=\frac{5}{9}\times \frac{4}{9}=\frac{20}{81}$ $...........................1$

Let the first ball is red and the second ball is black.

The probability of getting a red ball in the first draw

$=\frac{8}{18}=\frac{4}{9}$

The probability of getting a black ball in the second draw

$=\frac{10}{18}=\frac{5}{9}$

the probability that the first ball is red and the second is black

$=\frac{4}{9}\times \frac{5}{9}=\frac{20}{81}$ $...........................2$

Thus,

The probability that one of them is black and the other is red = the probability that the first ball is black and the second is red + the probability that the first ball is red and the second is black $=\frac{20}{81}+\frac{20}{81}=\frac{40}{81}$

(i) the problem is solved

$P(A)=\frac{1}{2}$ and $P(B)=\frac{1}{3}$

Since, problem is solved independently by A and B,

$\therefore$ $P(A\cap B)=P(A).P(B)$

$P(A\cap B)=\frac{1}{2}\times \frac{1}{3}$

$P(A\cap B)=\frac{1}{6}$

probability that the problem is solved $= P(A\cup B)$

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$P(A\cup B)=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}$

$P(A\cup B)=\frac{5}{6}-\frac{1}{6}$

$P(A\cup B)=\frac{4}{6}=\frac{2}{3}$

(ii) exactly one of them solves the problem

$P(A)=\frac{1}{2}$ and $P(B)=\frac{1}{3}$

$P(A')=1-P(A)$, $P(B')=1-P(B)$

$P(A')=1-\frac{1}{2}=\frac{1}{2}$ , $P(B')=1-\frac{1}{3}=\frac{2}{3}$

probability that exactly one of them solves the problem $=P(A\cap B') + P(A'\cap B)$

probability that exactly one of them solves the problem $=P(A).P(B')+P(A')P(B)$

$=\frac{1}{2}\times \frac{2}{3}+\frac{1}{2}\times \frac{1}{3}$

$= \frac{2}{6}+\frac{1}{6}$

$= \frac{3}{6}=\frac{1}{2}$

(i) E : ‘the card drawn is a spade’

F : ‘the card drawn is an ace’

One card is drawn at random from a well shuffled deck of $52$ cards

Total ace = 4

E : ‘the card drawn is a spade

F : ‘the card drawn is an ace’

$P(E)=\frac{13}{52}=\frac{1}{4}$

$P(F)=\frac{4}{52}=\frac{1}{13}$

$E\cap F :$ a card which is spade and ace = 1

$P(E\cap F)=\frac{1}{52}$

$P(E).P(F)=\frac{1}{4}\times \frac{1}{13}=\frac{1}{52}$

$\Rightarrow P(E\cap F)=P(E).P(F)=\frac{1}{52}$

Hence, E and F are indepentdent events .

(ii) E : ‘the card drawn is

F : ‘the card drawn is a king’

One card is drawn at random from a well shuffled deck of $52$ cards

Total black card = 26

total king =4

E : ‘the card drawn is black’

F : ‘the card drawn is a king’

$P(E)=\frac{26}{52}=\frac{1}{2}$

$P(F)=\frac{4}{52}=\frac{1}{13}$

$E\cap F :$ a card which is black and king = 2

$P(E\cap F)=\frac{2}{52}=\frac{1}{26}$

$P(E).P(F)=\frac{1}{2}\times \frac{1}{13}=\frac{1}{26}$

$\Rightarrow P(E\cap F)=P(E).P(F)=\frac{1}{26}$

Hence, E and F are indepentdent events .

(iii) E : ‘the card drawn is a king or queen’

F : ‘the card drawn is a queen or jack’.

One card is drawn at random from a well shuffled deck of $52$ cards

Total king or queen = 8

total queen or jack = 8

E : ‘the card drawn is a king or queen’

F : ‘the card drawn is a queen or jack’.

$P(E)=\frac{8}{52}=\frac{2}{13}$

$P(F)=\frac{8}{52}=\frac{2}{13}$

$E\cap F :$ a card which is queen = 4

$P(E\cap F)=\frac{4}{52}=\frac{1}{13}$

$P(E).P(F)=\frac{2}{13}\times \frac{2}{13}=\frac{4}{169}$

$\Rightarrow P(E\cap F)\neq P(E).P(F)$

Hence, E and F are not indepentdent events

(a) Find the probability that she reads neither Hindi nor English newspapers

H : $\inline 60\; ^{o}/_{o}$ of the students read Hindi newspaper,

E : $\inline 40\; ^{o}/_{o}$ read English newspaper and

$H \cap E :$ $\inline 20\; ^{o}/_{o}$ read both Hindi and English newspapers.

$P(H)=\frac{60}{100}=\frac{6}{10}=\frac{3}{5}$

$P(E)=\frac{40}{100}=\frac{4}{10}=\frac{2}{5}$

$P(H\cap E)=\frac{20}{100}=\frac{2}{10}=\frac{1}{5}$

the probability that she reads neither Hindi nor English newspapers $=1-P(H\cup E)$

$=1-(P(H)+P(E)-P(H\cap E))$

$=1-(\frac{3}{5}+\frac{2}{5}-\frac{1}{5})$

$=1-\frac{4}{5}$

$=\frac{1}{5}$

(b) If she reads Hindi newspaper, find the probability that she reads English newspaper.

H : $\inline 60\; ^{o}/_{o}$ of the students read Hindi newspaper,

E : $\inline 40\; ^{o}/_{o}$ read English newspaper and

$H \cap E :$ $\inline 20\; ^{o}/_{o}$ read both Hindi and English newspapers.

$P(H)=\frac{60}{100}=\frac{6}{10}=\frac{3}{5}$

$P(E)=\frac{40}{100}=\frac{4}{10}=\frac{2}{5}$

$P(H\cap E)=\frac{20}{100}=\frac{2}{10}=\frac{1}{5}$

The probability that she reads English newspape if she reads Hindi newspaper $=P(E|H)$

$P(E|H)=\frac{P(E\cap H)}{P(H)}$

$P(E|H)=\frac{\frac{1}{5}}{\frac{3}{5}}$

$P(E|H)=\frac{1}{3}$

(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.

H : $\inline 60\; ^{o}/_{o}$ of the students read Hindi newspaper,

E : $\inline 40\; ^{o}/_{o}$ read English newspaper and

$H \cap E :$ $\inline 20\; ^{o}/_{o}$ read both Hindi and English newspapers.

$P(H)=\frac{60}{100}=\frac{6}{10}=\frac{3}{5}$

$P(E)=\frac{40}{100}=\frac{4}{10}=\frac{2}{5}$

$P(H\cap E)=\frac{20}{100}=\frac{2}{10}=\frac{1}{5}$

the probability that she reads Hindi newspaper if she reads English newspaper $= P(H |E)$

$P(H |E)=\frac{P(H\cap E)}{P(E)}$

$P(H |E)=\frac{\frac{1}{5}}{\frac{2}{5}}$

$P(H |E)=\frac{1}{2}$

(A) $0$

(B) $\frac{1}{3}$

(C) $\frac{1}{12}$

(D) $\frac{1}{36}$

when a pair of dice is rolled, total outcomes $=6^2=36$

Even prime number $=\left \{ 2 \right \}$

$n(even \, \, prime\, \, number)=1$

The probability of obtaining an even prime number on each die $=P(E)$

$P(E)=\frac{1}{36}$

Option D is correct.

Question:18 Two events A and B will be independent, if

(A) $\inline A$ and $\inline B$ are mutually exclusive

(B) $\inline P(A'B')=\left [ 1-P(A) \right ]\left [ 1-P(B) \right ]$

(C) $P(A)=P(B)$

(D) $P(A)+P(B)=1$

Two events A and B will be independent, if

$P(A\cap B)=P(A).P(B)$

Or $P(A'\cap B')=P(A'B')=P(A').P(B')=(1-P(A)).(1-P(B))$

Option B is correct.

## More About NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.2:-

Class 12 Maths chapter 13 exercise 13.2 solutions are consist of questions related to multiplication theorem on probability. There are 7 examples given before the NCERT syllabus exercise questions. You can solve them before solving the NCERT problems. There are 18 questions questions given in this exercise which you can solve. All these questions are solved in NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.2 in a detailed manner. You can go through these solutions for reference.

Also Read| Probability Class 12th Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.2:-

• Class 12th Maths chapter 13 exercise 13.2 solutions are helpful for the students get conceptual clarity and hence performing well in the in the board exams.
• You must be thorough with the NCERT textbook problems.
• These Class 12 Maths chapter 13 exercise 13.2 solutions will give you conceptual clarity about probability multiplication rule and independent events.
• There are 18 questions in this exercise which are solved by the subject matter experts on which you can rely.
• You can quickly revise the important concepts before the final exam.

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%

## Key Features Of NCERT Solutions for Exercise 13.2 Class 12 Maths Chapter 13

• Comprehensive Coverage: The solutions encompass all the topics covered in ex 13.2 class 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 maths ex 13.2, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 ex 13.2 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this 12th class maths exercise 13.2 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for ex 13.2 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for class 12 maths ex 13.2 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

## Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. What is the number of questions in Probability exercise 13.2 ?

There are 18 questions in NCERT Class 12 Maths chapter 13 exercise 13.2.

2. What does it means by Independent Events ?

If the occurrence of one event doesn’t affect the probability of the other event such events are called independent events.

3. What is the probability of getting a tail when we toss a fair coin ?

The probability of getting a tail is 0.5 when we toss a fair coin.

4. Can I get free NCERT Exemplar solutions for Class 12 Maths probability?

You will get NCERT Exemplar Solutions for Class 12 Maths Probability by clicking on the given link.

5. What is an Impossible Event ?

An event is called an impossible event if the number of favorable outcomes is zero.

6. What is the probability of impossible event ?

The probability of an impossible event is zero.

7. Can I get free NCERT solutions for Class 12 Maths?

Click here to get NCERT Solutions for Class 12 Maths for other subjects.

8. If the probability of getting tail on tossing of a biased coin is 0.47 then what is the probability of getting head ?

The probability of getting head of this biased coin is 0.53.

## Upcoming School Exams

#### National Institute of Open Schooling 12th Examination

Application Date:07 June,2024 - 06 July,2024

#### National Institute of Open Schooling 10th examination

Application Date:07 June,2024 - 06 July,2024

#### Jammu and Kashmir State Board of School Education 12th Examination

Others:11 June,2024 - 02 July,2024

#### Gujarat Board Secondary School Certificate Examination

Admit Card Date:15 June,2024 - 06 July,2024

#### Goa Board Secondary School Certificate Examination

Exam Date:18 June,2024 - 28 June,2024

## Certifications By Top Providers

Full-Stack Web Development
Via Masai School
Introduction to Psychology
Via San Jose State University, San Jose
Edx
1111 courses
Coursera
792 courses
Udemy
327 courses
Futurelearn
147 courses
IBM
85 courses

## Explore Top Universities Across Globe

University of Essex, Colchester
Wivenhoe Park Colchester CO4 3SQ
University College London, London
Gower Street, London, WC1E 6BT
The University of Edinburgh, Edinburgh
Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Bristol, Bristol
Beacon House, Queens Road, Bristol, BS8 1QU
University of Nottingham, Nottingham
University Park, Nottingham NG7 2RD
Lancaster University, Lancaster
Bailrigg, Lancaster LA1 4YW

### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hello,

Yes, you can give improvement exams for 2-3 subjects in CBSE (Central Board of Secondary Education). CBSE allows students who have appeared for their Class 12 board exams to improve their scores by re-appearing for exams in up to five subjects the following year.

Hope this helps,

Thank you

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

• Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

• Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

• Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

• Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.

In India, the design and coding fields offer exciting career options that can leverage your interest in both. Here's how you can navigate this path:

• Graphic Design Focus: Consider a Bachelor's degree in Graphic Design or a design diploma. Build a strong portfolio showcasing your creative skills. Learn the basics of HTML, CSS, and JavaScript to understand web development better. Many online resources and bootcamps offer these introductory courses.

• Coding Focus: Pursue a Computer Science degree or a coding bootcamp in India. These programs are intensive but can equip you with strong coding skills quickly. While building your coding prowess, take online courses in graphic design principles and UI/UX design.

Engineering Subjects (for a Degree):

• Information Technology (IT): This offers a good mix of web development, networking, and database management, all valuable for web design/development roles.

• Human-Computer Interaction (HCI): This is a specialized field that bridges the gap between design and computer science, focusing on how users interact with technology. It's a perfect choice if you're interested in both aspects.

• Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

Here's why 2025 is more likely:

• JEE Main 2024 Admissions: The application process for NITs through JEE Main 2024 is likely complete by now (May 2024). They consider your 2023 Class 12th marks (CBSE in this case).
• NIOS Results: Since NIOS results typically come out after the NIT admission process, your October 2024 NIOS marks wouldn't be available for JEE Main 2024.

• Focus on JEE Main: Since you have a computer science background, focus on preparing for JEE Main 2025. This exam tests your knowledge in Physics, Chemistry, and Mathematics, crucial for engineering programs at NITs.
• NIOS Preparation: Utilize the time between now and October 2024 to prepare for your NIOS exams.
• Eligibility Criteria: Remember, NITs typically require a minimum of 75% marks in Class 12th (or equivalent) for general category students (65% for SC/ST). Ensure you meet this criteria in your NIOS exams.

Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore,  with your academic background. Here's why:

• High CAT Score: A score exceeding  99.9 percentile is exceptional and puts you amongst the top candidates vying for admission. IIM Bangalore prioritizes  CAT scores heavily in the shortlisting process.

• Strong Academics: Your 96% in CBSE 12th and a B.Tech degree demonstrate a solid academic foundation, which IIM Bangalore also considers during shortlisting.

However, the shortlisting process is multifaceted:

• Other Factors: IIM Bangalore considers other factors beyond CAT scores, such as your work experience (if any), XAT score (if you appear for it), academic diversity, gender diversity, and performance in the interview and Written Ability Test (WAT) stages (if shortlisted).

Here's what you can do to strengthen your application:

• Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

• Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9