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NCERT Solutions for Exercise 13.3 Class 12 Maths Chapter 13 - Probability

NCERT Solutions for Exercise 13.3 Class 12 Maths Chapter 13 - Probability

Edited By Ramraj Saini | Updated on Dec 04, 2023 11:24 AM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.3

NCERT Solutions for Exercise 13.3 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. In NCERT solutions for exercise 13.3 class 12 maths chapter 13 article, you will find most of the questions related to the total probability theorem, conditional probability, and various properties of conditional probability. These questions are solved in a step-by-step manner which is helpful for the students to perform well in the board exams. NCERT Solutions for exercise 13.3 class 12 maths are prepared by the subject matter experts as per CBSE guidelines which are reliable. You don't need to be but any other reference books as most of the questions in board exams are directly asked from the NCERT textbook. You just need to be thorough with the NCERT textbook.

12th class Maths exercise 13.3 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.

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Access NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.3

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Probability Class 12 Chapter 13-Exercise: 13.3

Question:1 An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

Answer:

Black balls = 5

Red balls = 5

Total balls = 10

CASE 1 Let red ball be drawn in first attempt.

P(drawing\, red\, ball)=\frac{5}{10}=\frac{1}{2}

Now two red balls are added in urn .

Now red balls = 7, black balls = 5

Total balls = 12

P(drawing\, red\, ball)=\frac{7}{12}

CASE 2

Let black ball be drawn in first attempt.

P(drawing\, black\, ball)=\frac{5}{10}=\frac{1}{2}

Now two black balls are added in urn .

Now red balls = 5, black balls = 7

Total balls = 12

P(drawing\, red\, ball)=\frac{5}{12}

the probability that the second ball is red =

=\frac{1}{2}\times \frac{7}{12}+\frac{1}{2}\times \frac{5}{12}

= \frac{7}{24}+ \frac{5}{24}

= \frac{12}{24}=\frac{1}{2}

Question:2 A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn frome bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Answer:

BAG 1 : Red balls =4 Black balls=4 Total balls = 8

BAG 2 : Red balls = 2 Black balls = 6 Total balls = 8

B1 : selecting bag 1

B2 : selecting bag 2

P(B1)=P(B2)=\frac{1}{2}

Let R be a event of getting red ball

P(R|B1) = P(drawing\, \, red\, \, ball\, \, from \, first \, \, bag)= \frac{4}{8}=\frac{1}{2}

P(R|B2) = P(drawing\, \, red\, \, ball\, \, from \, second \, \, bag)= \frac{2}{8}=\frac{1}{4}

probability that the ball is drawn from the first bag,

given that it is red is P(B1|R).

Using Baye's theorem, we have

P(B1|R) = \frac{P(B1).P(R|B1)}{P(B1).P(R|B1)+P(B2).P(R|B2)}

P(B1|R) = \frac{\frac{1}{2}\times \frac{1}{2}}{\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{4}}

P(B1|R) =\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{8}}

P(B1|R) =\frac{\frac{1}{4}}{\frac{3}{8}}

P(B1|R) = \frac{2}{3}

Question:3 Of the students in a college, it is known that 60^{o}/_{o} reside in hostel and 40^{o}/_{o} are day scholars (not residing in hostel). Previous year results report that 30^{o}/_{o} of all students who reside in hostel attainA grade and 20^{o}/_{o} of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier?

Answer:

H : reside in hostel

D : day scholars

A : students who attain grade A

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

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

P(A|H)=\frac{30}{100}=\frac{3}{10}

P(A|D)=\frac{20}{100}=\frac{2}{10}= \frac{1}{5}

By Bayes theorem :

P(H|A)=\frac{P(H).P(A|H)}{P(H).P(A|H)+P(D).P(A|D)}

P(H|A)=\frac{\frac{3}{5}\times \frac{3}{10}}{\frac{3}{5}\times \frac{3}{10}+\frac{2}{5}\times \frac{1}{5}}

P(H|A)=\frac{\frac{9}{50}}{\frac{9}{50}+\frac{2}{25}}

P(H|A)=\frac{\frac{9}{50}}{\frac{13}{50}}

P(H|A)=\frac{9}{13}

Question:4 In answering a question on a multiple choice test, a student either knows the answer or guesses. Let \frac{3}{4} be the probability that he knows the answer and\frac{1}{4} be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability \frac{1}{4} . What is the probability that the student knows the answer given that he answered it correctly?

Answer:

A : Student knows answer.

B : Student guess the answer

C : Answer is correct

P(A)=\frac{3}{4} P(B)=\frac{1}{4}

P(C|A)=1

P(C|B)=\frac{1}{4}

By Bayes theorem :

P(A|C)=\frac{P(A).P(C|A)}{P(A).P(C|A)+P(B).P(C|B)}

P(A|C)=\frac{\frac{3}{4}\times 1}{\frac{3}{4}\times 1+\frac{1}{4}\times \frac{1}{4}}

=\frac{\frac{3}{4}}{\frac{3}{4}+\frac{1}{16}} =\frac{\frac{3}{4}}{\frac{13}{16}}

P(A|C)=\frac{12}{13}

Question:5 A laboratory blood test is 99^{o}/_{o} effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for0.5^{o}/_{o} of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1^{o}/_{o} of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive ?

Answer:

A : Person selected is having the disease

B : Person selected is not having the disease.

C :Blood result is positive.

P(A)= 0.1 \%=\frac{1}{1000}=0.001

P(B)= 1 -P(A)=1-0.001=0.999

P(C|A)=99\%=0.99

P(C|B)=0.5\%=0.005

By Bayes theorem :

P(A|C)=\frac{P(A).P(C|A)}{P(A).P(C|A)+P(B).P(C|B)}

=\frac{0.001\times 0.99}{0.001\times 0.99+0.999\times 0.005}

=\frac{0.00099}{0.00099+0.004995}

=\frac{0.00099}{0.005985} =\frac{990}{5985}

=\frac{22}{133}

Question:6 There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75^{o}/_{o} of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

Answer:

Given : A : chossing a two headed coin

B : chossing a biased coin

C : chossing a unbiased coin

P(A)=P(B)=P(C)=\frac{1}{3}

D : event that coin tossed show head.

P(D|A)=1

Biased coin that comes up heads 75^{o}/_{o} of the time.

P(D|B)=\frac{75}{100}=\frac{3}{4}

P(D|C)=\frac{1}{2}

P(B|D)=\frac{P(B).P(D|B)}{P(B).P(D|B)+P(A).P(D|A)+P(C).P(D|C)}

P(B|D)=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\times 1+\frac{1}{3}\times \frac{3}{4}+\frac{1}{3}\times \frac{1}{2}}

P(B|D)=\frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{4}+\frac{1}{6}}

P(B|D)=\frac{\frac{1}{3}}{\frac{9}{12}}

P(B|D)={\frac{1\times 12}{3\times 9}}

P(B|D)={\frac{4}{9}}

Question:7 An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01,0.03 , and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Answer:

Let A : scooter drivers = 2000

B : car drivers = 4000

C : truck drivers = 6000

Total drivers = 12000

P(A)=\frac{2000}{12000}=\frac{1}{6}=0.16

P(B)=\frac{4000}{12000}=\frac{1}{3}=0.33

P(C)=\frac{6000}{12000}=\frac{1}{2}=0.5

D : the event that person meets with an accident.

P(D|A)= 0.01

P(D|B)= 0.03

P(D|C)= 0.15

P(A|D)=\frac{P(A).P(D|A)}{P(B).P(D|B)+P(A).P(D|A)+P(C).P(D|C)}

P(A|D)= \frac{0.16\times 0.01}{0.16\times 0.01+0.33\times 0.03+0.5\times 0.15}

P(A|D)= \frac{0.0016}{0.0016+0.0099+0.075}

P(A|D)= \frac{0.0016}{0.0865}

P(A|D)= 0.019

Question:8 A factory has two machines A and B. Past record shows that machine A produced 60^{o}/_{o} of the items of output and machine B produced 40^{o}/_{o} of the items. Further, 2^{o}/_{o} of the items produced by machine A and 1^{o}/_{o} produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

Answer:

A : Items produced by machine A =60\%

B : Items produced by machine B=40\%

P(A)= \frac{60}{100}=\frac{3}{5}

P(B)= \frac{40}{100}=\frac{2}{5}

X : Produced item found to be defective.

P(X|A)= \frac{2}{100}=\frac{1}{50}

P(X|B)= \frac{1}{100}

P(B|X)= \frac{P(B).P(X|B)}{P(B).P(X|B)+P(A).P(X|A)}

P(B|X)= \frac{\frac{2}{5}\times \frac{1}{100}}{\frac{2}{5}\times \frac{1}{100}+\frac{3}{5}\times \frac{1}{50}}

P(B|X)= \frac{\frac{1}{250}}{\frac{1}{250}+\frac{3}{250}}

P(B|X)= \frac{\frac{1}{250}}{\frac{4}{250}}

P(B|X)= \frac{1}{4}

Hence, the probability that defective item was produced by machine B =

P(B|X)= \frac{1}{4}.

Question:9 Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Answer:

A: the first groups will win

B: the second groups will win

P(A)=0.6

P(B)=0.4

X: Event of introducing a new product.

Probability of introducing a new product if the first group wins : P(X|A)=0.7

Probability of introducing a new product if the second group wins : P(X|B)=0.3

P(B|X)= \frac{P(B).P(X|B)}{P(B).P(X|B)+P(A).P(X|A)}

p(B|X) = \frac{0.4\times 0.3}{0.4\times 0.3+0.6\times 0.7}

p(B|X) = \frac{0.12}{0.12+0.42}

p(B|X) = \frac{0.12}{0.54}

p(B|X) = \frac{12}{54}

p(B|X) = \frac{2}{9}

Hence, the probability that the new product introduced was by the second group :

p(B|X) = \frac{2}{9}

Question:10 Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1,2,3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1,2,3 or 4with the die?

Answer:

Let, A: Outcome on die is 5 or 6.

B: Outcome on die is 1,2,3,4

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

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

X: Event of getting exactly one head.

Probability of getting exactly one head when she tosses a coin three times : P(X|A)=\frac{3}{8}

Probability of getting exactly one head when she tosses a coin one time : P(X|B)=\frac{1}{2}

P(B|X)= \frac{P(B).P(X|B)}{P(B).P(X|B)+P(A).P(X|A)}

P(B|X)= \frac{\frac{2}{3}\times \frac{1}{2}}{\frac{2}{3}\times \frac{1}{2}+\frac{1}{3}\times \frac{3}{8}}

P(B|X)= \frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{8}}

P(B|X)= \frac{\frac{1}{3}}{\frac{11}{24}}

P(B|X)= \frac{1\times 24}{3\times 11}=\frac{8}{11}

Hence, the probability that she threw 1,2,3 or 4 with the die =

P(B|X)=\frac{8}{11}

Question:11 A manufacturer has three machine operators A,B and C. The first operator A produces 1^{o}/_{o} defective items, where as the other two operators B and C produce 5^{o}/_{o} and 7^{o}/_{o} defective items respectively. A is on the job for 50^{o}/_{o} of the time, B is on the job for 30^{o}/_{o} of the time and C is on the job for 20^{o}/_{o} of the time. A defective item is produced, what is the probability that it was produced by A?

Answer:

Let A: time consumed by machine A =50\%

B: time consumed by machine B=30\%

C: time consumed by machine C =20\%

Total drivers = 12000

P(A)=\frac{50}{100}=\frac{1}{2}

P(B)=\frac{30}{100}=\frac{3}{10}

P(C)=\frac{20}{100}=\frac{1}{5}

D: Event of producing defective items

P(D|A)= \frac{1}{100}

P(D|B)= \frac{5}{100}

P(D|C)= \frac{7}{100}

P(A|D)=\frac{P(A).P(D|A)}{P(B).P(D|B)+P(A).P(D|A)+P(C).P(D|C)}

P(A|D)=\frac{\frac{1}{2}\times \frac{1}{100}}{\frac{1}{2}\times \frac{1}{100}+\frac{3}{10}\times \frac{5}{100}+\frac{1}{5}\times \frac{7}{100}}

P(A|D)=\frac{\frac{1}{2}\times \frac{1}{100}}{\frac{1}{100} (\frac{1}{2}+\frac{3}{2}+\frac{7}{5})}

P(A|D)=\frac{\frac{1}{2}}{ (\frac{17}{5})}

P(A|D)= \frac{5}{34}

Hence, the probability that defective item was produced by A =

P(A|D)= \frac{5}{34}

Question:12 A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Answer:

Let A : Event of choosing a diamond card.

B : Event of not choosing a diamond card.

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

P(B)=\frac{39}{52}=\frac{3}{4}

X : The lost card.

If lost card is diamond then 12 diamond cards are left out of 51 cards.

Two diamond cards are drawn out of 12 diamond cards in ^{12}\textrm{C}_2 ways.

Similarly, two cards are drawn out of 51 cards in ^{51}\textrm{C}_2 ways.

Probablity of getting two diamond cards when one diamond is lost : P(X|A)= \frac{^{12}\textrm{C}_2}{^{51}\textrm{C}_2}

P(X|A)=\frac{12!}{10!\times 2!}\times \frac{49!\times 2!}{51!}

P(X|A)=\frac{11\times 12}{50\times 51}

P(X|A)=\frac{22}{425}

If lost card is not diamond then 13 diamond cards are left out of 51 cards.

Two diamond cards are drawn out of 13 diamond cards in ^{13}\textrm{C}_2 ways.

Similarly, two cards are drawn out of 51 cards in ^{51}\textrm{C}_2 ways.

Probablity of getting two diamond cards when one diamond is not lost : P(X|B)= \frac{^{13}\textrm{C}_2}{^{51}\textrm{C}_2}

P(X|B)=\frac{13!}{11!\times 2!}\times \frac{49!\times 2!}{51!}

P(X|B)=\frac{13\times 12}{50\times 51}

P(X|B)=\frac{26}{425}

The probability of the lost card being a diamond : P(B|X)

P(B|X)= \frac{P(B).P(X|B)}{P(B).P(X|B)+P(A).P(X|A)}

P(B|X)= \frac{\frac{1}{4}\times \frac{22}{425}}{\frac{1}{4}\times \frac{22}{425}+\frac{3}{4}\times \frac{26}{425}}

P(B|X)= \frac{\frac{11}{2}}{25}

P(B|X)= \frac{11}{50}

Hence, the probability of the lost card being a diamond :

P(B|X)= \frac{11}{50}

Question:13 Probability that A speaks truth is \frac{4}{5} . A coin is tossed. A reports that a head appears. The probability that actually there was head is

(A) \frac{4}{5}

(B) \frac{1}{2}

C) \frac{1}{5}

(D) \frac{2}{5}

Answer:

Let A : A speaks truth

B : A speaks false

P(A)=\frac{4}{5}

P(B)=1-\frac{4}{5}=\frac{1}{5}

X : Event that head appears.

A coin is tossed , outcomes are head or tail.

Probability of getting head whether A speaks thruth or not is \frac{1}{2}

P(X|A)=P(X|B)=\frac{1}{2}

P(A|X)= \frac{P(A).P(X|A)}{P(B).P(X|B)+P(A).P(X|A)}

P(A|X)= \frac{\frac{4}{5}\times \frac{1}{2}}{\frac{4}{5}\times \frac{1}{2}+\frac{1}{5}\times \frac{1}{2}}

P(A|X)= \frac{\frac{4}{5}}{\frac{4}{5}+\frac{1}{5}}

P(A|X)= \frac{\frac{4}{5}}{\frac{1}{1}}

P(A|X)={\frac{4}{5}}

The probability that actually there was head is P(A|X)={\frac{4}{5}}

Hence, option A is correct.

Question:14 If A and B are two events such that A\subset B and P(B)\neq 0, then which of the following is correct?

(A) P(A\mid B)=\frac{P(B)}{P(A)}

(B) P(A\mid B)< P(A)

(C) P(A\mid B)\geq P(A)

(D) None of these

Answer:

If A\subset B and P(B)\neq 0,then

\Rightarrow \, \, \, (A\cap B) = A

Also, P(A)< P(B)

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

We know that P(B)\leq 1

1\leq \frac{1}{P(B)}

P(A)\leq \frac{P(A)}{P(B)}

P(A)\leq P(A|B)

Hence, we can see option C is correct.

More About NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.3:-

Exercise 13.3 Class 12 Maths solutions are consist of questions related to Bayes' Theorem and its applications. There are 7 solved examples before the NCERT exercise which you can try to solve before solving the NCERT problems. The NCERT syllabus exercises questions are very similar to the examples given in the textbook. If you have solved examples, you must be finding solving NCERT problems very easy. You can take help from Class 12 Maths chapter 13 exercise 13.3 solutions solved in a very detailed manner which could be understood very easily.

Also Read| Probability Class 12th Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.3:-

  • These Class 12 Maths chapter 13 exercise 13.3 solutions are useful for the students who are stuck while solving NCERT problems.
  • You are advised to solve the NCERT textbook example first then try to solve the NCERT exercise problems on your own.
  • Solving more problems will give the conceptual clarity therefore you will be able to solve new problems on your own.
  • You can take help from NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.3.
  • Here you can find Class 12th Maths chapter 13 exercise 13.3 questions with solutions and all the Class 12 Maths chapter solutions at one place.
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Key Features Of NCERT Solutions for Exercise 13.3 Class 12 Maths Chapter 13

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 13.3 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 13.3, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 13.3 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 13.3 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 13.3 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 13.3 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Questions (FAQs)

1. What is an certain event ?

When all possible outcomes are favorable to the event such events are called a certain event.

2. what is the probability of the certain event ?

The probability of a certain event is always 1.

3. What is a Complementary Events ?

The complementary event 'E' only occurs when event E doesn't occur.

4. How many questions are covered in Probability Exercise 13.3 ?

There are 14 questions in NCERT Class 12 Maths chapter 13 exercise 13.3.

5. If the probability of event E is P(E) then the probability of it's complementary event is ?

The probability of the complementary event is  P(E') = 1-P(E).

6. what is the random experiment ?

If an experiment has more than one possible outcome and we can't predict the outcome then such an experiment is called a random experiment.

7. where can I get NCERT solutions ?

Here you will get NCERT Solutions.

8. Can I get NCERT solutions for Class 12?

Click here to get NCERT Solutions for Class 12. 

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quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer
Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

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    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Read Complete Answer
Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer
Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
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Popular CBSE Class 12th Questions

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)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read More

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

 
Read More

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)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read More

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

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read More

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read More

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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
Read More

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.
Read More

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.
Read More

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
Read More

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
Read More

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