Category: Class 11 Mathematics

NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series. https://www.cbselabs.com/ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series/

NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series

Short Answer Type Questions:

Q1. The first term of an A.P. is terms is zero, show that the sum of its next q terms
\(\frac { -a(p+q)q }{ p-1 }  \)

Sol: Let the common differeence of the given A.P be d

Calculates the n-th term and sum of the geometric progression with the common ratio calculator.

Q2. A man saved Rs. 66000 in 20 years. In each succeeding year after the first year, he saved Rs. 200 more than what he saved in the previous year. How much did he save in the first year?
Sol: Let us assume that the man saved Rs.a in the first year.
In each succeeding year, an increment of Rs. 200 is made. So, it forms an A.P. whose
First term = a, Common difference, d = 200 and n=20

=> 6600 = 2a + 19 x 200 => 2a = 2800
∴a = 1400

 Q3. A man accepts a position with an initial salary of Rs. 5200 per month. It is understood that he will receive an automatic increase of Rs. 320 in the very next month and each month thereafter.
(i) Find his salary for the tenth month.
(ii) What is his total earnings during the first year?

Q5. A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?
Sol: Here, a = 5 and d = 2

Q6. The sum of interior angles of a triangle is 180°. Show that the sum of the interior angles of polygons with 3, 4, 5, 6, … sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.
Sol: We know that, sum of interior angles of a polygon of side n is (n – 2) x 180°.

Q7. A side of an equilateral triangle is 20 cm long. A second equilateral triangle is inscribed in it by joining the mid-points of the sides of the first triangle. This process is continued for third, fourth, fifth, triangles. Find the perimeter of the sixth inscribed equilateral triangle.
Sol: Let the given equilateral triangle be ∆ ABC with each side of 20 cm.
By joining the mid-points of this triangle, we get another equilateral triangle of side equal to half of the length of side of ∆ABC.
Continuing in this way, we get a set of equilateral triangles with side equal to half of the side of the previous triangle.
Now,
Perimeter of first triangle = 20 x 3 = 60 cm;
Perimeter of second triangle = 10 x 3 = 30 cm;
Perimeter of third triangle = 5×3 = 15 cm;

Q8. In a potato race 20 potatoes are placed in a line at intervals of 4 m with the first potato 24 m from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?
Sol: Distance travelled to bring first potato = 24 + 24 = 2 x 24 = 48 m
Distance travelled to bring second potato = 2(24 + 4) = 2 x 28 = 56 m
Distance travelled to bring third potato = 2(24 + 4 + 4) = 2 X 32 = 64 m; and so on…
Clearly, 48, 56, 64,… is an A.P. with first term 48 and common difference 8. Also, number of terms is 20.
Total distance run in bringing back all the potatoes,

Q9. In a cricket tournament 16 school teams participated. A sum of Rs. 8000 is to be awarded among themselves as prize money. If the last placed team is awarded Rs. 275 in prize money and the award increases by the same amount for successive finishing places, how much amount will the first place team receive?
Sol: Let the first place team get Rs. a as the prize money.
Since award money increases by the same amount for successive finishing places, we get an A.P.

Q11. Find the sum of the series (3³ – 2³) + (5³ – 4³) + (7³ – 6³) + … to
(i) n terms                                               
(ii) 10 terms
Sol: Given series is: (3³ — 2³) + (5³ – 4³) + (7³ – 6³) + … n terms

Q12. Find the rth term of an A.P. sum of whose first n terms is 2n +3n²
Sol: Sum of k terms of A.P., S_n = 2n + 3n²

Long Answer Type Questions

Q13. If A is the arithmetic mean and G₁ , G₂ be two geometric means between any two numbers, then prove that 2A = \(\frac { { G }_{ 1 }^{ 2 } }{ { G }_{ 2 } } +\frac { { G }_{ 2 }^{ 2 } }{ { G }_{ 2 } } \quad   \)
Sol: Let the numbers be a and b.

Q15. 1f the sum of p terms of an AP. is q and the sum of q terms isp, then show that the sum ofp + q terms is —(p + q). Also, find the sum of first p — q terms (where, p > q).
Sol: Let first term and common difference of the A.P. be a and d, respectively. Given, S_p = q

Q16. If pth, qth and rth terms of an A.P. and G.P. are both a, b and c, respectively, then show that a^b-c b^c-a-c^a-b = 1.
Sol: Let A and d be the first term and common difference of A.P., respectively. Also, let B and R be the first term and common ratio of G.P., respectively.
It is given that,

Objective Type Questions:

Q17. If the sum of n terms of an A.P. is given by S_n = 3n + 2n², then the common difference of the A.P. is
(a) 3                       
(b) 2                         
(c) 6                        
(d) 4
Sol: (d) Given, S_n = 3n + 2n²

Q18. If the third term of G.P. is 4, then the product of its first 5 terms is
(a) 4³
(b) 4⁴
(c) 4⁵
(d) none of these
Sol: (c) Let a and r the first term and common ratio, respectively.

Q19. If 9 times the 9th term of an A.P. is equal to 13 times the 13th term, then the 22nd term of the A.P. is
(a) 0
(b) 22
(c) 198
(d) 220
Sol: (a) Let the first term and coiqmon difference of given A.P. be a and d, respectively.

Q20. If x, 2y and 3z are in A.P. where the distinct numbers x, y and z are in G.P., then the common ratio of the G.P.is
Sol: Since x, 2y and 3z are in A.P., we get

Q21. If in an A.P., S_n = qn² and S_m = qm², where S_r denotes the sum of r terms of the AP, then S_q equals

Q22. Let S_n denote the sum of the first n terms of an A.P. If S_2n = 3S_n, then S_3n : S_n  is equal to
(a) 4
(b) 6
(c) 8
(d) 10
Sol: (b) Let first term be a and common difference be d.
Then, S_2n = 3S_n

Q25. If t_n denotes the nth term of the series 2 + 3+ 6+11 + 18+…, then t₅₀ is
(a) 49² – 1     
(b) 49²   
(c) 50²+l
(d) 49² +2

Q26. The lengths of three unequal edges of a rectangular solid block are in G.P. If the volume of the block is 216 cm³ and the total surface area is 252 cm², then the length of the longest edge is
(a) 12 cm               
(b) 6 cm                   
(c) 18 cm                
(d) 3 cm
Sol: (a) Let the length, breadth and height of rectangular solid block be a/r, a and ar, respectively.

Fill in the Blanks

Q27. If a, b and c are in G.P., then the value of \(\frac { a-b }{ b-c }   \) is equal to ________

Q28. The sum of terms equidistant from the beginning and end in an A.P. is equal to _____.
Sol: Let a be the first term and d be ihe common difference of the A.P.

Q29. The third term of a G.P. is 4. The product of the first five terms is  
Sol: Let a and r the first term and common ratio, respectively.

True/False Type Questions

Q30. Two sequences cannot be in both A.P. and G.P. together.
Sol: False
Consider the sequence 3,3,3; which is A.P. and G.P. both.

Q31. Every progression is a sequence but the converse, i.e., every sequence is also a progression need not necessarily be true.
Sol: True                                                            –
Consider the progression a, a + d, a + 2d, … and sequence of prime number 2, 3, 5, 7, 11,…
Clearly, progression is a sequence but sequence is not progression because it does not follow a specific pattern.

Q32. Any term of an A.P. (except first) is equal to half the sum of terms which are equidistant from it.
Sol: True

Q33. The sum or difference of two G.P.s, is again a G.P.

Q34. If the sum of n terms of a sequence is quadratic expression, then it always represents an A.P.
Sol: False

Match the questions given under Column I with their appropriate answers given under the column II.

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We hope the NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series help you. If you have any query regarding NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Maths Chapter 8 Binomial Theorem are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 8 Binomial Theorem. https://www.cbselabs.com/ncert-exemplar-problems-class-11-mathematics-chapter-8-binomial-theorem/

NCERT Exemplar Class 11 Maths Chapter 8 Binomial Theorem

Short Answer Type Questions:

Expanding Binomial Calculator is a free online tool that lets you solve the expansion of a binomial in the blink of an eye.

Q1. Find the term independent of x, where x≠0, in the expansion of \({ \left( \frac { 3{ x }^{ 2 } }{ 2 } -\quad \frac { 1 }{ 3x }  \right)  }^{ 15 }\)

Q2. If the term free from x is the expansion of  \({ \left( \sqrt { x } -\frac { k }{ { x }^{ 2 } }  \right)  }^{ 10 }\) is 405, then find the value of k.

Sol: Given expansion is \({ \left( \sqrt { x } -\frac { k }{ { x }^{ 2 } }  \right)  }^{ 10 }\)

Q3. Find the coefficient of x in the expansion of (1 – 3x + 1x²)( 1 -x)¹⁶.

Sol: (1 – 3x + 1x²)( 1 -x)¹⁶

Q4. Find the term independent of x in the expansion of \({ \left( 3x-\frac { 2 }{ { x }^{ 2 } }  \right)  }^{ 15 }\)

Sol: Given Expression \({ \left( 3x-\frac { 2 }{ { x }^{ 2 } }  \right)  }^{ 15 }\)

Q5. Find the middle term (terms) in the expansion of

Q6. Find the coefficient of x¹⁵ in the expansion of \({ \left( x-{ x }^{ 2 }\quad  \right)  }^{ 10 }\)

Sol: Given expression is   \({ \left( x-{ x }^{ 2 }\quad  \right)  }^{ 10 }\)

Q7. Find the coefficient of \(\frac { 1 }{ { x }^{ 17 } } \) in the expansion of \({ \left( { x }^{ 4 }-\frac { 1 }{ { x }^{ 3 } } \quad  \right)  }^{ 15 } \)

Q8. Find the sixth term of the expansion (y^1/2 + x^1/3)ⁿ, if the binomial coefficient of the third term from the end is 45.

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Q9. Find the value of r, if the coefficients of (2r + 4)th and (r – 2)th terms in the expansion of (1 + x)¹⁸ are equal.

Q10. If the coefficient of second, third and fourth terms in the expansion of (1 + x)²” are in A.P., then show that 2n² – 9n + 7 = 0.

Q11. Find the coefficient of x⁴ in the expansion of (1 + x + x² + x³)¹¹.

Long Answer Type Questions

Q12. If p is a real number and the middle term in the expansion \({ \left( \frac { p }{ 2 } +2\quad \right) }^{ 8 } \) is 1120, then find the value of p.

Q15. In the expansion of (x + a)ⁿ, if the sum of odd term is denoted by 0 and the sum of even term by Then, prove that

Q17. Find the term independent ofx in the expansion of (1 +x + 2x³)\({ \left( \frac { 3 }{ 2 } { x }^{ 2 }-\frac { 1 }{ 3x } \quad \quad  \right)  }^{ 9 } \)

Objective Type Questions

Q18. The total number of terms in the expansion of (x + a)¹⁰⁰ + (x – a)¹⁰⁰ after simplification is

(a) 50
(b) 202
(c) 51
(d) none of these

Q19. If the integers r > 1, n > 2 and coefficients of (3r)th and (r + 2)nd terms in the binomial expansion of (1 + x)²ⁿ are equal, then
(a) n = 2r             
(b) n = 3r            
(c) n = 2r + 1       
(d) none of these

Q20. The two successive terms in the expansion of (1 + x)²⁴ whose coefficients are in the ratio 1 : 4 are
(a) 3^rd and 4^th
(b) 4^th and 5^th
(c) 5^th and 6^th
(d) 6^th and 7^th

Q21. The coefficients of xⁿ in the expansion of (1 + x)²ⁿ and (1 + x)²ⁿ ~¹ are in the ratio
(a) 1 : 2                   
(b) 1 : 3                  
(c) 3 : 1
(d) 2:1

Q22. If the coefficients of 2^nd, 3^rd and the 4^th terms in the expansion of (1 + x)ⁿ are in A.P., then the value of n is
(a) 2           
(b) 7 
(c) 11               
(d) 14

Q23. If A and B are coefficients of  xⁿ   in the expansions of (1 + x)²ⁿ and (1 + x)²ⁿ–¹  respectively, then A/B  equals to

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We hope the NCERT Exemplar Class 11 Maths Chapter 8 Binomial Theorem help you. If you have any query regarding NCERT Exemplar Class 11 Maths Chapter 8 Binomial Theorem, drop a comment below and we will get back to you at the earliest.

Solved CBSE Sample Paper 2021-2022 Class 11 Applied Mathematics with Solutions Term 1 & Term 2: Solving Pre Board CBSE Sample Papers for Class 11 Applied Mathematics with Solutions Answers 2021-2022 Pdf Download to understand the pattern of questions ask in the board exam. Know about the important concepts to be prepared for CBSE Class 11 Applied Mathematics board exam and Score More marks. Here we have given CBSE Class 11 Applied Mathematics Sample Papers 2022 Term 1 & Term 2.

CBSE Sample Paper 2022 Class 11 Applied Mathematics with Solutions Term 1 & Term 2

According to new CBSE Exam Pattern, MCQ Questions for Class 11 Applied Mathematics Carries 20 Marks. Click on the link below to access the CBSE Class 11 Applied Mathematics Sample Papers 2022 Solved Term 1 & Term 2.

CBSE Sample Paper 2022 Class 11 Applied Mathematics with Solutions Term 2

• CBSE Sample Papers for Class 11 Applied Mathematics Term 2 Set 1
• CBSE Sample Papers for Class 11 Applied Mathematics Term 2 Set 2
• CBSE Sample Papers for Class 11 Applied Mathematics Term 2 Set 3 for Practice
• CBSE Sample Papers for Class 11 Applied Mathematics Term 2 Set 4 for Practice
• CBSE Sample Papers for Class 11 Applied Mathematics Term 2 Set 5 for Practice

CBSE Class 11 Applied Mathematics Question Paper Design 2022

Sections	Typology of Questions	No. of Questions	Marks	Total
Section – A	Short Answer Type Questions – 1	Q1 – Q6	2 × 6	12
Section – B	Short Answer Type Questions – 2	Q7 – Q10	3 × 4	12
Section – C	Long Answer Type Questions	Q11 – Q14	4 × 4	16
Total		14 Questions		40

We hope these CBSE Sample Papers for Class 11 Applied Mathematics with Solutions 2021-2022 Term 1 & Term 2 will help in self-evaluation. Stay tuned for further updates on CBSE Sample Paper of Applied Mathematics Class 11 2022 Term 1 & Term 2 for their exam preparation.

NCERT Exemplar Class 11 Maths Chapter 16 Probability are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 16 Probability. https://www.cbselabs.com/ncert-exemplar-problems-class-11-mathematics-chapter-16-probability/

NCERT Exemplar Class 11 Maths Chapter 16 Probability

Short Answer Type Questions 

Q1. If the letters of the word ALGORITHM are arranged at random in a row what is the probability the letters GOR must remain together as a unit?
Sol: We have word ALGORITHM Number of letters = 9

Q2. Six new employees, two of whom are married to each other, are to be assigned six desks that are lined up in a row. If the assignment of employees to desks is made randomly, what is the probability that the married couple will have nonadjacent desks?
Sol: Six employees can be arranged in 6! ways.
n(S) = 6!
Two adjacent desks for married couple can be selected in 5 ways viz.,(l, 2), (2, 3), (3,4), (4, 5), (5,6).
This couple can be arranged in the two desks in 2! ways.
Other four persons can be arranged in 4! ways.
So, number of ways in which married couple occupy adjacent desks
= 5×2! x4! =2×5!
So, number of ways in which married couple occupy non-adjacent desks = 6! – 2 x 5! = 4 x 5! = n(E)

Q3. Suppose an integer from 1 through 1000 is chosen at random, find the probability that the integer is a multiple of 2 or a multiple of 9.
Sol: We have integers 1,2, 3,…1000
We have integers 1,2, 3,…1000
n(S) = 1000
Number of integers which are multiple of 2 = 500 Let the number of integers which are multiple of 9 be n.
nth term = 999 =>   9 + (n -1)9 = 999
=>             9 + 9n – 9 = 999
=>          n = 111
From 1 to 1000, the number of multiples of 9 is 111.
The multiple of 2 and 9 both are 18, 36,…, 990.
Let m be the number of terms in above series.
.’.               mth term = 990
=>             18 + (m- 1)18 = 990
=>             18+18m-18 = 990
=>        m = 55
Number of multiples of 2 or 9 = 500 +111-55 = 556 = n(E)

Q4. An experiment consists of rolling a die until a 2 appears.
(i) How many elements of the sample space correspond to the event that the 2 appears on the Ath roll of the die?
(ii) How many elements of the sample space correspond to the event that the 2 appears not later than the Ath roll of the die?
Sol: Number of outcomes when die is thrown is ‘6’.
(i) If 2 appears on the Ath roll of the die.
So, first (k -1) roll have 5 outcomes each and Kth roll results 2
Number of outcomes = 5^k-1

(ii) If we consider that 2 appears not later than K th roll of the die, then 2 comes before Ath roll.
If 2 appears in first roll, number of ways = 1 If 2 appears in second roll, number of ways
= 5 x 1 (as first roll does not result in 2)
If 2 appears in third roll, number of ways
= 5 x 5 x 1 (as first two rolls do not result in 2)
Similarly if 2 appears in (k – l)th roll, number of ways = [5x5x5… (k- 1) times] x 1 = 5^k-1 Possible outcomes if 2 appears before kth roll = 1 +5 + 5² + 5³+ … +5^k-l

Q5. A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find P(G), where G is the event that a number greater than 3 occurs on a single roll of the die.
Sol: If is given that 2 x Probability of even number = Probability of odd

Q6. In a large metropolitan area, the probabilities are .87, .36, .30 that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either anyone or both kinds of sets?
Sol: Let C be the even that family own colour television set and B be the event that family owns a black and white television set It is given that,
P(C) = 0.87, P{B) = 0.36 and P(C∩B) = 0.30 We have to find probability that a family owns either anyone or both kind of sets i.e., P(B ∪ C)
Now, P(B∪C) = P(B) + P(C)-P(C∩ B)
= 0.87 + 0.36-0.30= 0.93

Q7. If A and B are mutually exclusive events, P(A) =35 and P(B) = 0.45, find
(a) P(A’)
(b) P(B’)
(c) P(A∪ B)
(d) P(A∩ B)
(e) P(A∪ B’)                                           
(f) P(A’∩B’)

Q8. A team of medical students doing their internship have to assist during surgeries at a city hospital. The probabilities of surgeries rated as very complex, complex, routine, simple or very simple are respectively, 0.15,0.20, 0.31, 0.26, .08. Find the probabilities that a particular surgery will be rated
(a) complex or very complex;
(b) neither very complex nor very simple;
(c) routine or complex
(d) routine or simple

Q9. Four candidates A, B, C, ZJhave applied for the assignment to coach a school cricket team. If A is twice as likely to be selected as B, and B and C are given about the same chance of being selected, while C is twice as likely to be selected as D, what are the probabilities that
(a) C will be selected? (b) A will not be selected?
Sol: It is given that A is twice as likely to be selected as B.
P(A) = 2P(B)
B and C are given about the same chance of being selected.
P(B) = P(C)
C is twice as likely to be selected as D.
P(C) = 2 P(D)

Q10. One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Consequently the sample space consists of four elementary outcomes S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}. You are told that the chances of John’s promotion is same as that of Gurpreet, Rita’s chances of promotion are twice as likely as Johns. Aslam’s chances are four times that of John.
(a) Determine
P(John promoted)
P(Rita promoted)
P(Aslam promoted)
P(Gurpreet promoted)
(b) If A = {John promoted or Gurpreet promoted}, find P(A).
Sol: Let Event: J = John promoted
R = Rita promoted
A = Aslam promoted
G = Gurpreet promoted
Given sample space, S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}
i.e. S={J,R,A,G)
It is given that, chances of John’s promotion is same as that of Gurpreet.
P(J) = P(G)
Rita’s chances of promotion are twice as likely as John.
P(R) = 2P(J)
And Aslam’s chances of promotion are four times that of John.
P(A) = 4P(J)
Now, P(J) + P(R) + P(A) + P(G) = 1 => P(J) + 2P(J) + 4P(J) + P(J) = 1
=> 8P(J) = 1
P(J) = 1/8

Q11. The accompanying Venn diagram shows three events, A, B, and C, and also the probabilities of the various intersections (for instance, P(A∩B) = .07).

Q12. One urn contains two black balls (labelled Bx and B2) and one white ball. A second urn contains one black ball and two white balls (labelled W₁ and W₂). Suppose the following experiment is performed. One of the two urns is chosen at random. Next a ball is randomly chosen from the urn. Then a second ball is chosen at random from the same urn without replacing the first ball.
(a) Write the sample space showing all possible outcomes
(b) What is the probability that two black balls are chosen?
(c) What is the probability that two balls of opposite colour are chosen?
Sol:It is given that one of the two urn is chosen, then a ball is randomly chosen
from the urn, then a second ball is chosen at random from the same urn without replacing the first ball.

Q13. A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the Probability that
(a) All the three balls are white
(b) All the three balls are red
(c) One ball is red and two balls are white

Q15. A card is drawn from a deck of 52 cards. Find the probability of getting a king or a heart or a red card.
Sol: Number of cards = 52 .-.            n(S) = 52
4 king + 13 heart + 26 red – 13 – 2 = 28 = n{E)
.’.             Required probability = 28/52 = 7/13

Q17. Determine the probability p, for each of the following events.
(a) An odd number appears in a single toss of a fair die.
(b) At least one head appears in two tosses of a fair coin.
(c) The sum of 6 appears in a single toss of a pair of fair dice.
Sol: (a) When a die is thrown the possible outcomes are
S = {1, 2, 3,4, 5, 6} out of which 1, 3, 5 are odd,

Objective type Questions

Q18. In a non-leap year, the probability of having 53 Tuesdays or 53 Wednesdays is

Q19. Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive

Q20. While shuffling a pack of 52 playing cards, 2 are accidentally dropped. Find the probability that the missing cards to be of different colours

Q21. Seven persons are to be seated in a row. The probability that two particular persons sit next to each other is

Q22. Without repetition of the numbers, four digit numbers are formed with the numbers 0, 2, 3, 5. The probability of such a number divisible by 5 is

Q23. If A and B are mutually exclusive events, then

Q24. If P(A ∪B) = P(A n B) for any two events A and B, then
(a) P(A) = P(B) (b) P (A) > P (B)
(c) P(A ) < P(B) (d) none of these
Sol: (a) We have, P(A ∪ B) = P(A n B)
P(A) + P(B) – P(A ∩ B) = P(A ∩ B)

Q25. If 6 boys and 6 girls sit in a row at random. The probability that all the girls sit together is

Q26. A single letter is selected at random from the word ‘PROBABILITY’. The probability that it is a vowel is

Q27. If the probabilities for A to fail in an examination is 0.2 and that for B is 0.3, then the probability that either A or B fails is

Q29. If M and N are any two events, tlie probability that at least one of them occurs is                            .
(a) P(M) + P(N) – 2 P(M ∩N)         
(b) P(M) + P(N) – P(M ∩ N)
(c) P(M) + P(N) + P(M ∩ N)
(d) P(M) + P(N) + 2P(M∩N)
Sol: (B) If M and N are any two events.
.-. P(M ∪N) = P(M) + P(N) – P(M ∩ N) .

True/False Type Questions

Q30. The probability that a person visiting a zoo will see the giraffe is 0.72, the probability that he will see the bears is 0.84 and the probability that he will see both is 0.52.
Sol: False
P(to see giraffe or bear) = P (giraffe) + P (bear) – P(giraffe and bear)

=0.72 + 0.84-0.52= 1.04
which is not possible.

Q31. The probability that a student will pass his examination is 0.73, the probability of the student getting a compartment is 0.13, and the probability that the student will either pass or get compartment is 0.96.
Sol: False
Let A = Student will pass examination
B = Student will getting compartment
P(A) = 0.73, P(B) = 0.13 and P(A or B) = 0.96
P(A or B) = P(A) + P(B) = 0.73 + 0.13 = 0.86
But P(A or B) = 0.96
Hence, given statement is false.

Q32.The probabilities that a typist will make 0, 1,2, 3, 4, 5 or more mistakes in typing a report are, respectively, 0.12, 0.25, 0.36, 0.14, 0.08, 0.11.
Sol: False
Sum of these probabilities must be equal to 1.
P(0) + P( 1) + P( 2) + P(3) + P{ 4) + P(5)
= 0.12 + 0.25+0.36 + 0.14 + 0.08 + 0.11 = 1.06 which is greater than 1,
So, it is false statement.

Q33. If A and B are two candidates seeking admission in an engineering College. The probability that A is selected is .5 and the probability that both A and B are selected is at most .3. Is it possible that the probability of B getting selected is 0.7?

Sol. False
Given that, P(A ) = 0.5, P(A ∩B)< 0.3
Now, P(A) x P(B) ≤ 0.3
=>0.5 x P(B) ≤0.3
=> P(B) ≤0.6
Hence, it is false statement

Q34. The probability of intersection of two events A and B is always less than or equal to those favourable to the event
Sol: True
We know that A ∩ B ⊂ A
P(A ∩ B) ≤ P(A)
Hence, it is a true statement.

Q35. The probability of an occurrence of event A is .7 and that of the occurrence of event B is .3 and the probability of occurrence of both is .4.
Sol: False
A ∩B⊆ A, B
P(A ∩B) ≤ P(A), P(B)
But given that P(B) = 0.3 and P(A ∩B) = 0.4, which is not possible.

Q36. The sum of probabilities of two students getting distinction in their final examinations is 1.2.
Sol: True
Probability of each student getting distinction in their final examination is less than or equal to 1, sum of the probabilities of two may be 1.2.
Hence, it is true statement.

Fill in the Blanks Type Questions

Q37. The probability that the home team will win an upcoming football game is 0.77, the probability that it will tie the game is 0.08, and the probability that it will lose the game is ______.
Sol: P(losing) = 1 – (0.77 + 0.08) = 0.15

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We hope the NCERT Exemplar Class 11 Maths Chapter 16 Probability help you. If you have any query regarding NCERT Exemplar Class 11 Maths Chapter 16 Probability, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Maths Chapter 15 Statistics are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 15 Statistics. https://www.cbselabs.com/ncert-exemplar-problems-class-11-mathematics-chapter-15-statistics/

NCERT Exemplar Class 11 Maths Chapter 15 Statistics

Short Answer Type Questions

Q1. Find the mean deviation about the mean of the distribution:

Size	20	21	22	23	24
Frequency	6	4	5	1	4

Q2. Find the mean deviation about the median of the following distribution:

Marks obtained	10	11	12	14	15
Number of students	2	3	8	3	4

 

Q3. Calculate the mean deviation about the mean of the set of first n natural numbers when n is an odd number.
Sol: Consider first natural number when n is an odd i.e., 1, 2, 3,4,… , n [odd].

Q4. Calculate the mean deviation about the mean of the set of first n natural numbers when n is an even number.
Sol: Consider first n natural number, when n is even i.e., 1, 2, 3,4..n.

Q5. Find the standard deviation of the first n natural numbers.

Q6. The mean and standard deviation of some data for the time taken to complete . a test are calculated with the following results:
Number of observations = 25, mean = 18.2 seconds, standard deviation = 3.25 s.

Q8. Two sets each of 20 observations, have the same standard derivation 5. The first set has a mean 17 and the second a mean 22. Determine the standard deviation of the set obtained by combining the given two sets.

 

Q9. The frequency distribution:

X	A	2A	3 A	4A	5 A	6A
f	2	1	1	1	1	1

where A is a positive integer, has a variance of 160. Determine the value of A.

Q10. For the frequency distribution:

X	2	3	4	5	6	7
f	4	9	16	14	11	6

Find the standard deviation.

Q11. There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test:

Marks	0	i	2	3	4	5
Frequency	x – 2	X	x2	(x+1)2	2x	x + 1

where x is a positive integer. Determine the mean and standard deviation of the marks.

Q12. The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. Find the overall standard deviation.

Q13. Mean and standard deviation of 100 items are 50 and 4, respectively. Then find the sum of all the item and the sum of the squares of the items.

Q14. If for a distribution Σ (x -5)= 3,Σ (x -5)²= 43 and the total number of item is 18, find the mean and standard deviation.
Sol: Given, n = 18, Σ (x – 5) = 3 and Σ (x – 5)² = 43

Q15. Find the mean and variance of the frequency distribution given below:

Long Answer Type Questions
Q16. Calculate the mean deviation about the mean for the following frequency distribution:

Class interval	0-4	4-8	8-12	12-16	16-20
Frequency	4	6	8	5	2

Q17. Calculate the mean deviation from the median of the following data

Class interval	0 – 6	6 – 12	12 -18	18 -24	24 -30
Frequency	4	5	3	6	2

Q18. Determine the mean and standard deviation for the following distribution:

Marks	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16
Frequency	1	6	6	8	8	2	2	3	0	2	1	0	0	0	1

Q19. The weights of coffee in 70 jars are shown in the following table:

Weight (in grams)	Frequency
200-201	13
201-202	27
202 – 203	18
203-204	10
204-205	1
205-206	1

Determine variance and standard deviation of the above distribution.

Q20. Determine mean and standard deviation of first n terms of an A.P. whose first term is a and common difference is d.

Q21. Following are the marks obtained, out of 100, by two students Ravi and Hashinain 10 tests.

Ravi	25	50	45	30	70	42	36	48	35	60
Hashina	10	70	50	20	95	55	42	60	48	80

Who is more intelligent and who is more consistent?

Q22. Mean and standard deviation of 100 observations were found to be 40 and 10,respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.

Q23. While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.

Objective Type Questions
Q24. The mean deviation of the data 3,10, 10,4, 7, 10, 5 from the mean is (a) 2 (b) 2.57 (c) 3 (d) 3.75
Sol: (b) Given, observations are 3, 10, 10, 4, 7, 10 and 5.

Q26. When tested, the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623 The mean deviations (in hours) from their mean is (a) 178 (b) 179 (c) 220 (d) 356

Q27. Following are the marks obtained by 9 students in a mathematics test:
50, 69,20, 33, 53, 39,40, 65, 59 The mean deviation from the median is:
(a) 9 (b) 10.5 (c) 12.67 (d) 14.76
Sol: (c) Since, marks obtained by 9 students in Mathematics are 50,69,20,33,53, 39,40, 65 and 59.
Rewrite the given data in ascending order.
20, 33, 39,40, 50, 53, 59, 65, 69,
Here, n = 9 [odd]

Q30. The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is
(a) 50000 (b) 250000 (c) 252500 (d) 255000

Q31. Let a, b, c, d, e be the observations with mean m and standard deviation V. The standard deviation of the observations a + k,b + k,c + k,d+k,e + k is

Q34. Standard deviations for first 10 natural numbers is
(a) 5.5 (b) 3.87 (c) 2.97 (d) 2.87

Q35. Consider the numbers 1,2, 3,4, 5, 6, 7, 8,9,10. If 1 is added to each number, the variance of the numbers so obtained is
(a) 6.5 (b) 2.87 (c) 3.87 (d) 8.25
Sol: (d) Given numbers are 1, 2, 3,4, 5, 6, 7, 8, 9 and 10
If 1 is added to each number, then observations will be 2, 3,4, 5, 6,7, 8, 9, 10 and 11.

Q36. Consider the first 10 positive integers. If we multiply each number by -1 and then add 1 to each number, the variance of the numbers so obtained is (a) 8.25 (b) 6.5 (c) 3.87 (d) 2.87
Sol: (a) Since, the first 10 positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.
On multiplying each number by -1, we get
-1, -2, -3, -4, -5, -6, -7, -8, -9, -10 On adding 1 in each number, we get
0, -1, -2, -3, -4, -5, -6, -7, -8, -9

Q38. Coefficient of variation of two distributions are 50 and 60, and their arithmetic means are 30 and 25 respectively. Difference of their standard deviation is
(a) 0 (b) 1 (c) 1.5 (d) 2.5

Q39. The standard deviation of some temperature data in °C is 5. If the data were converted into °F, the variance would be
(a) 81 (b) 57 (c) 36 (d) 25

Fill in the Blanks

Q43. The standard deviation of a data is _____ of any change in origin, but is ________ on the change of scale.
Sol: The standard deviation of a data is independent of any change in origin but is dependent of charge of scale.
Q44. The sum of the squares of the deviations of the values of the variable is ________ when taken about their arithmetic mean.
Sol: The sum of the squares of the deviations of the values of the variable is minimum when taken about their arithmetic mean.
Q45. The mean deviation of the data is ________ when measured from the median.
Sol: The mean deviation of the data is least when measured from the median.
Q46. The standard deviation is________ to the mean deviation taken from the arithmetic mean.
Sol: The standard deviation is greater than or equal to the mean deviation taken from the arithmetic mean.

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We hope the NCERT Exemplar Class 11 Maths Chapter 15 Statistics help you. If you have any query regarding NCERT Exemplar Class 11 Maths Chapter 15 Statistics, drop a comment below and we will get back to you at the earliest.