# NCERT Solutions Class 11th Maths Chapter 1: Sets

### Exercise 1.1

1. The collection of all months of a year beginning with the letter J.
2. The collection of ten most talented writers of India.
3. A team of eleven best-cricket batsmen of the world.
4. The collection of all boys in your class.
5. The collection of all natural numbers less than 100.
6. A collection of novels written by the writer Munshi Prem Chand.
7. The collection of all even integers.
8. The collection of questions in this Chapter.
9. A collection of most dangerous animals of the world.

SOLUTION

1. The collection of all months of a year beginning with the letter J is a well-defined collection of objects as one can identify a month which belongs to this collection. Hence, this collection is a set.
2. The collection of ten most talented writers of India is not a well-defined collection as the criteria to determine a writer's talent may differ from one person to another and be subjective for different people. Hence, this collection cannot be considered a set.
3. A team of eleven best-cricket batsmen of the world is not a well-defined collection as the criteria to determine a batsman's talent may vary from one person to another and be subjective for different people. Hence, this collection cannot be considered a set.
4. The collection of all boys in your class is a well-defined collection as one can identify a boy who belongs to this collection. Hence, this collection is a set.
5. The collection of all natural numbers less than 100 is a well-defined collection as one can find a number which belongs to this collection.
Therefore, this collection is a set.
6. A collection of novels written by the writer Munshi Prem Chand is a well-defined collection as one can find any book which belongs to this collection. Therefore, this collection is a set.
7. The collection of all even integers is a well-defined collection as one can find an integer which belongs to this collection. Therefore, this collection is a set.
8. The collection of questions in this chapter is a well-defined collection as one can find a question which belongs to this chapter. Therefore, this collection is a set.
9. A collection of most dangerous animals of the world is not a well-defined collection as the criteria to find the dangerousness of an animal can differ from one animal to another. Therefore, this collection is not a set.

2. Let A = {1, 2, 3, 4, 5, 6}. Insert the appropriate symbol ∈ or ∉ in the blank spaces:

(i) 5. . .A (ii) 8 . . . A (iii) 0. . .A

(iv) 4. . . A (v) 2. . .A (vi) 10. . .A

SOLUTION

1. 5 ∈ A
2. 8 ∉ A
3. 0 ∉ A
4. 4 ∈ A
5. 2 ∈ A
6. 10 ∉ A

3. Write the following sets in roster form:

1. A = {x:xis an integer and -3 <x< 7}.
2. B = {x:xis a natural number less than 6}.
3. C = {x:xis a two-digit natural number such that the sum of its digits is 8}
4. D = {x:xis a prime number which is divisor of 60}.
5. E = The set of all letters in the word TRIGONOMETRY.
6. F = The set of all letters in the word BETTER.

SOLUTION

(i) Integers greater than -3 and smaller than 7 are -2, -1, 0, 1, 2, 3, 4, 5, 6.

Therefore, A = {-2, -1, 0, 1, 2, 3, 4, 5, 6}

(ii) Natural numbers less than 6 are 1, 2, 3, 4, 5

Therefore, B = {1, 2, 3, 4, 5}

(iii) Two-digit numbers whose digits add up to 8 are 17, 26, 35, 44, 53, 62, 71, 80.

Therefore, C = {17, 26, 35, 44, 53, 62, 71, 80}

(iv) We know that, 60 = 2 × 2 × 3 × 5

Prime numbers that are divisors of 60 are 2, 3, 5.

Therefore, D = {2, 3, 5}

(v) All unique letters in the word TRIGONOMETRY are T, R, I, G, O, N, M, E, Y.

Therefore, E = {T, R, I, G, O, N, M, E, Y}

(vi) All unique letters in the word BETTER are B, E, T, R.

Therefore, F = {B, E, T, R}

4. Write the following sets in the set-builder form :

(i) (3, 6, 9, 12} (ii) {2,4,8,16,32} (iii) {5, 25, 125, 625}

(iv) {2, 4, 6, . . .} (v) {1,4,9, . . .,100}

SOLUTION

(i) It can be observed that the elements of this set are multiples of 3.

Therefore, it can be written as {x: x = 3n, where n ∈ N and 1 ≤n ≤ 4}

(ii) It can be observed that the elements of this set are two times that of their respective previous ones.

Therefore, it can be written as {x: x = 2n, where n ∈ N and 1 ≤ n ≤ 5}

(iii) It can be observed that the elements of this set are five times that of their respective previous ones.

Therefore, it can be written as {x: x = 5n, where n ∈ N and 1 ≤ n ≤ 4}

(iv) It can be observed that the elements of this set are even natural numbers.

Therefore, it can be written as {x: x is an even natural number}

(v) It can be observed that the elements of this set are in the series 12, 22, 32 and so on upto 102.

Therefore, it can be written as {x: x = n2, where n ∈ N and 1 ≤ n ≤ 10}

5. List all the elements of the following sets:

1. A = {x:xis an odd natural number}
2. B = {x:xis an integer, -1/2 < x < 9/2}
3. C = {x:xis an integer, x2≤ 4}
4. D = {x:xis a letter in the word "LOYAL"}
5. E = {x:xis a month of a year not having 31 days}
6. F = {x:xis a consonant in the English alphabet which preceedsk}.

SOLUTION

(i) Odd natural numbers are 1, 3, 5, 7, ... and so on. Therefore,

A = {1, 3, 5, 7, ...}

(ii) Integers between -1/2 and 9/2 will be 0, 1, 2, 3, 4. Therefore,

B = {0, 1, 2, 3, 4}

(iii) Integers whose squares are less than or equal to 4 include -2, -1, 0, 1, 2. Therefore,

C = {-2, -1, 0, 1, 2}

(iv) The unique letters in the word LOYAL are L, O, Y, A. Therefore,

D = {L, O, Y, A}

(v) Months of the year not having 31 days are February, April, June, September, November. Therefore,

E = {February, April, June, September, November}

(vi) Consonants of the English alphabet that occur before k are b, c, d, f, g, h, j. Therefore,

F = {b, c, d, f, g, h, j}

6. Match each of the set on the left in the roster form with the same set on the right described in set-builder form:

(i) {1, 2, 3, 6} (a) {x : x is a prime number and a divisor of 6}

(ii) {2, 3} (b) {x : x is an odd natural number less than 10}

(iii) {M,A,T,H,E,I,C,S} (c) {x : x is natural number and divisor of 6}

(iv) {1, 3, 5, 7, 9} (d) {x : x is a letter of the word MATHEMATICS}.

SOLUTION

1. It can be observed that the elements of this set are natural numbers and divisors of 6. Therefore, correct option is (c).
2. It can be observed that the elements of this set prime numbers and divisors of 6. Therefore, correct option is (a).
3. It can be observed that elements of this set can be used to comprise the word MATHEMATICS. Therefore, correct option is (d).
4. It can be observed that elements of this set are odd natural numbers and are less than 10. Therefore, correct option is (b).

### Exercise 1.2

1. Which of the following are examples of the null set

1. Set of odd natural numbers divisible by 2
2. Set of even prime numbers
3. { x : x is a natural numbers, x < 5 and x > 7 }
4. { y : y is a point common to any two parallel lines}

SOLUTION

1. We know that only even numbers can be divisible by 2. Hence, this is an example of a null set.
2. 2 is a prime number as well as an even number. Hence, this is not an example of a null set.
3. It is impossible for a natural number to be less than 5 and greater than 7 at the same time. Hence, this is an example of a null set.
4. We know that two parallel lines never intersect each other. Hence, this is an example of a null set.

2. Which of the following sets are finite or infinite

1. The set of months of a year
2. {1, 2, 3, . . .}
3. {1, 2, 3, . . .99, 100}
4. The set of positive integers greater than 100
5. The set of prime numbers less than 99

SOLUTION

1. We know that there are only 12 months in a year. Therefore, this set is finite.
2. The number of natural numbers are infinite. Therefore, this set is infinite.
3. The given set has natural numbers upto 100 only. Therefore, this set is finite.
4. There exists an infinite number of positive integers greater than 100. Therefore, this set is infinite.
5. The number of primes numbers less than 99 are finite. Therefore, this set is finite.

3. State whether each of the following set is finite or infinite:

1. The set of lines which are parallel to the x-axis
2. The set of letters in the English alphabet
3. The set of numbers which are multiple of 5
4. The set of animals living on the earth

SOLUTION

1. An infinite number of lines can be parallel to x-axis. Therefore, this set is infinite.
2. We know that there are only 26 letters in the English alphabet. Therefore, this set is finite.
3. An infinite number of multiples of 5 exist. Therefore, this set is infinite.
4. The number of animals living in the earth are finite. Therefore, this set is finite.
5. It is possible to draw an infinite number of circles that pass through the origin. Therefore, this set is infinite.

4. In the following, state whether A = B or not:

1. A = { a, b, c, d } B = { d, c, b, a }
2. A = { 4, 8, 12, 16 } B = { 8, 4, 16, 18}
3. A = {2, 4, 6, 8, 10} B = { x : x is positive even integer and x ≤ 10}
4. A = { x : x is a multiple of 10}, B = { 10, 15, 20, 25, 30, . . . }

SOLUTION

1. The order of elements is different but the elements present in A are same as the elements present in B. Hence, A = B.
2. It can be observed that 12 ∈ A but 12 ∉ B, and 18 ∈ B but 18 ∉ A. Hence, A ≠ B.
3. A is the set of even positive integers ≤ 10 whereas B is the set of all positive integers ≤ 10. Hence, A ≠ B.
4. A is the set of multiples of 10 whereas B is the set of multiples of 5 ≥ 10. Hence, A ≠ B.

5. Are the following pair of sets equal ? Give reasons.

1. A = {2, 3}, B = {x : x is solution of x2 + 5x + 6 = 0}
2. A = { x : x is a letter in the word FOLLOW}
B = { y : y is a letter in the word WOLF}

SOLUTION

(i) First, we need to find the solution of x2 + 5x + 6 = 0.

x2 + 5x + 6 = 0

x2 + 2x + 3x + 6 = 0

x(x + 2) + 3(x + 2) = 0

(x + 2)(x + 3) = 0

x = -2 and x = -3.

Thus, B = {-2, -3}.

-2 ∉ A and -3 ∉ A.

Hence, A ≠ B.

(ii) Letters in the word FOLLOW are F, O, L, W. Thus, A = {F, O, L, W}

Letters in the word WOLF are W, O, L, F. Thus, B = {W, O, L, F}

All elements of A are present in B and all elements of B are present in A. Hence, A = B.

6. From the sets given below, select equal sets :

A = { 2, 4, 8, 12}, B = { 1, 2, 3, 4}, C = { 4, 8, 12, 14}, D = { 3, 1, 4, 2}

E = {-1, 1}, F = { 0, a}, G = {1, -1}, H = { 0, 1}

SOLUTION

All elements of B are present in D and all elements of D are present in A.

Therefore, B = D.

All elements of E are present in G and all elements of G are present in E.

Therefore, E = G.

None of the other sets have all their elements present in the other.

### Exercise 1.3

1. Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces :

1. { 2, 3, 4 } . . . { 1, 2, 3, 4,5 }
2. { a, b, c } . . . { b, c, d }
3. {x : x is a student of Class XI of your school}. . .{x : x student of your school}
4. {x : x is a circle in the plane} . . .{x : x is a circle in the same plane with radius 1 unit}
5. {x : x is a triangle in a plane} . . . {x : x is a rectangle in the plane}
6. {x : x is an equilateral triangle in a plane} . . . {x : x is a triangle in the same plane}
7. {x : x is an even natural number} . . . {x : x is an integer}

SOLUTION

1. {2, 3, 4} ⊂ {1, 2, 3, 4, 5}
2. {a, b, c} ⊄ {b, c, d}
3. {x:xis a student of Class XI of your school} ⊂ {x:xstudent of your school}
4. {x:x is a circle in the plane} ⊄ {x:x is a circle in the same plane with radius 1 unit}
5. {x:x is a triangle in a plane} ⊄ {x:xis a rectangle in the plane}
6. {x:xis an equilateral triangle in a plane} ⊂ {x:xis a triangle in the same plane}
7. {x:xis an even natural number} ⊂ {x:xis an integer}

2. Examine whether the following statements are true or false:

1. { a, b } ⊄ { b, c, a }
2. { a, e } ⊂ { x : x is a vowel in the English alphabet}
3. { 1, 2, 3 } ⊂ { 1, 3, 5 }
4. { a } ⊂ { a, b, c }
5. { a } ∈ { a, b, c }
6. { x : x is an even natural number less than 6} ⊂ { x : x is a natural number which divides 36}

SOLUTION

1. It can be observed that each element of {a, b} is present in {b, c, a}. Hence, the statement is false.
2. a and e of the set {a, e} are vowels in the English alphabet. Hence, the statement is true.
3. It can be observed that 2 ∉ {1, 3, 5}. Hence, the statement is false.
4. It can be observed that each element of {a} is present in {a, b, c}. Hence, the statement is true.
5. {a} is not an element in {a, b, c}. Hence, the statement is false.
6. First set = {2, 4}, Second set = {2, 3, 4, 6, 9, 12, 18, 36}.

Hence, the statement is true.

3. Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?

(i) {3, 4} ⊂ A (ii) {3, 4} ∈ A (iii) {{3, 4}} ⊂ A

(iv) 1 ∈ A (v) 1 ⊂ A (vi) {1, 2, 5} ⊂ A

(vii) {1, 2, 5} ∈ A (viii) {1, 2, 3} ⊂ A (ix) φ ∈ A

(x) φ ⊂ A (xi) {φ} ⊂ A

SOLUTION

1. This statement is incorrect as the set A does not contain 3 and 4 as its elements.
2. This statement is correct as the set A does contain {3, 4} as an element.
3. This statement is correct as elements of {{3, 4}} are present in the set A.
4. This statement is correct as the set A does contain 1 as an element.
5. This statement is incorrect as 1 is an element of set A and not a subset of itself.
6. This statement is correct as elements of {1, 2, 5} are present in the set A.
7. This statement is incorrect as the set A does not contain {1, 2, 5} as an element.
8. This statement is incorrect as the set A does not contain 3 as an element.
9. This statement is incorrect as the set A does not contain φ as an element.
10. This statement is correct as φ is the subset of all sets.
11. This statement is incorrect as the set A does not contain {φ} as an element.

4. Write down all the subsets of the following sets

1. {a}
2. {a, b}
3. {1, 2, 3}
4. φ

SOLUTION

1. Possible subsets of {a} are φ and {a}.
2. Possible subsets of {a, b} are φ, {a}, {b}, and {a, b}.
3. Possible subsets of {1, 2, 3} are φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, and {1, 2, 3}.
4. φ is the only possible subset of φ.

5. How many elements has P(A), if A = φ?

SOLUTION

Let m be the number of elements in the set A.

Now, n(A) = m. Therefore,

n[P(A)] = 2m

It is given that A = φ. So,

n(A) = m = 0

n[P(A)] = 20 = 1

Hence, P(A) has 1 element.

6. Write the following as intervals :

(i) {x : x ∈ R, - 4 < x ≤ 6} (ii) {x : x ∈ R, - 12 < x < -10}

(iii) {x : x ∈ R, 0 ≤ x < 7} (iv) {x : x ∈ R, 3 ≤ x ≤ 4}

SOLUTION

1. Since, -4 < x ≤ 6. Therefore, the intervals for it will be (-4, 6].
2. Since, -12 < x < -10. Therefore, the intervals for it will be (-12, -10).
3. Since, 0 ≤ x < 7. Therefore, the intervals for it will be [0, 7).
4. Since, 3 ≤ x ≤ 4. Therefore, the intervals for it will be [3, 4].

7. Write the following intervals in set-builder form :

(i) (- 3, 0) (ii) [6 , 12] (iii) (6, 12] (iv) [-23, 5)

SOLUTION

1. {x: x ∈ R, -3 < x < 0}
2. {x: x ∈ R, 6 ≤ x ≤ 12}
3. {x: x ∈ R, 6 < x ≤ 12}
4. {x: x ∈ R, -23 ≤ x < 5}

8. What universal set(s) would you propose for each of the following :

(i) The set of right triangles. (ii) The set of isosceles triangles.

SOLUTION

1. A set of polygons or triangles will be a universal set as all the right triangles will be elements of it.
2. A set of polygons or triangles will be a universal set as all the isosceles triangles will be elements of it.

9. Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the following may be considered as universal set (s) for all the three sets A, B and C

1. {0, 1, 2, 3, 4, 5, 6}
2. φ
3. {0,1,2,3,4,5,6,7,8,9,10}
4. {1,2,3,4,5,6,7,8}

SOLUTION

1. {0, 1, 2, 3, 4, 5, 6} cannot be considered a universal set for sets A, B and C because the element 8 of the set C is not present in {0, 1, 2, 3, 4, 5, 6}.
2. φ cannot be considered a universal set for sets A, B and C because none of the elements of any of the three sets is present in φ.
3. {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} can be considered a universal set for sets A, B and C as all elements of all the three sets are present in {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
4. {1, 2, 3, 4, 5, 6, 7, 8} cannot be considered a universal set for sets A, B and C because the element 0 of the set C is not present in {1, 2, 3, 4, 5, 6, 7, 8}.

### Exercise 1.4

1. Find the union of each of the following pairs of sets :

1. X = {1, 3, 5} Y = {1, 2, 3}
2. A = [ a, e, i, o, u} B = {a, b, c}
3. A = {x : x is a natural number and multiple of 3}
B = {x : x is a natural number less than 6}
4. A = {x : x is a natural number and 1 < x ≤6 }
B = {x : x is a natural number and 6 < x < 10 }
5. A = {1, 2, 3}, B = φ

SOLUTION

(i) X ∪ Y = {1, 2, 3, 5}

A ∪ B = {a, b, c, e, i, o , u}

(iiI) A = {3, 6, 9, ...}

B = {1, 2, 3, 4, 5}

A ∪ B = {1, 2, 3, 4, 5, 6, 9, 12, ...}

(iv) A = {2, 3, 4, 5, 6}

B = {7, 8, 9}

A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9}

(v) A ∪ B = {1, 2, 3}

2. Let A = { a, b }, B = {a, b, c}. Is A ⊂ B ? What is A ∪ B ?

SOLUTION

Each element of the set A is present in set B. Therefore, A ⊂ B.

A ∪ B = {a, b, c} = B as A is the subset of B.

3. If A and B are two sets such that A ⊂ B, then what is A ∪ B ?

SOLUTION

If A ⊂ B, then A ∪ B will be the same as set B as it is the superset.

4. If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8 }and D = { 7, 8, 9, 10 }; find

(i) A ∪ B (ii) A ∪ C (iii) B ∪ C (iv) B ∪ D

(v) A ∪ B ∪ C (vi) A ∪ B ∪ D (vii) B ∪ C ∪ D

SOLUTION

1. A∪B = {1, 2, 3, 4, 5, 6}
2. A∪C = {1, 2, 3, 4, 5, 6, 7, 8}
3. B∪C = {3, 4, 5, 6, 7, 8}
4. B∪D = {3, 4, 5, 6, 7, 8, 9, 10}
5. A ∪ B ∪ C= {1, 2, 3, 4, 5, 6} ∪ {5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8}
6. A ∪ B ∪ D= {1, 2, 3, 4, 5, 6} ∪ {7, 8, 9, 10} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
7. B ∪ C ∪ D= {3, 4, 5, 6, 7, 8} ∪ {7, 8, 9, 10} = {3, 4, 5, 6, 7, 8, 9, 10}

5. Find the intersection of each pair of sets of question 1 above.

[Pairs of Sets From Question 1:

1. X = {1, 3, 5} Y = {1, 2, 3}
2. A = [ a, e, i, o, u} B = {a, b, c}
3. A = {x : x is a natural number and multiple of 3}
B = {x : x is a natural number less than 6}
4. A = {x : x is a natural number and 1 < x ≤6 }
B = {x : x is a natural number and 6 < x < 10 }
5. A = {1, 2, 3}, B = φ]

SOLUTION

(i) X ∩ Y = {1, 3}

(ii) A ∩ B = {a}

(iiI) A = {3, 6, 9, ...}

B = {1, 2, 3, 4, 5}

A ∩ B = {3}

(iv) A = {2, 3, 4, 5, 6}

B = {7, 8, 9}

A ∩ B = φ

(v) A ∩ B = φ

6. If A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, C = {11, 13, 15}and D = {15, 17}; find

(i) AB (ii) B ∩ C (iii) A ∩ C ∩ D

(iv) A ∩ C (v) B ∩ D (vi) A ∩ (B ∪ C)

(vii) A ∩ D (viii) A ∩ (B ∪ D) (ix) ( A ∩ B ) ∩ ( B ∪ C )

(x) ( A ∪ D) ∩ ( B ∪ C)

SOLUTION

1. A∩B = {7, 9, 11}
2. B∩C = {11, 13}
3. A∩C∩D = {11}∩{15, 17} =φ
4. A∩C = {11}
5. B∩D =φ
6. A∩(B∪C) = {7, 9, 11}∪{11} = {7, 9, 11}
7. A∩D =φ
8. A∩(B∪D) = {7, 9, 11}∪φ = {7, 9, 11}
9. (A∩B)∩(B∪C) = {7, 9, 11}∩{7, 9, 11, 13, 15}
= {7, 9, 11}
10. (A∪D)∩(B∪C) = {3, 5, 7, 9, 11, 15, 17)∩{7, 9, 11, 13, 15}
= {7, 9, 11, 15}

7. If A = {x : x is a natural number }, B = {x : x is an even natural number} C = {x : x is an odd natural number} and D = {x : x is a prime number }, find

(i) A ∩ B (ii) A ∩ C (iii) A ∩ D

(iv) B ∩ C (v) B ∩ D (vi) C ∩ D

SOLUTION

A = {1, 2, 3, 4, ...}

B = {2, 4, 6, 8, ...}

C = {1, 3, 5, 7, ...}

D = {2, 3, 5, 7, ...}

1. A ∩ B = {2, 4, 6, 8, ...} = B
2. A ∩ C = {1, 3, 5, 7, ...} = C
3. A ∩ D = {2, 3, 5, 7, ...} = D
4. B ∩ C = φ
5. B ∩ D = {2}
6. C ∩ D = {3, 5, 7, 11, ...} = {x : x is an odd prime number}

8. Which of the following pairs of sets are disjoint

1. {1, 2, 3, 4} and {x : x is a natural number and 4 ≤ x ≤ 6 }
2. { a, e, i, o, u } and { c, d, e, f }
3. {x : x is an even integer } and {x : x is an odd integer}

SOLUTION

(i) {x:xis a natural number and 4≤x≤6} = {4, 5, 6}

{1, 2, 3, 4} ∩ {4, 5, 6} = {4}

Since, the intersection of the two sets is not φ. Therefore, the given pair of sets are not disjoint.

(ii) {a, e, I, o, u} ∩ {c, d, e, f} = {e}

Since, the intersection of the two sets is not φ. Therefore, the given pair of sets are not disjoint.

(iii) {x : x is an even integer} = {..., -4, -2, 0, 2, 4, ...}

{x : x is an odd integer} = {..., -3, -1, 1, 3, ...}

{..., -4, -2, 0, 2, 4, ...} ∩ {..., -3, -1, 1, 3, ...} = φ

Since, the intersection of the two sets is φ. Therefore, the given pair of sets are disjoint.

9. If A = {3, 6, 9, 12, 15, 18, 21}, B = { 4, 8, 12, 16, 20 }, C = { 2, 4, 6, 8, 10, 12, 14, 16 }, D = {5, 10, 15, 20 }; find

(i) A - B (ii) A - C (iii) A - D (iv) B - A

(v) C - A (vi) D - A (vii) B - C (viii) B - D

(ix) C - B (x) D - B (xi) C - D (xii) D - C

SOLUTION

1. A - B = {3, 6, 9, 15, 18, 21}
2. A - C = {3, 9, 15, 18, 21}
3. A - D = {3, 6, 9, 12, 18, 21}
4. B - A = {4, 8, 16, 20}
5. C - A = {2, 4, 8, 10, 14, 16}
6. D - A = {5, 10, 20}
7. B - C = {20}
8. B - D = {4, 8, 12, 16}
9. C - B = {2, 6, 10, 14}
10. D - B = {5, 10, 15}
11. C - D = {2, 4, 6, 8, 12, 14, 16}
12. D - C = {5, 15, 20}

10. If X = { a, b, c, d } and Y = { f, b, d, g}, find

(i) X - Y (ii) Y - X (iii) X ∩ Y

SOLUTION

1. X - Y = {a,c}
2. Y - X = {f,g}
3. X∩Y = {b,d}

11. If R is the set of real numbers and Q is the set of rational numbers, then what is R - Q?

SOLUTION

The set of real numbers is broadly made up of rational and irrational numbers. Therefore, R - Q will be the set of irrational numbers.

12. State whether each of the following statement is true or false. Justify your answer.

1. { 2, 3, 4, 5 } and { 3, 6} are disjoint sets.
2. { a, e, i, o, u } and { a, b, c, d }are disjoint sets.
3. { 2, 6, 10, 14 } and { 3, 7, 11, 15} are disjoint sets.
4. { 2, 6, 10 } and { 3, 7, 11} are disjoint sets.

SOLUTION

(i) {2, 3, 4, 5} ∩ {3, 6} = {3}

Since, the intersection of the two sets is not φ. Therefore, the given pair of sets are not disjoint.

Hence, the statement is false.

(ii) {a, e, i, o, u} ∩ {a, b, c, d} = {a}

Since, the intersection of the two sets is not φ. Therefore, the given pair of sets are not disjoint.
Hence, the statement is false.

(iii) {2, 6, 10, 14} ∩ {3, 7, 11, 15} = φ

Since, the intersection of the two sets is φ. Therefore, the given pair of sets are disjoint.
Hence, the statement is true.

(iv) {2, 6, 10} ∩ {3, 7, 11} = φ

Since, the intersection of the two sets is φ. Therefore, the given pair of sets are disjoint.
Hence, the statement is true.

### Exercise 1.5

1. Let U = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = { 1, 2, 3, 4}, B = { 2, 4, 6, 8 } and C = { 3, 4, 5, 6 }. Find

1. A?
2. B?
3. (A ∪ C)?
4. (A ∪ B)?
5. (A?)?
6. (B - C)?

SOLUTION

(i) A' = {5, 6, 7, 8, 9}

(ii) B' = {1, 3, 5, 7, 9}

(iii) A U C = {1, 2, 3, 4, 5, 6}

(A U C)' = {7, 8, 9}

(iv) A U B = {1, 2, 3, 4, 6, 8}

(A U B)' = {5, 7, 9}

(v) (A')' = A = {1, 2, 3, 4}

(vi) B - C = {2, 8}

(B - C)' = {1, 3, 4, 5, 6, 7, 9}

2. If U = { a, b, c, d, e, f, g, h}, find the complements of the following sets :

(i) A = {a, b, c} (ii) B = {d, e, f, g}

(iii) C = {a, c, e, g} (iv) D = { f, g, h, a}

SOLUTION

(i)A = {a, b, c}

A' = {d, e, f, g, h}

(ii)B = {d, e, f, g}

B' = {a, b, c, h}

(iii)C = {a, c, e, g}

C' = {b, d, f, h}

(iv)D = {f,g,h,a}

D' = {b, c, d, e}

3. Taking the set of natural numbers as the universal set, write down the complements of the following sets:

(i) {x : x is an even natural number} (ii) { x : x is an odd natural number }

(iii) {x : x is a positive multiple of 3} (iv) { x : x is a prime number }

(v) {x : x is a natural number divisible by 3 and 5}

(vi) { x : x is a perfect square } (vii) { x : x is a perfect cube}

(viii) { x : x + 5 = 8 } (ix) { x : 2x + 5 = 9}

(x) { x : x ≥ 7 } (xi) { x : x ∈ N and 2x + 1 > 10 }

SOLUTION

(i) {x : x is an even natural number} = {2, 4, 6, 8, ...}

{2, 4, 6, 8, ...}' = {1, 3, 5, 7, ...}

= {x : x is an odd natural number}

(ii) {x : x is an odd natural number} = {1, 3, 5, 7, ...}

{1, 3, 5, 7, ...}' = {2, 4, 6, 8, ...}

= {x : x is an even natural number}

(iii) {x : x is a positive multiple of 3} = {3, 6, 9, 12, ...}

{3, 6, 9, 12, ...}' = {1, 2, 4, 5, 7, 8, ...}

= {x : x is not a multiple of 3}

(iv) {x : x is a prime number} = {2, 3, 5, 7, ...}

{2, 3, 5, 7, ...}' = {1, 4, 6, 8, ...}

= {x : x is a composite number or x = 1}

(v) {x : x is a natural number divisible by 3 and 5} = {15, 30, 45, 60, ...}

{15, 30, 45, 60, ...}' = {1, 2, 3, 4, ..., 13, 14, 16, ...}

= {x : x is a natural number not divisible by 3 or 5}

(vi) {x : x is a perfect square} = {1, 4, 9, 16, ...}

{1, 4, 9, 16, ...}' = {2, 3, 5, 6, ...}

= {x : x is not a perfect square}

(vii) {x : x is a perfect cube} = {1, 8, 27, 64, ...}

{1, 8, 27, 64, ...}' = {2, 3, 4, 5, 6, 7, 9, ...}

(viii) x + 5 = 8

x = 3

{x : x + 5 = 8}' = {x : x ∈ N where x ≠ 3}

(ix) 2x + 5 = 9

2x = 4

x = 2

{x : 2x + 5 = 9}' = {x : x ∈ N where x ≠ 2}

(x) {x : x ≥ 7}' = {x : x < 7}

(xi) 2x + 1 > 10

2x > 9

x > 9/2

{x :x∈ N and 2x+ 1 > 10}´ = {x:x∈ N wherex≤ 9/2}

4. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8} and B = { 2, 3, 5, 7}. Verify that

(i) (A ∪ B)? = A? ∩ B? (ii) (A ∩ B)? = A? ∪ B?

SOLUTION

(i) LHS = (A ∪ B)'

= {2, 3, 4, 5, 6, 7, 8}' = {1, 9}

RHS = A' ∩ B'

= {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9}

= {1, 9}

LHS = RHS. Hence, verified.

(ii) LHS = (A ∩ B)'

= {2}' = {1, 3, 4, 5, 6, 7, 8, 9}

RHS = A' ∪ B'

= {1, 3, 5, 7, 9} ∪ {1, 4, 6, 8, 9}

= {1, 3, 4, 5, 6, 7, 8, 9}

LHS = RHS. Hence, verified.

5. Draw appropriate Venn diagram for each of the following :

(i) (A ∪ B)?, (ii) A? ∩ B?, (iii) (A ∩ B)?, (iv) A? ∪ B?

SOLUTION

(i)

(ii)

(iii)

(iv)

6. Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A??

SOLUTION

A is the set of all triangles with at least one angle different from 60°.

Then, A' will be the set of all triangles with no angle different from 60°.

Hence, A' will be the set of all equilateral triangles.

7. Fill in the blanks to make each of the following a true statement :

(i) A ∪ A? = . . . (ii) φ? ∩ A = . . .

(iii) A ∩ A? = . . . (iv) U? ∩ A = . . .

SOLUTION

1. A ∪ A' = U
2. φ? ∩ A = U ∩ A = A
3. A ∩ A' = φ
4. U' ∩ A = φ ∩ A = φ

### Exercise 1.6

1. If X and Y are two sets such that n ( X ) = 17, n ( Y ) = 23 and n ( X ∪ Y ) = 38, find n ( X ∩ Y ).

SOLUTION

We know that

n (X U Y) = n (X) + n (Y) - n (X ∩ Y)

38 = 17 + 23 - n (X ∩ Y)

38 = 40 - n (X ∩ Y)

-2 = -n (X ∩ Y)

n (X ∩ Y) = 2

2. If X and Y are two sets such that X ∪ Y has 18 elements, X has 8 elements and Y has 15 elements ; how many elements does X ∩ Y have?

SOLUTION

We know that

n (X U Y) = n (X) + n (Y) - n (X ∩ Y)

18 = 8 + 15 - n (X ∩ Y)

18 = 23 - n (X ∩ Y)

-5 = -n (X ∩ Y)

n (X ∩ Y) = 5

3. In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?

SOLUTION

Let the set of people who can speak hindi be X and let the set of people who can speak English be Y.

n (X) = 250

n (Y) = 200

n (X ∪ Y) = 400

We know that

n (X U Y) = n (X) + n (Y) - n (X ∩ Y)

400 = 250 + 200 - n (X ∩ Y)

400 = 450 - n (X ∩ Y)

-50 = -n (X ∩ Y)

n (X ∩ Y) = 50

Hence, 50 people can speak both Hindi and English.

4. If S and T are two sets such that S has 21 elements, T has 32 elements, and S ∩ T has 11 elements, how many elements does S ∪ T have?

SOLUTION

We know that

n (S U T) = n (S) + n (T) - n (S ∩ T)

n (S U T) = 21 + 32 - 11

n (S ∪ T) = 42

5. If X and Y are two sets such that X has 40 elements, X ∪ Y has 60 elements and X ∩ Y has 10 elements, how many elements does Y have?

SOLUTION

We know that

n (X U Y) = n (X) + n (Y) - n (X ∩ Y)

60 = 40 + n (Y) - 10

60 = 30 + n (Y)

n (Y) = 30

6. In a group of 70 people, 37 like coffee, 52 like tea and each person likes at least one of the two drinks. How many people like both coffee and tea?

SOLUTION

Let the set of people who like coffee be X and let the set of people who like tea be Y.

n (X) = 37

n (Y) = 52

n (X ∪ Y) = 70

We know that

n (X U Y) = n (X) + n (Y) - n (X ∩ Y)

70 = 37 + 52 - n (X ∩ Y)

70 = 89 - n (X ∩ Y)

-19 = -n (X ∩ Y)

n (X ∩ Y) = 19

Hence, 19 people like both tea and coffee.

7. In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?

SOLUTION

Let the set of people who like cricket be X and let the set of people who like tennis be Y.

n (X) = 40

n (X ∪ Y) = 65

n (X ∩ Y) = 10

We know that

n (X U Y) = n (X) + n (Y) - n (X ∩ Y)

65 = 40 + n (Y) - 10

65 = 30 + n (Y)

n (Y) = 35

Therefore, 35 people like tennis.

Number of people who like tennis only and not cricket = n (Y - X)

n (Y - X) = n (Y) - n (X ∩ Y)

n (Y - X) = 35 - 10 = 25

Hence, 25 people like only tennis and not cricket, whereas 35 people like tennis.

8. In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speak at least one of these two languages?

SOLUTION

Let the set of people who can speak French be X and let the set of people who can speak Spanish be Y.

n (X) = 50

n (Y) = 20

n (X ∩ Y) = 10

We know that

n (X U Y) = n (X) + n (Y) - n (X ∩ Y)

n (X U Y) = 50 + 20 - 10

n (X ∪ Y) = 60

Hence, 60 people can speak at least one of the two languages.

### Miscellaneous Exercise

1. Decide, among the following sets, which sets are subsets of one and another:

A = { x : x ∈ R and x satisfy x2 - 8x + 12 = 0 },

B = { 2, 4, 6 }, C = { 2, 4, 6, 8, . . . }, D = { 6 }.

SOLUTION

x2 - 8x + 12 = 0

x2 - 6x - 2x + 12 = 0

x(x - 6) - 2(x - 6) = 0

(x - 6)(x - 2) = 0

x = 6 and x = 2.

Therefore, A = {2, 6}.

Now, B = {2, 4, 6 }, C = {2, 4, 6, 8, . . . }, and D = {6}.

Hence, we can conclude:

D ⊂ A, D ⊂ B, D ⊂ C,

A ⊂ B, A ⊂ C

B ⊂ C

2. In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.

1. If x ∈ A and A ∈ B , then x ∈ B
2. If A ⊂ B and B ∈ C , then A ∈ C
3. If A ⊂ B and B ⊂ C , then A ⊂
4. If A ⊄ B and B ⊄ C , then A ⊄ C
5. If x ∈ A and A ⊄ B , then x ∈ B
6. If A ⊂ B and x ∉ B , then x ∉ A

SOLUTION

(i) This statement is false.

Let us assume that A = {1, 2, 3} and B = {{1, 2, 3}, 4}

Here, 1 ∈ A and A ∈ B. However, 1 ∉ B.

(ii) This statement is false.

Let us assume that A = {1, 2}, B = {1, 2, 3} and C = {{1, 2, 3}, 4}

Here, A ⊂ B and B ∈ C. However, A ∉ C.

(iii) This statement is true.

It is given that A ⊂ B and B ⊂ C. Therefore,

For any element x of the set A: x ∈ A, then x ∈ B and x ∈ C.

Thus, A ⊂ C.

(iv) This statement is false.

Let us assume that A = {1, 2}, B = {3, 4} and C = {1, 2, 3}.

Here, A ⊄ B and B ⊄ C. However, A ⊂ C.

(v) This statement is false.

Let us assume that A = {1, 2, 3} and B = {4, 5, 6}.

Here, 1 ∈ A and A ⊄ B. However, 1 ∉ B.

(vi) This statement is true.

It is given that A ⊂ B and x ∉ B.

If x ∈ A is true, then x ∈ B is also true as A ⊂ B.

But x ∉ B.

Hence, x ∉ A.

3. Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C.

SOLUTION

If x ∈ B for any element x, then

x ∈ A ∪ B

Since, A ∪ B = A ∪ C. Therefore,

x ∈ A ∪ C

This means that either set A or set C have x.

If x ∈ A, then

x ∈ A ∩ B as x ∈ B.

Since, A ∩ B = A ∩ C. Therefore,

x ∈ A ∩ C

This means that set C has x.

Since, x ∈ B and x ∈ C. Therefore, B ⊂ C.

Similarly, C ⊂ B.

Since, B ⊂ C and C ⊂ B.

Hence, B = C.

4. Show that the following four conditions are equivalent :

(i) A ⊂ B (ii) A - B = φ (iii) A ∪ B = B (iv) A ∩ B = A

SOLUTION

Equivalence in (i) and (ii):

Let A ⊂ B be true.

Now, let us assume that A - B ≠ φ.

This means that, an element x exists such that x ∈ A and x ∉ B. But this is not possible as A ⊂ B, so every element in A has to be in B.

Therefore, A - B = φ

Hence, (i) and (ii) conditions are equivalent.

Equivalence in (i) and (iii):

Let A ⊂ B be true.

We know that, B ⊂ A ∪ B.

For any element x of the set A ∪ B, we have

x ∈ A ∪ B

So, x ∈ A or x ∈ B.

I. If x ∈ A, then x ∈ B because A ⊂ B.

So, A ∪ B ⊂ B

II. If x ∈ B, then A ∪ B = B as B is the superset.

Now, let us assume that A ∪ B = B.

If x ∈ A, then

x ∈ A ∪ B because clearly A ⊂ A ∪ B.

and x ∈ B because A ∪ B = B.

Therefore, A ⊂ B.

Hence, (i) and (iii) conditions are equivalent.

Equivalence in (i) and (iv):

Let A ⊂ B be true.

A ∩ B ⊂ A

If x ∈ A then x ∈ B as A ⊂ B.

So, x ∈ A ∩ B

This means that A ⊂ A ∩ B.

Since, A ∩ B ⊂ A and A ⊂ A ∩ B. Therefore, A = A ∩ B.

Now, let us assume that A ∩ B = B.

If x ∈ A, then

x ∈ A ∩ B

So, x is an element of A as well as B.

x ∈ A and x ∈ B.

Therefore, A ⊂ B.

Hence, (i) and (iv) conditions are equivalent.

Hence, (i), (ii), (iii), and (iv) conditions are all equivalent.

5. Show that if A ⊂ B, then C - B ⊂ C - A.

SOLUTION

Let there be an element x in the set C - B.

This means that x ∈ C and x ∉ B.

Since, A ⊂ B. Therefore,

x ∈ C and x ∉ A.

This implies that

x ∈ C - A.

Hence, C - B ⊂ C - A.

6. Assume that P ( A ) = P ( B ). Show that A = B.

SOLUTION

Let there be an element x in the set A.

We know that P (A) contains all the subsets of set A.

So, A ∈ P (A).

P (A) = P (B)

Thus, A ∈ P (B)

Let there be a subset C of the set B such that x ∈ C.

We know that P (B) contains all the subsets of set B.

So, C ∈ P (B).

This implies that C ⊂ B.

Therefore,

x ∈ B which further implies that A ⊂ B.

Similarly, we can prove that B ⊂ A.

Hence, A = B.

7. Is it true that for any sets A and B, P ( A ) ∪ P ( B ) = P ( A ∪ B )? Justify your answer.

SOLUTION

The given statement is false for any sets A and B.

P (A) ∪ P (B) ≠ P (A ∪ B)

Let us assume that there are two sets A = {1, 2, 3} and B = {3, 4, 5}.

A ∪ B = {1, 2, 3, 4, 5}

P (A) = {φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}

P (B) = {, {3}, {4}, {5}, {3, 4}, {4, 5}, {3, 5}, {3, 4, 5}}

LHS = P (A) ∪ P (B)

= {φ, {1}, {2}, {3}, {4}, {5}, {1, 2}, {2, 3}, {1, 3}, {3, 4}, {4, 5}, {3, 5}, {1, 2, 3}, {3, 4, 5}}

RHS = P (A ∪ B)

= {φ, {1}, {2}, {3}, {4}, {5}, {1, 2}, {2, 3}, {3, 4}, {4, 5}, {1, 3}, {1, 4}, {1, 5}, {2, 4}, {2, 5}, {3, 5}, {1, 2, 3}, {2, 3, 4}, {3, 4, 5}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 5}, {2, 4, 5}, {1, 2, 3, 4}, {2, 3, 4, 5}, {1, 2, 3, 5}, {1, 3, 4, 5}, {1, 2, 4, 5}, {1, 2, 3, 4, 5}}

LHS ≠ RHS.

Hence, proved that the given statement is not true.

8. Show that for any sets A and B,

A = ( A ∩ B ) ∪ ( A - B ) and A ∪ ( B - A ) = ( A ∪ B )

SOLUTION

Let there be an element x such that x ∈ A.

CASE I: If x ∈ A ∩ B

Then,

x ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A - B)

CASE II: If x ∉ A ∩ B

Then,

Either x ∉ A or x ∉ B. But we know that x ∈ A. Thus,

x ∉ B

x ∉ (A - B) ⊂ (A ∪ B) ∪ (A - B)

Therefore,

A ⊂ (A ∩ B) ∪ (A - B)

Since, A ∩ B ⊂ A and (A - B) ⊂ A.

Then, (A ∩ B) ∪ (A - B) ⊂ A

From the above two obtained statements, we can conclude that

A = (A ∩ B) ∪ (A - B)

The first statement has been proven.

Now, let us assume that x ∈ A ∪ (B - A).

This implies that x ∈ A or x ∈ (B - A)

x ∈ A or [x ∈ B and x ∉ A]

x ∈ B ∪ A

Therefore, A ∪ (B - A) ⊂ (A ∪ B)

Let there be an element y such that y ∈ A ∪ B

y ∈ A or y ∈ B

y ∈ A or [y ∈ B and y ∉ A]

y ∈ A ∪ (B - A)

Therefore, A ∪ B ⊂ A ∪ (B - A)

From the above two obtained statements, we can conclude that

A ∪ (B - A) = (A ∪ B)

Hence, both the required statements have been proven.

9. Using properties of sets, show that

(i) A ∪ ( A ∩ B ) = A (ii) A ∩ ( A ∪ B ) = A

SOLUTION

(i) It is obvious that A ⊂ A. Then,

A ∩ B ⊂ A

A ∪ (A ∩ B) ⊂ A

Also, A ⊂ A ∪ (A ∩ B)

Therefore, A = A ∪ (A ∩ B)

(ii) A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B)

= A ∪ (A ∩ B)

= A

Therefore, A = A ∪ (A ∩ B)

10. Show that A ∩ B = A ∩ C need not imply B = C.

SOLUTION

Let us assume that there are three sets A = {1, 2}, B = {1, 3, 4} and C = {1, 5, 6}.

Now,

A ∩ B = {1}

A ∩ C = {1}

Therefore, A ∩ B = A ∩ C

3 and 4 ∈ B but 3 and 4 ∉ C.

Hence, B ≠ C.

11. Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B.

(Hints A = A ∩ ( A ∪ X ) , B = B ∩ ( B ∪ X ) and use Distributive law )

SOLUTION

To Prove: A = B

Proof:

A = A ∩ (A ∪ X) = A ∩ (B ∪ X) [A ∪ X = B ∪ X]

A = (A ∩ B) ∪ (A ∩ X) (Using the Distributive law)

A = (A ∩ B) ∪ Φ [A ∩ X = Φ]

A = A ∩ B ... (I)

Now, B = B ∩ (B ∪ X)

B = B ∩ (A ∪ X) [A ∪ X = B ∪ X]

B = (B ∩ A) ∪ (B ∩ X) (Using the Distributive law)

B = (B ∩ A) ∪ Φ [B ∩ X = Φ]

B = A ∩ B ... (II)

From Equations (I) and (II)

A = B

Hence, proved.

12. Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = φ.

SOLUTION

Let there be three sets A = {1, 2}, B = {2, 3}, and C = {1, 3}.

A ∩ B = {2}

B ∩ C = {3}

A ∩ C = {1}

A ∩ B ∩ C = φ

Therefore, sets A = {1, 2}, B = {2, 3} and C = {1, 3} are example of three sets such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = φ.

13. In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?

SOLUTION

Let the set of students who were taking tea be T and let the set of students who were taking coffee be C.

The 600 students who took part in the survey form the universal set denoted by U.

n (U) = 600

n (T) = 150

n (C) = 225

n (T ∩ C) = 100

We know that

n (T ∪ C) = n (T) + n (C) - n (T ∩ C)

= 150 + 225 - 100

n (T ∪ C) = 275

Number of students who are not taking either tea or coffee = Total number of students -

Number of students who are taking either tea or coffee

n (T' ∩ C')= n (U) - n (T ∪ C)

= 600 - 275

= 325

Hence, number of students who take neither tea nor coffee = 325

14. In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?

SOLUTION

Let the set with all students in the group be U.

Let the set of students who know Hindi be X and those who know English be Y.

n (X) = 100

n (Y) = 50

n (X ∩ Y) = 25

n (X ∪ Y) = n (U) (Given)

We know that

n (X ∪ Y) = n (X) + n (Y) - n (X ∩ Y)

= 100 + 50 - 25

= 125

Hence, there were a total of 125 students in the group.

15. In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find:

1. the number of people who read at least one of the newspapers.
2. the number of people who read exactly one newspaper.

SOLUTION

Let the set of people who read newspaper H be A, those who read T be B and those who read I be C.

n (A) = 25

n (B) = 26

n (C) = 26

n (A ∩ C) = 9

n (B ∩ C) = 8

n (A ∩ B) = 11

n (A ∩ B ∩ C) = 3

(i) We know that

n (A ∪ B ∪ C) = n (A) + n (B) + n (C) - n (A ∩ B) - n (B ∩ C) - n (A ∩ C) + n (A ∩ B ∩ C)

= 25 + 26 + 26 - 9 - 8 - 11 + 3

= 52

Hence, there are 52 people who read at least one newspaper.

(ii) n (A ∩ B) = People who read H and T only + People who read all three

n (B ∩ C) = People who read H and I only + People who read all three

n (A ∩ C) = People who read I and T only + People who read all three

n (A ∩ B) + n (B ∩ C) + n (A ∩ C) = Number of people who read more than one newspaper + 2 × n (A ∩ B ∩ C)

11 + 8 + 9 = Number of people who read more than one newspaper + 6

Number of people who read more than one newspaper = 22

Number of people who read only one newspaper = 52 - 22 = 30

Hence, 30 people read exactly one newspaper.

16. In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.

SOLUTION

n (A) = 21

n (B) = 26

n (C) = 29

n (A ∩ B) = 14

n(C ∩ A) = 12

n (B ∩ C) = 14

n (A ∩ B ∩ C) = 8

We need to draw the venn diagram.

Number of people who only like product C = 29 - (4 + 8 + 6)

= 11

Hence, 11 people like product C only.