Sets Solution of Bihar Board Class 11 Mathematics

Set: Collection of well defined objects.
There are three months of a year which begins with the letter J, rest of the month`s name begin with different letter. So, the given collection has well-defined objects namely January, June and July.

Hence, this collection is a set.

{x: x=months of a year beginning with letter J}

Alternatively

{x:x=January, June, July where January, June, July are month of a year}

Question 2.

Which of the following are sets? Justify your answer.

The collection of ten most talented writers of India.

Answer:

The object of the given collection is quiet subjective and vary with person`s personal choice. Some people may be biased toward some as the most talented writers of India whereas for other the list of most talented writers may be altogether different. Hence the given collection doesn’t have well defined objects with universal standards.

Hence, this collection is not a set.

Question 3.

Which of the following are sets? Justify your answer.

A team of eleven best-cricket batsmen of the world.

Answer:

A team of eleven best cricket batsmen of the world is not very clearly defined collection. The cricket team of a nation could be well defined set but best cricket batsmen team would be judgemental and subjective concept varying from person to person.

Hence, this collection is not a set.

Question 4.

Which of the following are sets? Justify your answer.

The collection of all boys in your class.

Answer:

The collection of all boys in your class is a well-defined collection because one can definitely differentiate between boys who could belongs to this collection on the basis of appearance itself. The criteria for judging the gender would not vary from person to person. The set has very well defined object which is boys in this class.
Hence, this collection is a set.

Question 5.

Which of the following are sets? Justify your answer.

The collection of all natural numbers less than 100.

Answer:

The collection of all natural numbers less than 100 is a well-defined collection because there is no ambiguity in regard of numbers and there values greater than or less than 100. There are specific numbers which belongs to the given collection. Such numbers choice doesn’t vary from person to person. The clearly defined objects make the collection a set.

{x:x={0, 1, 2, 3, ...99} where n∈N and 0≤n<100}

N is natural number

Question 6.

Which of the following are sets? Justify your answer.

A collection of novels written by the writer Munshi Prem Chand.

Answer:

A collection of novels written by the writer Munshi Prem Chand is a well-defined collection because there are finite numbers of books which Munshi Prem Chand has written. The names of the book could not vary from person to person on the basis of personal choice. The well-defined objects of the collection make it a set.

Hence, this collection is a set.

Question 7.

Which of the following are sets? Justify your answer.

The collection of all even integers.

Answer:

The collection of all even integers is a well-defined collection because there couldn’t be difference between choosing the even integers from the collection of given numbers or all the numbers taken at a large. The choice will not vary from person to person. The given collection has well-defined objects which meet the universal criteria.

Hence, this collection is a set.

{x:x=….-6,-4,-2, 0,2, 4, 6…where n is even integer 2k for some unique interger k}

Question 8.

Which of the following are sets? Justify your answer.

The collection of questions in this Chapter.

Answer:

The collection of the questions in this chapter is a well-defined collection and the set will have well defined objects. One can definitely identify the question that would belong to the collection and could easily number them. The number of questions and the type of questions will not be different for different persons.

Hence the collection is a set.

Question 9.

Which of the following are sets? Justify your answer.

A collection of most dangerous animals of the world.

Answer:

A collection of most dangerous animals of the world is not a very clearly defined set as the ranking of the animals keep on altering and their ranking vary from countries to countries. The objects of the collection are not well –defined and doesn`t have universal acceptance as it is .

Hence the collection is not set.

Question 10.

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

(i) 5. . .A

(ii) 8 . . . A

(iii) 0. . .A

(iv) 4. . . A

(v) 2. . .A

(vi) 10. . .A

Answer:

Here ∈ =belongs to

∉ = does not belongs to

(i) 5 ∈ A, since 5 is in the set A

(ii) 8 A, since 8 is not in the set A

(iii) 0 A, since 0 is not in the set A

(iv) 4 ∈ A, since 4 is in the set A

(v) 2 ∈ A, since 2 is in the set A

(vi) 10 A, since 10 is not in set A

Question 11.

Write the following sets in roster form:
A = {x : x is an integer and –3 ≤ x < 7}

Answer:

Integers are …..-5, -4, -3,-2,-1 0,1, 2, 3, 4, 5, 6, 7, 8, 9, …………….

The elements of this set are -3,-2,-1 0, 1, 2, 3, 4, 5 and 6

The roaster form is

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

Question 12.

Write the following sets in roster form:

B = {x : x is a natural number less than 6}

Answer:

Here natural numbers are = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10………………

The elements of the set are 1, 2, 3, 4 and 5

The roaster form is

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

Question 13.

Write the following sets in roster form:

C = {x : x is a two-digit natural number such that the sum of its digits is 8}

Answer:

The elements of the set are numbers 17, 26, 35, 44, 53, 62, 71, 80

0+8 = 8 but putting 0 in tens place makes it single digit number so will not consider it.

8+0 = 80, number is 80

1+7 =8, number is 17

7+1 =8, number is 71

2+6 = 8, number is 26

6+2 = 8, number is 62

3+ 5 = 8, number is 35

5+3 = 8, number is 53

4+ 4 =8, number is 44

Set in roaster form is

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

Question 14.

Write the following sets in roster form:

D = {x : x is a prime number which is divisor of 60}

Answer:

Prime numbers are those which have factors as one and the number itself.

List of prime numbers is 2, 3, 5, 7, 11, 13, 17, 19, 21, 23 ……

Divisors of 60 are 1, 2, 3, 4, 5, 6 10, 12, 15, 20, 30, 60

So 1 × 60 = 60

2 × 30 = 60

3 × 20 = 60

4 × 15 =60

5 × 12 = 60

6 × 10 = 60

The elements of the set are prime numbers which are divisor of 60 that is 2, 3 and 5

The roaster form of set is

D = {2,3, 5}

Question 15.

Write the following sets in roster form:

E = The set of all letters in the word TRIGONOMETRY

Answer:

There are 12 letters in the word TRIGONOMETRY out of which T, R and O are repeating

So the element of the set are T, R, I, G,O, N, M, E and Y

The set in the roaster form is

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

Question 16.

Write the following sets in roster form:

F = The set of all letters in the word BETTER

Answer:

There are four letters in the word B, E, T, R . T and E are repeating in the word.

So the elements of the set are B, E, T, R.

The set in the roaster form is

F= {B, E, T, R}

Question 17.

Write the following sets in the set-builder form:

(3, 6, 9, 12}

Answer:

Here the element of the set can be disintegrated as

3 = 3 × 1

6 = 3 × 2

9= 3 × 3

12 = 3 × 4

The set in the set-builder form is

A= {x:x=3n where n∈N and 1≤n ≤4}

Question 18.

Write the following sets in the set-builder form:

{2,4,8,16,32}

Answer:

The elements of the set can be disintegrated as

2 = 2¹

4 = 2²

8 = 2³

16 = 2⁴

32 = 2⁵

The set in the set-builder form is

A= {x:x=2ⁿ, n ∈N and 1≤n ≤5}

Question 19.

Write the following sets in the set-builder form:

{5, 25, 125, 625}

Answer:

The elements of the set can be disintegrated as

5 = 5¹

25 = 5²

125 = 5³

625 = 5⁴

The set in the set-builder form is

A= {x :x= 5ⁿ, n ∈N and 1≤n ≤4}

Question 20.

Write the following sets in the set-builder form:

{2, 4, 6, . . .}

Answer:

Natural numbers are 1, 2, 3, 4, 5, 6……..

Even natural numbers are 2, 4, 6, 8……..

∴ the given set is set of all even natural numbers

The set in the set-builder form is

A= {x : x=2k for unique value of k, k > 0}

Question 21.

Write the following sets in the set-builder form:

{1,4,9, . . .,100}

Answer:

The element of the set can be disintegrated as

1 = 1²

4 = 2²

9 = 3²

…..

. 100 = 10²

The set in the set-builder form is

A= {x:x = n², n ∈N, 1≤n ≤10}

Alternatively

A = {x:x is the square of natural number less than or equal to 10}

Question 22.

List all the elements of the following sets:

A = {x : x is an odd natural number}

Answer:

Natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10…………………

Odd natural numbers are 1, 3, 5, 7, 9, 11, ………………

So the elements of the set are 1, 3, 5, 7, 9, 11, ………………

A = {1, 3, 5, 7, 9, 11,………………}

Question 23.

List all the elements of the following sets:

B = {x : x is an integer, –1/2 < x < 9/2}

Answer:

Here, Given

⇒ - 0.5 < x < 4.5

Integers = ……. -4,-3, -2, -1, 0, 1, 2, 3, 4, 5, 6 ………

Elements of the given set are 0, 1, 2, 3, 4

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

Question 24.

List all the elements of the following sets:

C = {x : x is an integer, x² ≤ 4}

Answer:

Integers are ……-4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8

x² = 4

x= 2

(0)²= 0

(1)² = 1

(2)²= 4

(-2)² = 4

(-1)² =1

The elements of the set are -2, -1, 0, 1, 2

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

Question 25.

List all the elements of the following sets:

D = {x : x is a letter in the word “LOYAL”}

Answer:

There are four letters in the word L,O, Y, A as L is repeating

The element of the sets are L, O, Y,A

D = {L, O, Y,A}

Question 26.

List all the elements of the following sets:

E = {x : x is a month of a year not having 31 days}

Answer:

The set would include months with 30 days and 28 days (February)

Now the elements of the set would be months: - February, April, June, September and November

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

Question 27.

List all the elements of the following sets:

F = {x : x is a consonant in the English alphabet which precedes k}.

Answer:

The vowels are a, e, i, o,u

The letters before k are a, b, c, d, e, f, g, h, i, j, k

Thus the consonant before k are b, c, d, f, g, h, j

So, the elements of the sets are b, c, d, f, g, h, j

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

Question 28.

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

Answer:

The elements of sets are

6× 1 = 6

3× 2 = 6

{1, 2, 3, 6} is a set of natural numbers and divisor of 6

6× 1 = 6

2 × 3 = 6

The divisors of 6 are 1, 6, 2 and 3 out of which 2 and 3 are prime

{2, 3} is the set of prime numbers which are divisor of 6

There are 11 letters in the word MATHEMATICS out of which 3 (M, T, A)are repeating .

{M, A, T, H, E, I, C, S} is a set of letters of the word MATHEMATICS

Natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9,…………..

Odd numbers less than 10 are 1, 3, 5, 7, 9

{1, 3, 5, 7, 9} is a set of odd natural numbers less than 10

Exercise 1.2

Question 1.

Which of the following are examples of the null set

(i) Set of odd natural numbers divisible by 2

(ii) Set of even prime numbers

(iii) {x : x is a natural numbers, x < 5 and x > 7}

(iv) {y : y is a point common to any two parallel lines}

Answer:

(i) Given: Set of odd natural numbers divisible by 2.

As we know set is a collection of well defined objects.

Let we represent the given set in roaster form:

⇒ Set of odd natural numbers divisible by 2 is {ϕ}.

Because no odd natural number can be divided by 2. Hence, it is a null set.

(ii) Given: Set of even prime numbers.

As we know set is a collection of well defined objects.

Let we represent the given set in roaster form:

⇒ Set of even prime numbers is {2}.

Because 2 is an even prime number. Hence, it is not a null set.

(iii) Given: {x : x is a natural numbers, x < 5 and x > 7}

As we know set is a collection of well defined objects.

Let we represent the given set in roaster form:

⇒ {x : x is a natural numbers, x < 5 and x > 7} is {ϕ}.

Because no number can be simultaneously less than 5 and greater than 7.Hence, it is a null set.

(iv) Given: {y : y is a point common to any two parallel lines}

As we know set is a collection of well defined objects.

Let we represent the given set in roaster form:

⇒ {y : y is a point common to any two parallel lines} is {ϕ}.

Because two parallel lines never meet at any of the point so they don’t have any common point. Hence, it is a null set.

Question 2.

Which of the following sets are finite or infinite

(i) The set of months of a year

(ii) {1, 2, 3, . . .}

(iii) {1, 2, 3, . . .99, 100}

(iv) The set of positive integers greater than 100.

(v) The set of prime numbers less than 99

Answer:

(i) Given: The set of months of a year.

As we know set is a collection of well defined objects.

Let we represent the given set in roaster form:

⇒ Set of months of a year is {January, February, march, April, May, June, July, August, September, October, November, December}.

Because the set contain 12 elements. Hence, it is a finite set.

(ii) Given: {1, 2, 3, . . .}.

As we know set is a collection of well defined objects.

As it is already represented in roaster form:

⇒ Set = {1, 2, 3, . . .}.

Because the set contain infinite number of natural numbers. Hence, it is an infinite set.

(iii) Given: {1, 2, 3, . . .99, 100}.

As we know set is a collection of well defined objects.

As it is already represented in roaster form:

⇒ Set = {1, 2, 3, . . .99, 100}.

Because the set contain finite number from 1 to 100.Hence, it is a finite set.

(iv) Given: The set of positive integers greater than 100.

As we know set is a collection of well defined objects.

Let we represent the set in roaster form:

⇒ Set of positive integers greater than 100 = {100,101,102,…}.

Because the set contain an infinite number from 100 to infinity. Hence, it is an infinite set.

(v) Given: The set of prime numbers less than 99.

As we know set is a collection of well defined objects.

Let we represent the set in roaster form:

⇒ The set of prime numbers less than 99 = {2, 3, . . .99}.

Because the set contain finite prime number from 2 to 99. Hence, it is a finite set.

Question 3.

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

(i) The set of lines which are parallel to the x-axis.

(ii) The set of letters in the English alphabet.

(iii) The set of numbers which are multiple of 5.

(iv) The set of animals living on the earth.

(v) The set of circles passing through the origin (0,0).

Answer:

(i) Given: The set of lines which are parallel to the x-axis.

As we know set is a collection of well defined objects.

Let we represent the set in set builder form:

⇒ S = {x:x is number of parallel lines to x-axis}.

Because the set of lines parallel to x-axis are infinite in number. Hence, I t is an infinite set.

(ii) Given: The set of letters in the English alphabet.

As we know set is a collection of well defined objects.

Let we represent the set in roaster form:

⇒ The set of letters in the English alphabet = {A,B,C,…,Z}.

Because the set contain finite alphabet series and having 26 element. Hence, it is a finite set.

(iii) Given: The set of numbers which are multiple of 5

As we know set is a collection of well defined objects.

Let we represent the set in roaster form:

⇒ The set of numbers which are multiple of 5 = {5,10,15,…}.

Because the set contain infinite numbers which are multiple of 5.Hence, it is an infinite set.

(iv) Given: The set of animals living on the earth.

As we know set is a collection of well defined objects.

Let we represent the set in set builder form:

⇒ S = {x:x is the set of animals living on the earth}.

Because the number of animal living on earth are though too large but they are finite in number. Hence, It is finite set.

(v) Given: The set of circles passing through the origin (0,0).

As we know set is a collection of well defined objects.

Let we represent the set in set builder form:

⇒ S = {x:x is the set of circles passing through the origin (0,0)}.

Because the number of circles passing through the origin are infinite in number. Hence, It is an infinite set.

Question 4.

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

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

(ii) A = {4, 8, 12, 16} B = {8, 4, 16, 18}

(iii) A = {2, 4, 6, 8, 10} B = {x : x is positive even integer and x ≤ 10}.

(iv) A = {x : x is a multiple of 10}, B = {10, 15, 20, 25, 30, . . .}

Answer:

(i) Given: A = {a, b, c, d} B = {d, c, b, a}.

Two set A and B are said to be equal if they have exactly same elements then we say A = B.

Because elements of set A and B do not have significant order but A and B have same element.

∴ A=B.

(ii) Given: A = {4, 8, 12, 16} B = {8, 4, 16, 18}.

Two set A and B are said to be equal if they have exactly same elements then we say A = B.

As 12 Є A but 12 does not belongs to B.

Because elements of set A and B do not have same element.

∴ A≠B.

(iii) Given: A = {2, 4, 6, 8, 10} B = {x : x is positive even integer and x ≤ 10}.

Let we represent the set B in roaster form:

⇒ x is positive even integer and x ≤ 10= {2,4,6,8,10}.

Two set A and B are said to be equal if they have exactly same elements then we say A = B.

Because elements of set A and B have same element.

∴ A=B.

(iv) Given: A = {x : x is a multiple of 10}, B = {10, 15, 20, 25, 30, . . .}

Let we represent the set A in roaster form:

⇒ Set A = x is a multiple of 10= {10, 20 , 30 ,…}.

And set B ={10, 15, 20, 25, 30, . . .}

Two set A and B are said to be equal if they have exactly same elements then we say A = B.

As 15 Є B but 15 does not belongs to B

Because elements of set A and B do not have same element.

∴ A≠B.

Question 5.

Are the following pair of sets equal? Give reasons.

(i) A = {2, 3}, B = {x : x is solution of x² + 5x + 6 = 0}

(ii) A = {x : x is a letter in the word FOLLOW}

B = {y : y is a letter in the word WOLF}

Answer:

(i) Given: A = {2, 3}, B = {x : x is solution of x² + 5x + 6 = 0}

Solution of equation x² + 5x + 6 = 0

⇒ x² + 3x + 2x + 6 = 0

⇒ x(x + 3) + 2(x + 3) = 0

⇒ (x + 3)(x + 2) = 0

⇒ x = {-3,-2}

Let we represent the set B in roaster form:

⇒ Set B = {x is solution of x² + 5x + 6 = 0}={-3,-2}.

And set A ={2,3}

Two set A and B are said to be equal if they have exactly same elements then we say A = B.

Because the elements of set A and B do not have same numbers.

∴ A≠B.

(ii) Given: A = {x : x is a letter in the word FOLLOW},B = {y : y is a letter in the word WOLF}.

Let we represent the set A in roaster form:

⇒ Set A = {x : x is a letter in the word FOLLOW}={F,O,L,W}.

Let we represent the set B in roaster form:

⇒ Set B = {y : y is a letter in the word WOLF}={W,O,L,F}.

Two set A and B are said to be equal if they have exactly same elements then we say A = B.

Because elements of set A and B do not have significant order, but A and B have same element.

∴ A=B.

Question 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}

Answer:

Given: 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}

As we see,

8 Є A, but 8 does not belong to B,D,E,F,G and H

⇒ A≠B, A≠D, A≠E, A≠F, A≠G, A≠H

And 2 Є A but 2 does not belong to C

⇒ A≠C.

Now, 3 Є B, but 3 does not belong to C,E,F,G and H

⇒ B≠C, B≠E, B≠F, B≠G, B≠H

Also, 12 Є C, but 12 does not belong to D,E,F,G and H

⇒ C≠D, C≠E, C≠F, C≠G, C≠H

Also, 4 Є D, but 4 does not belong to E,F,G and H

⇒ D≠E, D≠F, D≠G, D≠H

Also, -1 Є E, but -1 does not belong to F,G and H

⇒ E≠F, E≠G, E≠H

Also, a Є F, but a does not belong to G and H

⇒F≠G, F≠H

Also, -1 Є G, but -1 does not belong to H

⇒ G≠H

But B = D and E = G.

As, Two set A and B are said to be equal if they have exactly same elements then we say A = B.

Because elements of set (B and D) and (E and G) do not have significant order but (B and D) and (E and G) have same element.

∴ B = D and E = G.

Exercise 1.3

Question 1.

Make correct statements by filling in the symbols â or ⊄ in the blank spaces:

(i) {2, 3, 4} . . . {1, 2, 3, 4, 5}

(ii) {a, b, c} . . . {b, c, d}

(iii) {x : x is a student of Class XI of your school}. . .{x : x student of your school}

(iv) {x : x is a circle in the plane} . . .{x : x is a circle in the same plane with radius 1 unit}

(v) {x : x is a triangle in a plane} . . . {x : x is a rectangle in the plane}

(vi) {x : x is an equilateral triangle in a plane} . . . {x : x is a triangle in the same plane}

(vii) {x : x is an even natural number} . . . {x : x is an integer}

Answer:

The blanks are filled below:

(i) {2, 3, 4} ⊂ {1, 2, 3, 4, 5}

Since, 2, 3, 4 comes in the second set.

(ii) {a, b, c} ⊄ {b, c, d}

Since, ‘a’ is not in the second set.

(iii) {x : x is a student of Class XI of your school}⊂{x : x student of your school}

Since, x is a part of the school too.

(iv) {x : x is a circle in the plane} ⊄ {x : x is a circle in the same plane with radius 1 unit}

Since, first set has no fixed radius circles whereas the second set has only circles with radius 1 unit

(v) {x : x is a triangle in a plane} ⊄ {x : x is a rectangle in the plane}

Since, set 1 has triangle whereas set 2 has rectangle.

(vi) {x : x is an equilateral triangle in a plane} ⊂ {x : x is a triangle in the same plane}

Since, equilateral triangle is a type of triangle itself.

(vii) {x : x is an even natural number} ⊂ {x : x is an integer}

Since, all the even natural numbers are a type of integers.

Question 2.

Examine whether the following statements are true or false:

(i) {a, b} ⊄ {b, c, a}

(ii) {a, e} ⊂ {x : x is a vowel in the English alphabet}

(iii) {1, 2, 3} ⊂ {1, 3, 5}

(iv) {a} ⊂ {a, b, c}

(v) {a} ∈ {a, b, c}

(vi) {x : x is an even natural number less than 6} ⊂ {x : x is a natural number which divides 36}

Answer:

(i) Let us assume that A = {a, b} and B = {b, c, a}

Now, we can observe that every element of A is an element of B.

Thus, A⊂ B

∴ The statement is false.

(ii) Let us assume that A = {a, e} and

B = {x: x is a vowel in the English alphabets}

={a, e, i, o, u}

Now, we can observe that every element of A is an element of B.

Thus, A⊂ B

∴ The statement is true.

(iii) Let us assume that A = {1, 2, 3} and B = {1, 3, 5},

Now, we can observe that 2 belongs to A but 2 does not belongs to B.

Thus, A B

∴ The statement is false.

(iv) Let us assume that A = {a} and B = {b, c, a}

Now, we can observe that every element of A is an element of B.

Thus, A⊂ B

∴ The statement is true.

(v) Let us assume that A = {a} and B = {b, c, a}

Now, we can observe that every element of A is an element of B.

Thus, A⊂ B

∴ The statement is false.

(vi) Let us assume that A = {x:x is an even natural number less than 6}

= {2, 4}

and B = {x:x is a natural number which divide 36}

= {1, 2, 3, 4, 6, 9, 12, 18, 36}

Now, we can observe that every element of A is an element of B.

Thus, A⊂ B

∴ The statement is true.

Question 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

Answer:

The parts of the question are solved below:

(i) Here, we know that {3, 4} is a member of set A.

= {3, 4} ∈ A

Thus, the given statement is incorrect.

(ii) Here, we know that {3, 4} is a member of set A.

Thus, the given statement is incorrect.

(iii) Here, we know that {3, 4} is a member of set A.

={{3, 4}} is a set.

Thus, the given statement is correct.

(iv) Here, we can observe that 1 is a member of set A.

Thus, the given statement is correct.

(v) Here, we can see that 1 is a member of set A but is not any set itself.

Thus, the given statement is incorrect.

(vi) Here, we can see that 1, 2, 5 is a member of set A

Thus, the given statement is correct.

(vii) Here, we can see that 1, 2, 5 is a member of set A

=, {1, 2, 5} is a subset of A

Thus, the given statement is incorrect.

(viii) Here, we can see that 3 is not a member of set A

={1, 2, 3} is not a subset of A

Thus, the given statement is incorrect.

(ix) Here, we can see that ϕ is not a member of set A

Thus, the given statement is correct.

(x) Here, as we know the null set is a subset of every set.

Thus, the given statement is correct.

(xi) {ϕ } is the set containing the null set.

{ϕ } ⊂ A is only possible if ϕ is in set A. But it is not there.
So, the statement is incorrect.

Question 4.

Write down all the subsets of the following sets:

(i) {a}

(ii) {a, b}

(iii) {1, 2, 3}

(iv) ϕ

Answer:

(i) Let A={a}

Now, number of elements in A = 1

Number of subsets of A = 2¹

∴ subsets of A are: ϕ, {a}

(ii) Let A={a, b}

Now, number of elements in A = 2

Number of subsets of A = 2² = 4

∴ subsets of A are: ϕ, {a}, {b}, {a, b}

(iii) Let A={1, 2, 3}

Now, number of elements in A = 3

Number of subsets of A = 2³ = 8

∴ subsets of A are:

ϕ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}

(iv) Let A=ϕ

Now, number of elements in A = 0

Number of subsets of A = 2⁰ = 1

∴ subset of A is: ϕ

Question 5.

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

Answer:

P(A) is the power set of set A.

Number of elemenst of P(A) = 2ⁿ

Where n is the number of elements of the set A.

Given A=ϕ

Then the numberv of elements of set A = 0

∴ Number of elements of P(A) = 2ⁿ

= 2⁰

= 1

Question 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}

Answer:

(i) Let us assume that A = {x: x ∈ R, – 4 < x ≤ 6}

∴ Set A can be written in the form of intervals as follows:

= (-4, 6]

(ii) Let us assume that A = {x : x ∈ R, – 12 < x < –10}

∴ Set A can be written in the form of intervals as follows:

= (-12, -10)

(iii) Let us assume that A = {x: x ∈ R, 0 ≤ x < 7}

∴ Set A can be written in the form of intervals as follows:

= [0, 7)

(iv) Let us assume that A = {x : x ∈ R, 3 ≤ x ≤ 4}

∴ Set A can be written in the form of intervals as follows:

= [3, 4]

Question 7.

Write the following intervals in set-builder form:

(i) (– 3, 0)

(ii) [6, 12]

(iii) (6, 12]

(iv) [–23, 5)

Answer:

(i) Here, the given interval = (-3, 0)

for finding the set builder form for the given set the values range from - 3 to 0

This means value of any variable x is greater than - 3 and less than 0

Now, set builder form of the given interval can be written as follows:

A = {x:x ϵ R, -3<x<0}

(ii) Here, the given interval = [6, 12]

Now the set of values of any variable for set given will be such that x is greater than 6 (including 6 as the bracket is closed) and less than 12(including 12)

set builder form of the given interval can be written as follows:

A = {x:x ϵR, 6≤x≤12}

(iii) Here, the given interval = (6, 12]

Now the set of values of any variable for set given is such that x is greater than 6 (excluding 6 as the bracket is open) and less than 12 (including 12, as the bracket is closed)

set builder form of the given interval can be written as follows:

A = {x:x ϵR, 6<x≤12}

(iv) Here, the given interval = [-23, 5)

Now the set of values of any variable for given set is such that x is greater than - 23( including 23 as the bracket is closed) and less than 5 (as the bracket is open)

set builder form of the given interval can be written as follows:

A = {x:x ϵR, -23≤x<5}

Question 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.

Answer:

(i) We know that right triangles are a type of triangle with an angle 90°

Thus, a set of triangles will contain all the right triangles.

∴universal set, U = {x: x is a triangle in plane}

(ii) We know that isosceles triangles are a type of triangle with any two of the angles equal in measure.

Thus, a set of triangles will contain all the isosceles triangles.

∴ universal set, U = {x: x is a triangle in plane}

Question 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 ?
(i) {0, 1, 2, 3, 4, 5, 6}

(ii) ϕ

(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(iv) {1, 2, 3, 4, 5, 6, 7, 8}

Answer:

A universal set is a set which contains all the elements of its subsets. So if there are two sets with some members then the universal set containing those two sets, will have elements will all the members of both sets without repetition.

(i) Given: A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}

Now, let D = {0, 1, 2, 3, 4, 5, 6}

Since, 8 belongs to C then its universal set must contain 8, but D does not contain 8.

∴ D is not a universal set for A, B, C.

(ii) Given: A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}

Now, let D = ϕ

Since, D is an empty set it does not contain any element

∴ D is not a universal set for A, B, C.

(iii) Given: A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}

Now, let D = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Since, all the members of A, B, C belongs to D.

∴D is a universal set for A, B, C.

(iv) Given: A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}

Now, let D = {1, 2, 3, 4, 5, 6, 7, 8}

Since, 0 belongs to C but it is not a member of D.

∴ D is not a universal set.

Exercise 1.4

Question 1.

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

(i) X = {1, 3, 5} Y = {1, 2, 3}

(ii) A = [ a, e, i, o, u} B = {a, b, c}

(iii) A = {x: x is a natural number and multiple of 3}

B = {x: x is a natural number less than 6}

(iv) A = {x: x is a natural number and 1 < x ≤ 6}

B = {x: x is a natural number and 6 < x < 10}

(v) A = {1, 2, 3}, B = ϕ

Answer:

Venn diagram shown above represents that the union of any two sets will give all the members of the two sets together, and not replicating the common member twice.

(i) It is given in the question that,

X = {1, 3, 5}

And, Y = {1, 2, 3}

All the members of both sets together are 1, 3, 5, 1, 2, 3.

Now we don't have to replicate members. So, the members of union of these two sets are 1, 2, 3, 5

∴

(ii) It is given in the question that,

A = {a, e, i, o, u}

And, B = {a, b, c}

All the members of both sets together are a, e, i, o, u, a, b, c

Now we don't have to replicate members. So, the members of union of these two sets are a, b, c, e, i, o, u

∴

(iii) It is given in the question that,

A = {3, 6, 9, 12, 15, 18,……}

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

All the members of both sets together are 3, 6, 9, 12, 15, 18, ..........., 1, 2, 3, 4, 5

Now we don't have to replicate members. So, the members of union of these two sets are 1, 2, 3, 4, 5, 6, 9, 12, 15, 18, ..........

∴

(iv) It is given in the question that,

A = {2, 3, 4, 5, 6}

And, B = {7, 8, 9}

All the members of both sets together are 2, 3, 4, 5, 6, 7, 8, 9.

Now we don't have to replicate members. So, the members of union of these two sets are 2, 3, 4, 5, 6, 7, 8, 9

∴

(v) It is given in the question that,

A = {1, 2, 3}

And,

All the members of both sets together are 1, 2, 3

Now we don't have to replicate members. So, the members of union of these two sets are 1, 2, 3

∴

Question 2.

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

Answer:

It is given in the question that,

A = {a, b}

And, B = {a, b, c}

Here, it is clearly seen that all the elements of set A are present in set B

And,

Question 3.

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

Answer:

It is given in the question that, A and B are two sets such that,

Let us take, A = {a, b}

And, B = {a, b, c} we have:

Question 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

Answer:

(i) It is given in the question that,

A = {1, 2, 3, 4}

And, B = {3, 4, 5, 6}

∴

(ii) It is given in the question that,

A = {1, 2, 3, 4}

And, C = {5, 6, 7, 8}

∴

(iii) It is given in the question that,

B = {3, 4, 5, 6}

And, C = {5, 6, 7, 8}

∴

(iv) It is given in the question that,

B = {3, 4, 5, 6}

And, D = {7, 8, 9, 10}

∴

(v) It is given in the question that,

A = {1, 2, 3, 4}

B = {3, 4, 5, 6}

And, C = {5, 6, 7, 8}

∴

(vi) It is given in the question that,

A = {1, 2, 3, 4}

B = {3, 4, 5, 6}

And, D = {7, 8, 9, 10}

∴

(vii) It is given in the question that,

B = {3, 4, 5, 6}

C = {5, 6, 7, 8}

And, D = {7, 8, 9, 10}

∴

Question 5.

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

Answer:

The intersection of each pair of sets of question 1 is as follows:

(i) It is given in the question that,

X = {1, 3, 5}

And, Y = {1, 2, 3}

∴

(ii) It is given in the question that,

A = {a, e, i, o, u}

And, B = {a, b, c}

∴

(iii) It is given in the question that,

A = {3, 6, 9, 12, 15, 18,……}

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

∴ A∩B = {3}

(iv) It is given in the question that,

A = {2, 3, 4, 5, 6}

And, B = {7, 8, 9}

∴

(v) It is given in the question that,

A = {1, 2, 3}

And,

∴

Question 6.

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

(i) A ∩ B

(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)

Answer:

(i) It is given in the question that,

A = {3, 5, 7, 9, 11}

And,

∴

A∩B will give the members of set A and B that are common. Therefore,

(ii) It is given in the question that,

B = {7, 9, 11, 13}

And,

∴

Intersection of two sets give the members of set that are common to both sets.

(iii) It is given in the question that,

A = {3, 5, 7, 9, 11}

And, D = {15, 17}

∴

As there are no members of sets that are common, the intersection of A, B and C will be a null set.

(iv) It is given in the question that,

A = {3, 5, 7, 9, 11}

And,

∴

Only 11 is common to sets A and C. Therefore,

= {11}

(v) It is given in the question that,

B = {7, 9, 11, 13}

And, D = {15, 17}

∴

No member of set B is common with set D, Therefore intersection of B and D will be a null set.

(vi) It is given in the question that,

A = {3, 5, 7, 9, 11}

B = {7, 9, 11, 13}

And,

∴

B U C will give all the members of both sets combined and not replicating the common members of the set.

After that intersection of set A and B U C is calculated as the members that are common to these two sets.

= {7, 9, 11}

(vii) It is given in the question that,

A = {3, 5, 7, 9, 11}

And, D = {15, 17}

∴

No member of the two sets is common, therefore the intersection of these two sets will be a null set.

(viii) It is given in the question that,

A = {3, 5, 7, 9, 11}

B = (7, 9, 11, 13}

And, D = {15, 17}

∴

= {7, 9, 11}

(ix) It is given in the question that,

A = {3, 5, 7, 9, 11}

B = (7, 9, 11, 13}

And,

∴

= {7, 9, 11}

(x) It is given in the question that,

A = {3, 5, 7, 9, 11}

B = (7, 9, 11, 13}

And, D = {15, 17}

∴

Calculate A U B and B U C first and then intersection of these two sets is calculated.

= {7, 9, 11, 15}

Question 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

Answer:

(i) It is given in the question that,

A = {x: x is a natural number}

And, B = {x: x is an even natural number}

∴

= B

(ii) It is given in the question that,

A = {x: x is a natural number}

And, C = {x: x is an odd natural number}

∴

= C

(iii) It is given in the question that,

A = {x: x is a natural number}

And, D = {x: x is a prime number}

∴

= D

(iv) It is given in the question that,

B = {x: x is an even natural number}

And, C = {x: x is an odd natural number}

∴

(v) It is given in the question that,

B = {x: x is an even natural number}

And, D = {x: x is a prime number}

∴

= {2}

(vi) It is given in the question that,

C = {x: x is an odd natural number}

And, D = {x: x is a prime number}

∴

Question 8.

Which of the following pairs of sets are disjoint?

(i) {1, 2, 3, 4} and {x: x is a natural number and 4 ≤ x ≤ 6}

(ii) {a, e, i, o, u} and {c, d, e, f}

(iii) {x: x is an even integer} and {x: x is an odd integer}

Answer:

Disjoint sets are those sets which have no element in common. So, for that A∩B = ø

(i) Let us assume, A = {1, 2, 3, 4}

And, B = {x: x is a natural number and 4 x 6}

Set B = {4, 5, 6}

∴

Hence, A and B are not disjoint.

(ii) Let us assume, A = {a, e, i, o, u}

And, B = {c, d, e, f}

∴ A ∩ B = {e}

Hence, A and B are not disjoint

(iii) Let us assume, A = {x: x is an even integer}

And, B = {x: x is an odd integer}

∴

Hence, A and B are disjoint

Question 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

Answer:

A - B shows the relative complement of the set, which represents members of set A which are not part of set B

(i) It is given in the question that,

A = {3, 6, 9, 12, 15, 18, 21}

And, B = {4, 8, 12, 16, 20}

∴ A – B = {3, 6, 9, 12, 15, 18, 21} - {4, 8, 12, 16, 20}

12 is present in both the sets, so A - B will not contain 12 and will contain rest of the elements of A

= {3, 6, 9, 15, 18, 21}

(ii) It is given in the question that,

A = {3, 6, 9, 12, 15, 18, 21}

And, C = {2, 4, 6, 8, 10, 12, 14, 16}

∴ A – C = {3, 6, 9, 12, 15, 18, 21} - {2, 4, 6, 8, 10, 12, 14, 16}

= {3, 9, 15, 18, 21}

(iii) It is given in the question that,

A = {3, 6, 9, 12, 15, 18, 21}

And, D = {5, 10, 15, 20}

∴ A – D = {3, 6, 9, 12, 15, 18, 21} - {5, 10, 15, 20}

= {3, 6, 9, 12, 18, 21}

(iv) It is given in the question that,

B = {4, 8, 12, 16, 20}

And, A = {3, 6, 9, 12, 15, 18, 21}

∴ B – A = {4, 8, 12, 16, 20} - {3, 6, 9, 12, 15, 18, 21}

= {4, 8, 16, 20}

(v) It is given in the question that,

C = {2, 4, 6, 8, 10, 12, 14, 16}

And, A = {3, 6, 9, 12, 15, 18, 21}

∴ C – A = {2, 4, 6, 8, 10, 12, 14, 16} - {3, 6, 9, 12, 15, 18, 21}

= {2, 4, 8, 10, 14, 16}

(vi) It is given in the question that,

D = {5, 10, 15, 20}

And, A = {3, 6, 9, 12, 15, 18, 21}

∴ D – A = {5, 10, 15, 20} - {3, 6, 9, 12, 15, 18, 21}

= {5, 10,20}

(vii) It is given in the question that,

B = {4, 8, 12, 16, 20}

And, C = {2, 4, 6, 8, 10, 12, 14, 16}

∴ B – C = {4, 8, 12, 16, 20} - {2, 4, 6, 8, 10, 12, 14, 16}

= {20}

(viii) It is given in the question that,

B = {4, 8, 12, 16, 20}

And, D = {5, 10, 15, 20}

∴ B – D = {4, 8, 12, 16, 20} - {5, 10, 15, 20}

= {4, 8, 12, 16}

(ix) It is given in the question that,

C = {2, 4, 6, 8, 10, 12, 14, 16}

And, B = {4, 8, 12, 16, 20}

∴ C – B = {2, 4, 6, 8, 10, 12, 14, 16} - {4, 8, 12, 16, 20}

= {2, 6, 10, 14}

(x) It is given in the question that,

D = {5, 10, 15, 20}

And, B = {4, 8, 12, 16, 20}

∴ D – B = {5, 10, 15, 20} - {4, 8, 12, 16, 20}

= {5, 10, 15}

(xi) It is given in the question that,

C = {2, 4, 6, 8, 10, 12, 14, 16}

And, D = {5, 10, 15, 20}

∴ C – D = {2, 4, 6, 8, 10, 12, 14, 16} - {5, 10, 15, 20}

= {2, 4, 6, 8, 12, 14, 16}

(xii) It is given in the question that,

D = {5, 10, 15, 20}

And, C = {2, 4, 6, 8, 10, 12, 14, 16}

∴ D – C = {5, 10, 15, 20} - {2, 4, 6, 8, 10, 12, 14, 16}

= {5, 15, 20}

Question 10.

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

(i) X – Y

(ii) Y – X

(iii) X ∩ Y

Answer:

(i) It is given in the question that,

X = {a, b, c, d}

And, Y = {f, b, d, g}

X - Y implies the elements which are in X but not in Y.

∴ X – Y = {a, b, c, d} - {f, b, d, g}

= {a, c}

(ii) It is given in the question that,

Y = {f, b, d, g}

And, X = {a, b, c, d}

Y - X implies the elements which are in Y but not in X.

∴ Y – X = {f, b, d, g} - {a, b, c, d}

= {f, g}

(iii) It is given in the question that,

X = {a, b, c, d}

And, Y = {f, b, d, g}

∴

= {b, d}

Question 11.

If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?

Answer:

It is given in the question that,

R = Set of real numbers

And, Q = Set of rational numbers

We know that, set of real numbers contain rational numbers and irrational numbers

∴ R – Q = Set of real numbers - Set of rational numbers

= Set of irrational numbers

Question 12.

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

(i) {2, 3, 4, 5} and {3, 6} are disjoint sets.

(ii) {a, e, i, o, u} and {a, b, c, d}are disjoint sets.

(iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.

(iv) {2, 6, 10} and {3, 7, 11} are disjoint sets.

Answer:

(i) Let us assume, A = {2, 3, 4, 5}

And, B = {3, 6}

∴

= {3}

Hence, A and B are not disjoint sets

∴ The given statement is false

(ii) Let us assume, A = {a, e, i, o, u}

And, B = {a, b, c, d}

∴

= {a}

Hence, A and B are not disjoint sets

∴ The given statement is false

(iii) Let us assume, A = {2, 6, 10, 14}

And, B = {3, 7, 11, 15}

∴

Hence, A and B are disjoint sets

∴ The given statement is true

(iv) Let us assume, A = {2, 6, 10}

And, B = {3, 7, 11}

∴

Hence, A and B are disjoint sets

∴ The given statement is true

Exercise 1.5

Question 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 (i) A’ (ii) B’ (iii) (A ∪ C)’ (iv) (A ∪ B)’ (v) (A’)’ (vi) (B – C)’

Answer:

Given that

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

A = {1, 2, 3, 4}

B = {2, 4, 6, 8}

C = {3, 4, 5, 6}

(i) We have to find complement of A, which is given by (U - A)

⇒ A’ = U - A

⇒ A’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4}

⇒ A’ = {5, 6, 7, 8, 9}

(ii) B’ = U - B

⇒ B’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 4, 6, 8}

⇒ B’ = {1, 3, 5, 7, 9}

(iii) (A ⋃ C)’ = U - (A ⋃ C)

First we find (A ⋃ C)

⇒ (A ⋃ C) = {1, 2, 3, 4} ⋃ {3, 4, 5, 6}

⇒ (A ⋃ C) = {1, 2, 3, 4, 5, 6}

Now,

(A ⋃ C)’ = U - (A ⋃ C)

⇒ (A ⋃ C)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4, 5, 6}

⇒ (A ⋃ C)’ = {7, 8, 9}

(iv) (A ⋃ B)’ = U - (A ⋃ B)

First we find (A ⋃ B),

(A ⋃ B) = {1, 2, 3, 4} ⋃ {2, 4, 6, 8}

⇒ (A ⋃ B) = {1, 2, 3, 4, 6, 8}

Now,

(A ⋃ B)’ = U - (A ⋃ B)

⇒ (A ⋃ B)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4, 6, 8}

⇒ (A ⋃ B)’ = {5, 7, 9}

(v)We want to find complement of A’, which is given by U - A’

From part (i), we have A’ = {5, 6, 7, 8, 9}

⇒ (A’)’ = U - A’

⇒ (A’)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {5, 6, 7, 8, 9}

⇒ (A’)’ = {1, 2, 3, 4} = A

We conclude that (A’)’ = A.

(vi) (B - C)’ = U - (B - C)

First we find (B - C),

⇒ (B - C) = {2, 4, 6, 8} - {3, 4, 5, 6}

⇒ (B - C) = {2, 8}

Now,

(B - C)’ = U - (B - C)

⇒ (B - C)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 8}

⇒ (B - C)’ = {1, 3, 4, 5, 6, 7, 9}

Question 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}

Answer:

(i) We have to find complement of A, which is given by (U - A)

⇒ A’ = U - A

⇒ A’ = {a, b, c, d, e, f, g, h} - {a, b, c}

⇒ A’ = {d, e, f, g, h}

(ii) B’ = U - B

⇒ B’ = {a, b, c, d, e, f, g, h} - {d, e, f, g}

⇒ B’ = {a, b, c , h}

(iii) C’ = U - C

⇒ C’ = {a, b, c, d, e, f, g, h} - {a, c, e, g}

⇒ C’ = {b, d , f, h}

(iv) D’ = U - D

⇒ D’ = {a, b, c, d, e, f, g, h} - {f, g , h , a}

⇒ D’ = {b, c, d, e}

Question 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}

Answer:

For all parts, given that
(i) Let A = {x : x is an even natural number}

We want to find complement of A , which is given by U - A

⇒ A’ = U - A

⇒ A’ = {x:x ϵ N} - {x : x is an even natural number}

⇒ A’ = {x : x is an odd natural number}

(ii) Let A = {x : x is an odd natural number}

⇒ A’ = U - A

⇒ A’ = {x:x ϵ N} - {x : x is an odd natural number}

⇒ A’ = {x : x is an even natural number}

(iii) Let A = {x : x is a positive multiple of 3}

⇒ A’ = U - A

⇒ A’ = {x:x ϵ N} - {x : x is a positive multiple of 3}

⇒ A’ = {x : x is not a positive multiple of 3}

(iv) Let A = {x : x is a prime number}

⇒ A’ = U - A

⇒ A’ = {x: x ϵ N} - {x : x is a prime number}

⇒ A’ = {x : x is not a prime number}

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

∴ A = {x : x is a natural number divisible by 15}

⇒ A’ = U - A

⇒ A’ = {x: x ϵ N} - {x : x is a natural number divisible by 15}

⇒ A’ = {x : x is a natural number not divisible by 15}

(vi) Let A = {x : x is a perfect square}

⇒ A’ = U - A

⇒ A’ = {x: x ϵ N} - {x : x is a perfect square}

⇒ A’ = {x : x is not a perfect square}

(vii) Let A = {x : x is a perfect cube}

⇒ A’ = U - A

⇒ A’ = {x: x ϵ N} - {x : x is a perfect cube}

⇒ A’ = {x : x is not a perfect cube}

(viii) Let A = {x : x + 5 = 8}

∴ A = {x : x = 3}

⇒ A’ = U - A

⇒ A’ = {x: x ϵ N} - {x : x = 3}

⇒ A’ = {x : x ϵ N and x ≠ 3}

(ix) Let A = {x : 2x + 5 = 9}

∴ A = {x : x = 2}

⇒ A’ = U - A

⇒ A’ = {x: x ϵ N} - {x : x = 2}

⇒ A’ = {x : x ϵ N and x ≠ 2}

(x) Let A = {x : x ≥ 7}

⇒ A’ = U - A

⇒ A’ = {x: x ϵ N} - {x : x ≥ 7}

⇒ A’ = {x : x < 7}

(xi) Let A = {x: x ϵ N} - {x : 2x + 1 > 10}

∴ A = {x: x ϵ N and x > 9/2}

⇒ A’ = U - A

A’ = {x:x ϵ N} - {x: x ϵ N and x > 9/2}

⇒ A’ = {x: x ϵ N and x < 9/2}

∴ A’ = {1, 2, 3, 4}

Question 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’

Answer:

(i) First solving for left hand side,

(A ⋃ B)’ = U - (A ⋃ B)

First we find (A ⋃ B),

(A ⋃ B) = {2, 4, 6, 8} ⋃ {2, 3, 5, 7}

⇒ (A ⋃ B) = {2, 3, 4, 5, 6, 7, 8}

Now,

(A ⋃ B)’ = U - (A ⋃ B)

⇒ (A ⋃ B)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 3, 4, 5, 6, 7, 8}

⇒ (A ⋃ B)’ = {1, 9}

Now, solving for right hand side,

A’ = U - A

A’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 4, 6, 8}

A’ = {1, 3, 5, 7, 9}

B’ = U - B

B’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 3, 5, 7}

B’ = {1, 4, 6, 8, 9}

A’ ⋂ B’ = {1, 3, 5, 7, 9} ⋂ {1, 4, 6, 8, 9}

⇒ A’ ⋂ B’ = {1, 9}

∴ LHS = RHS, Hence verified.

(ii) First solving for left hand side,

(A ⋂ B)’ = U - (A ⋂ B)

First we find (A ⋂ B),

(A ⋂ B) = {2, 4, 6, 8} ⋂ {2, 3, 5, 7}

⇒ (A ⋂ B) = {2}

Now,

(A ⋂ B)’ = U - (A ⋂ B)

⇒ (A ⋂ B)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2}

⇒ (A ⋂ B)’ = {1, 3, 4, 5, 6, 7, 8, 9}

Now, solving for right hand side,

A’ = U - A

A’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 4, 6, 8}

A’ = {1, 3, 5, 7, 9}

B’ = U - B

B’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 3, 5, 7}

B’ = {1, 4, 6, 8, 9}

A’ ⋃ B’ = {1, 3, 5, 7, 9} ⋃ {1, 4, 6, 8, 9}

⇒ A’ ⋃ B’ = {1, 3, 4, 5, 6, 7, 8, 9}

∴ LHS = RHS, Hence verified.

Question 5.

Draw appropriate Venn diagram for each of the following:

(i) (A ∪ B)’, (ii) A’ ∩ B’, (iii) (A ∩ B)’,

(iv) A’ ∪ B’

Answer:

(i) (A ∪ B)’

First we draw (A ⋃ B)

Yellow region is (A ⋃ B).

The shaded region represents (A ⋃ B).

We have to draw diagram for complement of (A ⋃ B) i.e.(A ∪ B)’, which is given by U - (A ⋃ B).

Green region is (A ∪ B)’.

(ii) A’ ∩ B’

Here we have to draw diagram of (A’ ⋂ B’)

So, first we draw A’( = U - A)

Now, we draw B’( = U - B)

Now the area common in both the shaded regions gives us (A’ ⋂ B’)

Here, we observe that the final result for (i) and (ii) is same.

⇒(A⋃B)’ = (A’⋂B’)

(iii) (A ∩ B)’

First we draw (A ⋂ B)

The shaded region represents (A ⋂ B).

We have to draw diagram for complement of (A ⋂ B) i.e.(A ∩ B)’ , which is given by U - (A ⋂ B)

(iv) A’ ∪ B’

Here we have to draw diagram of (A’ ⋃ B’)

So, first we draw A’( = U - A)

Now, we draw B’( = U - B)

Now the area present in both is added to give (A’ ⋃ B’)

Here, we observe that the final result for (iii) and (iv) is same.

⇒(A⋂B)’ = (A’⋃B’)

Question 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’?

Answer:

Given that,

U = {x: x is a triangle in the plane}

A = {x: x is a triangle with atleast one angle not equal to 60°}

A’ = U - A = {x : x is a triangle with all angles equal to 60°}

∴ A’ = {x: x is set of all equilateral triangle}.

Question 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 = . . .

Answer:

(i) A ⋃ A’ = U

(ii) ϕ’ ⋂ A = U ⋂ A = A (as ϕ’ = U)

(iii) A ⋂ A’ = ϕ

(iv) U’ ⋂ A = ϕ ⋂ A = ϕ (as U’ = ϕ)

Exercise 1.6

Question 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 ).

Answer:

Given-

n(X) = 17, n(Y) = 23, and n(X ∪ Y) = 38

We know that-

n(X ∪ Y) = n(X)+ n(Y) - n(X ∩ Y)

⇒ 38 = 17+23 - n(X ∩ Y)

⇒ 38 = 40 - n(X ∩ Y)

⇒ n(X ∩ Y) = 40-38

∴ n(X ∩ Y) = 2

Question 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?

Answer:

Given-

n(X) = 8, n(Y) = 15, and n(X ∪ Y) = 18

We know that-

n(X ∪ Y) = n(X)+ n(Y) - n(X ∩ Y)

⇒ 18 = 8+15 - n(X ∩ Y)

⇒ 18 = 23 - n(X ∩ Y)

⇒ n(X ∩ Y) = 23-18

∴ n(X ∩ Y) = 5

Question 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?

Answer:

Let H be the set of people who speak Hindi,

and E be the set of people who speak English.

Number of people who speak Hindi = n(H) = 250

Number of people who speak English = n(E) = 200

Total Number of people = n(H ∪ E) = 400

Number of people who can both speak Hindi and English

= n(H ∩ E)

We know that-

n(H ∪ E) = n(H)+ n(E) - n(H ∩ E)

⇒ 400 = 250+200 - n(H ∩ E)

⇒ 400 = 450 - n(H ∩ E)

⇒ n(H ∩ E) = 450-400

∴ n(H ∩ E) = 50

Thus, 50 people can speak both Hindi and English.

Question 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?

Answer:

Given-

n(S) = 21, n(T) = 32, and n(S ∩ T) = 11

We know that-

n(S ∪ T) = n(S)+ n(T) - n(S ∩ T)

= 21+32 - 11

= 53 - 11

= 42

∴ n(S ∪ T) = 42

Thus, the no. of elements in n(S ∪ T) is 42.

Question 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?

Answer:

Given-

n(X) = 40, n(X ∩ Y) = 10, and n(X ∪ Y) = 60

We know that-

n(X ∪ Y) = n(X)+ n(Y) - n(X ∩ Y)

⇒ 60 = 40+n(Y) - 10

⇒ 60 = 30+n(Y)

⇒ n(Y) = 60-30

∴ n(Y) = 30

Thus, the no. of elements in n(Y) is 30.

Question 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?

Answer:

Let T be the set of people who like Tea,

and C be the set of people who like Coffee.

Number of people who like Tea = n(T) = 52

Number of people who like Coffee = n(C) = 37

Number of people who like at least tea or Coffee = n(T ∪ C ) = 70

Number of people who like both tea and Coffee = n(T ∩ C )

We know that-

n(T ∪ C) = n(T)+ n(C) - n(T ∩ C)

⇒ 70 = 52+37 - n(T ∩ C)

⇒ 70 = 89 - n(H ∩ E)

⇒ n(H ∩ E) = 89-70

∴ n(H ∩ E) = 19

Thus, 19 people like both Tea and Coffee.

Question 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?

Answer:

Let T be the set of people who like Tennis,

and C be the set of people who like Cricket.

Number of people who like Cricket = n(C) = 40

Number of people who like at tennis or Cricket = n(T ∪ C ) = 65

Number of people who like both tennis and Cricket = n(T ∩ C )

= 10

Number of people who like Tennis = n(T)

We know that-

n(T ∪ C) = n(T)+ n(C) - n(T ∩ C)

⇒ 65 = n(T)+40 - 10

⇒ 65 = n(T)+30

⇒ n(T) = 65-30

∴ n(T) = 35

Thus, the number of people who like tennis = 35

Now,

The number of people who like tennis only and not cricket

= Number of people who like Tennis

- Number of people who like both tennis and Cricket

= n(T) - n(T ∩ C )

= 35-10

= 25

Question 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?

Answer:

Let F be the set of people who speak French,

and S be the set of people who speak Spanish.

Number of people who speak French = n(F) = 50

Number of people who speak Spanish = n(S) = 20

Number of people who can both speak French and Spanish

= n(F ∩ S)

= 10

Number of people who speak at least one of these two languages = n(F ∪ S)

We know that-

n(F ∪ S) = n(F) + n(S) - n(F ∩ S)

= 50+20 - 10

= 60

∴ n(H ∩ E) = 60

Thus, 60 people can speak at least one of French or Spanish.

Miscellaneous Exercise

Question 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}

Answer:

It is given in the question that,

A = {x : x R and x satisfy x² – 8x + 12 = 0}

As, 2 and 6 are the only solutions of x² – 8x + 12 = 0

∴ A = {2, 6}

Also it is given that,

B = {2, 4, 6}

C = {2, 4, 6, 8, …..}

And, D = {6}

∴

Hence,

Question 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:

(i) If x ∈ A and A ∈ B, then x ∈ B

(ii) If A ⊂ B and B ∈ C, then A ∈ C

(iii) If A ⊂ B and B ⊂ C, then A ⊂ C

(iv) If A ⊄ B and B ⊄ C, then A ⊄ C

(v) If x ∈ A and A ⊄ B, then x ∈ B

(vi) If A ⊂ B and x ∉ B, then x ∉ A

Answer:

(i) Let us assume A = {1, 2}

And, B = {1, {1, 2}, {3}}

So,

And,

∴

But,

Hence, the given statement is false

(ii) Let us assume, A {2}

B = {0, 2}

And, C = {1, {0, 2}, 3}

It is given in the question that,

∴

But,

Hence, the given statement is false

(iii) It is given in the question that,

Let us assume,

So,

And,

∴

Hence, the given statement is correct

(iv) It is given in the question that,

A ⊄ B

And, B ⊄ C

Let us now assume, A = {1, 2}

B = {0, 6, 8}

And, C = {0, 1, 2, 6, 9}

∴

Hence, the given statement is false

(v) It is given in the question that,

And, A ⊄ B

Let us now assume, A = {3, 5, 7}

And, B = {3, 4, 6}

As,

And,

∴

Hence, the given statement is false

(vi) It is given in the question that,

A ⊂ B

And, x ∉ B

Let us suppose, then we have:

But it is given in the question that, x ∉ B

∴ x ∉ A

Hence, the given statement is true

Question 3.

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

Answer:

It is given in the question that,

And,

We have to show that, B = C

Let us now assume,

So,

∴

Now, when then:

∴

As,

So,

∴

Similarly, it can be shown that C ⊂ B

Hence, B = C

Question 4.

Show that the following four conditions are equivalent:
(i) A ⊂ B

(ii) A – B = ϕ

(iii) A ∪ B = B
(iv) A ∩ B = A

Answer:

Here, first we will prove (i) ⬄ (ii)

Where, (i) = A ⊂ B and (ii) = A – B ≠ ϕ

Let us assume that A ⊂ B

Now, we need to prove A – B ≠ ϕ

If possible, let A – B ≠ ϕ

Thus, there exists X ϵ A, X ≠ B, but this is impossible as A⊂ B

∴ A – B = ϕ

And A⊂ B =>A – B ≠ ϕ

Let us assume that A – B ≠ ϕ

Now, to prove: A ⊂ B

Let Xϵ A

It can be concluded that X ϵ B (if X ∉ B, then A – B ≠ ϕ)

Thus, A – B = ϕ => A ⊂ B

∴(i) ⬄ (ii)

Let us assume that A ⊂ B

To prove: A ∪ B = B

⇒ B ⊂ A ∪ B

Let us assume that, x ϵ A∪ B

⇒ X ϵ A or X ϵ B

Taking Case I: X ϵ B

A ∪ B = B

Taking Case II: X ϵ A

⇒ X ϵ B (A ⊂ B)

⇒ A ∪ B ⊂ B

Let A ∪ B = B

Let us assume that X ϵ A

⇒ X ϵ A ∪ B (A ⊂ A ∪ B)

⇒ X ϵ B (A ∪ B = B)

∴A⊂ B

Thus, (i) ⬄ (iii)

Now, to prove (i) ⬄ (iv)

Let us assume that A ⊂ B

It can be observed that A ∩ B ⊂ A

Let X ϵ A

To show: X ϵ A∩ B

Since, A ⊂ B and X ϵ B

Thus, X ϵ A∩ B

⇒ A ⊂ A ∩ B

⇒ A = A ∩ B

Similarly, let us assume that A ∩ B = A

Let X ϵ A

⇒ X ϵ A ∩ B

⇒ X ϵ B and X ϵ A

⇒ A ⊂ B

∴ (i) ⬄ (iv)

Hence, proved that (i)⬄ (ii)⬄ (iii)⬄ (iv)

Question 5.

Show that if A ⊂ B, then C – B ⊂ C – A.

Answer:

To Prove: C – B ⊂ C – A

Given: A ⊂ B

Proof:

Let us now assume x is any element such that

∴

So,

∴ C – B ⊂ C – A

Hence, Proved.
Question 6.

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

Answer:

We have to show that: A = B

Given: P (A) = P (B)

Let x be any element of set A,

P(A) is the power set of set A, hence it contains all the subsets of set A. Thus set A is contained in set P(A).

Now let C be an element of set B

∴

Now, we have:

C ⊂ B

∴

∴ A ⊂ B

Similarly, we have:

B ⊂ A

Now if A ⊂ B and B ⊂ A

∴ A = B

Question 7.

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

Answer:

Let us assume, A = {0, 1}

And, B = {1, 2}

∴

Now, we have:

∴

Also,

∴

Hence, the given statement is false

Question 8.

Show that for any sets A and B,

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

Answer:

To Prove: A = (A ∩ B) ∪ (A – B)

Proof: Let X ϵ A

Now, we need to show that X ϵ (A ∩ B) ∪ (A – B)

In Case I,

X ϵ (A∩ B)

⇒ X ϵ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B)

In Case II,

X ∉A ∩ B

⇒ X ∉ B or X ∉ A

⇒ X ∉ B (X ∉ A)

⇒ X ∉ A – B ⊂ (A ∪ B) ∪ (A – B)

∴A ⊂ (A ∩ B) ∪ (A – B) (i)

It can be concluded that, A ∩ B ⊂ A and (A – B) ⊂ A

Thus, (A ∩ B) ∪ (A – B) ⊂ A (ii)

Equating (i) and (ii),

A = (A ∩ B) ∪ (A – B)

Now, we need to show, A ∪ (B – A) ⊂ A ∪ B

Let us assume that,

X ϵ A ∪ (B – A)

X ϵ A or X ϵ (B – A)

⇒ X ϵ A or (X ϵ B and X ∉A)

⇒ (X ϵ A or X ϵ B) and (X ϵ A and X ∉A)

⇒ X ϵ (B ∪A)

∴ A ∪ (B – A) ⊂ (A ∪ B) (iii)

Now, to prove: (A ∪ B) ⊂ A ∪(B – A)

Let y ϵ A∪B

yϵ A or y ϵ B

(y ϵ A or y ϵ B) and (X ϵ A and X ∉A)

⇒ y ϵ A or (y ϵ B and y ∉A)

⇒ y ϵ A ∪ (B – A)

Thus, A ∪ B ⊂ A ∪ (B – A) (iv)

∴Using (iii) and (iv), we get:

A ∪ (B – A) = A ∪ B

Question 9.

Using properties of sets, show that:

(i) A ∪ (A ∩ B) = A

(ii) A ∩ (A ∪ B) = A.

Answer:

(i) We have to show that:

Now, we know that:

∴ (i)

Also, we have:

(ii)

Hence, by using equation (i) and (ii) we have:

(ii) We have to show that:

Question 10.

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

Answer:

Let us assume, A = {0, 1}

B = {0, 2, 3}

And, C = {0, 4, 5}

Now, accordingly we have:

And,

∴

Also,

Question 11.

Let A and B be sets. If A ∩ X = B ∩ X = f 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)

Answer:

It is given in the question that,

A and B are sets such that

And, for some set X

Now, we have to show that A = B

As it can be seen that,

= (By using distributive law)

= (i)

Now, we have:

= (By using distributive law)

(ii)

∴ From the equation (i) and (ii), we have

A = B

Question 12.

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

Answer:

Let us assume, A {0, 1}

B = {1, 2}

And, C = {2, 0}

Now, accordingly we have:

And,

∴

Hence,

Question 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?

Answer:

Let us assume U be the set of all students who took part in the survey

Let T be the set of students taking tea

And, let C be the set of the students taking coffee

Total number of students in a school, n (U) = 600

Number of students taking tea, n (T) = 150

Number of students taking coffee, n (C) = 225

Also,

Now, we have to find that number of students taking neither coffee nor tea i.e.

∴

= 600 – [150 + 225 – 100]

= 600 – 275

= 325

∴ Number of students taking neither coffee nor tea = 325 students

Question 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?

Answer:

Let us assume U be the set of all students in the group

Let E be the set of students who know English

And, let H be the set of the students who know Hindi

∴

Given that, number of students who know Hindi n (H) = 100

Number of students who knew English, n (E) = 50

Number of students who know both,

We have to find the total number of students in the group i.e. n (U)

∴

= 100 + 50 – 25

= 125

∴ Total number of students in the group = 125 students

Question 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:

(i) The number of people who read at least one of the newspapers.

(ii) The number of people who read exactly one newspaper.

Answer:

(i) Let us assume A be the set of people who read newspaper H

Also, let B be the set of people who read newspaper T

And, let C be the set of people who read newspaper I

It is given in the question that,

Number of people who read newspaper H, n (A) = 25

Number of people who read newspaper T, n (B) = 26

Number of people who read the newspaper I, n (C) = 26

Number of people who read both newspaper H and I,

Number of people who read both newspaper H and T,

Number of people who read both newspaper T and I,

And, Number of people who read all three newspaper H, T and I,

Now, we have to find the number of people who read at least one of the newspaper

∴

= 25 + 26 + 26 – 11 – 8 – 9 + 3

= 80 – 28

= 52

∴ There are a total of 52 students who read at least one newspaper.

(ii) Let us assume a be the number of people who read newspapers H and T only

Let b denote the number of people who read newspapers I and H only

Let c denote the number of people who read newspapers T and I only

And, let d denote the number of people who read all three newspapers

It is given in the question that:

Now, we have:

And,

∴ a + d + c +d + b + d = 11 + 8 + 9

a + b + c + d = 28 – 2d

= 28 – 6

= 22

∴ Number of people read exactly one newspaper = 52 – 22

= 30 people

Question 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.

Answer:

Let us first assume A, B and C be the set of people who like product A, product B and product C respectively

Now, it is given in the question that:

Number of students who like product A, n (A) = 21

Number of students who like product B, n (B) = 26

Number of students who like product C, n (C) = 29

Number of students who like both products A and B,

Number of students who like both products A and C,

Number of students who like both product C and B,

Number of students who like all three product,

On the basis of given condition the venn diagram for the following question can be drawn as follows:

From the venn diagram, it can clearly be seen that:

Number of students who only like product C = {29 – (4 + 8 + 6)}

= {29 – 18}

= 11 students

Sets

Class 11th Mathematics Bihar Board Solution

Exercise 1.1

Exercise 1.2

Exercise 1.3

Exercise 1.4

Exercise 1.5

Exercise 1.6

Miscellaneous Exercise

Class 11^th Mathematics Bihar Board Solution