## Sets – Maharashtra Board Class 9 Solutions for Algebra

AlgebraGeometryScience and TechnologyHindi

**Exercise – 1.1**

**Solution 1:**

A well-defined collection of objects is called a set:

The collection of prime numbers is a well-defined collection of objects and is hence a set.

The collection of easy subtopics in this chapter is not a well-defined collection of objects and is hence not a set.

The collection of good teachers in your school is not a well-defined collection of objects and is hence not a set.

The collection of girls in your class is a well-defined collection of objects and is hence a set.

The collection of odd natural numbers is a well-defined collection of objects and is hence a set.

**Solution 2:**

**Solution 3:**

**Solution 4:**

Let ‘A’ be the set of first five integers whose square is odd.

A = {1,3,5,7,9}

**Exercise – 1.2**

**Solution 1:**

MP Scholarship

**Solution 2:**

An empty set has no elements.

i. Let ‘A’ be the set of all even prime numbers

A = {2}; Set A has only one element.

Thus, A is a singleton set.

Hence, A is not an empty set

ii. B = {x|x is a capital of India}

∴B = {Delhi}; Set B has only one element.

Thus, B is a singleton set.

Hence, B is not an empty set

iii. F = {y|y is a point of intersection of two parallel lines}

The two parallel lines do not intersect each other.

∴There are 0 elements in set F

∴Hence, F is an empty set.

iv. G = {z|z ∈ N, 3 < Z < 4}

There is no natural number between 3 and 4

∴G = ϕand hence G is an empty set.

v. H = {t|triangle having four sides}

A triangle always has three sides and it cannot have four sides.

Hence, H is an empty set.

**Solution 3:**

**Solution 4:**

Since the number of boys and girls in a class terminates at a certain stage, G and H are finite sets.

**Exercise – 1.3**

**Solution 1:**

i. The subset relation between the set A and set C is:

A = The set of all residents in Mumbai

C = The set of all residents in Maharashtra

Since Mumbai is in Maharashtra, every element of Set A is an element of Set C, hence, set A is a subset of set C.

But there are some elements in set C which are not in set A, therefore A is the proper subset of set C.

Hence, A ⊂ C

ii.The subset relation between the sets E and D is:

E ⊂ D

E = The set of all residents in Madhya-Pradesh.

D = The set of all residents in India

Since Madhya-Pradesh is inIndia, every element of Set E is an element of Set D, hence, set E is a subset of set D.

But there are some elements in set D which are not in set E, therefore D is the proper subset of set E.

Hence, E ⊂ D

iii. Mumbai, Bhopal, Maharashtra, and Madhya-Pradesh all are in India. Therefore, all the sets under consideration are the subsets of set D.

Hence, A ⊂ D, B ⊂ D, C ⊂ D, E ⊂ D

Therefore Set D can be chosen as the universal set.

**Solution 2:**

**Solution 3:**

**solution 4:**

Let x be an element of set A.

Thus, ∴ x ∊ A.

Given that A ⊂ B.

∴ Every element of set A is an element of B.

∴x ∊ B

Also given that B ⊂ C

∴ Every element of set B is an element of C.

∴x ∊ C

∴If x ∊ A, then x ∊ C for all x ∊ A.

∴ Every element of the set A is an element of the set C.

Hence, A ⊂ C.

**Solution 5:**

All possible subsets of X = {1, 2, 3} are:

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

**Exercise – 1.4**

**Solution 1:**

P = {x|x is a letter in the word ‘CATARACT’}

Let us avoid the repetition of letters and write the distinct letters in curly brackets, with comma as a separator.

∴ P = {C, A, T, R}

Q = {y|y is a letter in the word ‘TRAC’}

∴ Q = {T, R, A, C}

The order in which the letters are listed is immaterial.

∴Every element of set P is an element of set Q.

∴P Í Q.

Also, every element of set Q is an element of set P

∴Q Í P.

All the elements in set P and all the elements in set Q are equal, hence, P = Q.

**Solution 2:**

The union of pairs of sets A and B is the set of all elements which are in set A or in set B. It is denoted by A ∪ B.

i. A = {2, 3, 5, 6, 7}, B ={4, 5, 7, 8}

A ∪ B = {2, 3, 4, 5, 6, 7, 8}

ii. C = {a, e, i, o, u}, D ={a, b, c, d}

C ∪ D = { a, b, c, d, e, i, o, u }

iii. E= {x|x ∈ N and x is a divisor of 12}

∴E= {1, 2, 3, 4, 6, 12}

F= {y|y ∈ N and y is a divisor of 18}

∴F = {1, 2, 3, 6, 9, 18}

∴E ∪ F = {1, 2, 3, 4, 6, 9, 12, 18}

**Solution 3:**

**Solution 4:**

**Solution 5:**

**Exercise – 1.5**

**Solution 1:**

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

∴A ∪ B = {1, 3, 4, 5, 6, 7, 9}

The number of elements of the set A ∪ B is denoted by n(A ∪ B).

∴n(A ∪ B) = 7

∴L.H.S. = 7

Now consider the R.H.S, n (A) + n (B) – n (A ∩ B).

n (A) = 5, n (B) = 4, n (A ∩ B) = 2

∴n (A) + n (B) = 5 + 4 = 9

∴n (A) + n (B) – n (A ∩ B) = 5 + 4 – 2 = 7

∴L.H.S = R.H.S

∴n(A ∪ B) = n (A) + n (B) – n (A ∩ B)

Hence, verified.

**Solution 2:**

We know the identity,

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

Substituting the values n(A) = 5, n(A ∪ B) = 9 and n(A ∩ B) = 2 in the above identity, we have,

9= 5 + n(B) – 2

∴ n(B) = 9 + 2 – 5

∴ n(B) = 11 – 5

∴ n(B) = 6

**Solution 3:**

Let A be the set of students who drink tea.

∴ n (A) = 60

Let B be the set of students who drink coffee.

∴n (A) = 60

30 students drink both coffee and tea.

∴n (A ∩ B) = 30

(A ∪ B) will be the set of students who take at least one drink (tea or coffee or both)

n (A ∪ B) = n (A) + n (B) - n (A ∩ B)

∴ n (A ∪ B) = 60 + 50 - 30

∴ n (A ∪ B) = 80

Hence, there are 80 students who drink either tea or coffee or both. Consider the following Venn diagram.

∴ Number of students who neither drink tea nor coffee

= Total number of students in the class – number of students who drink either tea or coffee or both.

= 100 – 80 = 20

∴ Number of students who do neither drink tea nor coffee are 20

**Solution 4:**

Let A be the set of students who choose blue.

∴ n (A) = 60

Let B be the set of students who choose pink.

∴ n (A) = 70

(A ∩ B) will be the students who choose both the colours.

Every student has to choose at least one of the colour Therefore, n (A ∩ B) = n (A) + n (B) – total number of students

∴n (A ∩ B) = 60 + 70 – 110

∴n (A ∩ B) = 130 – 110 = 20

Hence, there are 20 students who choose both the colours as their favourite colour.

**Solution 5:**

Consider the given equation:

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

n (A) = 5, n (B) = 5, n (C) = 4

n (A ∩ B) = 2

n (B ∩ C) = 2

n (C ∩ A) = 2

n (A ∩ B ∩ C) = 1

Substituting the values in the given equation,

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

∴ n (A ∪ B ∪ C) = 5 + 5 + 4 – 2 – 2 – 2 + 1

∴ n (A ∪ B ∪ C) = 9

From the given figure it can be seen that there are 9 elements in all.

Hence, the equation is verified.

## Leave a Reply