Sets – Maharashtra Board Class 9 Solutions for Algebra

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.