Maharashtra Board Class 9 Maths Chapter 1 Set 1.3 Algebra Standard Part 1 Sets Solutions

Question 1. If \( A = \{a, b, c, d, e\} \), \( B = \{c, d, e, f\} \), \( C = \{b, d\} \), \( D = \{a, e\} \), then which of the following statements are true and which are false?
i. \( C \subseteq B \)
ii. \( A \subseteq D \)
iii. \( D \subseteq B \)
iv. \( D \subseteq A \)
Answer:
i. False. Element b of set C is not present in set B.
ii. False. Elements b, c, d of set A are not present in set D.
iii. False. Element a of set D is not present in set B.
iv. True. All elements of set D are present in set A. Understanding subset relationships helps in analyzing how different groups of elements interact with each other.
In simple words: A set is a subset of another only if every single member of the first set is also inside the second set. Here, only statement iv is true because all elements of D are also in A.

🎯 Exam Tip: To check if one set is a subset of another, carefully verify if every element of the first set is present in the second set. If even one element is missing, it is not a subset.

Question 1. If A = {a, b, c, d, e}, B = {c, d, e, f}, C = {b, d}, D = {a, e}, then decide which of the following statements are true and which are false:
(i) \( C \subseteq B \)
(ii) \( A \subseteq D \)
(iii) \( D \subseteq B \)
(iv) \( D \subseteq A \)
(v) \( B \subseteq A \)
(vi) \( C \subseteq A \)
Answer:
(i) False. Since \( C = \{b, d\} \) and \( B = \{c, d, e, f\} \), all the elements of C are not present in B.
(ii) False. Since \( A = \{a, b, c, d, e\} \) and \( D = \{a, e\} \), all the elements of A are not present in D.
(iii) False. Since \( D = \{a, e\} \) and \( B = \{c, d, e, f\} \), all the elements of D are not present in B.
(iv) True. Since \( D = \{a, e\} \) and \( A = \{a, b, c, d, e\} \), all the elements of D are present in A.
(v) False. Since \( B = \{c, d, e, f\} \) and \( A = \{a, b, c, d, e\} \), all the elements of B are not present in A.
(vi) True. Since \( C = \{b, d\} \) and \( A = \{a, b, c, d, e\} \), all the elements of C are present in A.
In simple words: A set is a subset of another set only if every single element of the first set is also found in the second set.

🎯 Exam Tip: To check if a subset relation is true, write down the elements of both sets and verify if every element of the first set is present in the second set.

Question 2. Take the set of natural numbers from 1 to 20 as universal set and show set X and Y using Venn diagram.
(i) \( X = \{x \mid x \in \mathbb{N}, \text{ and } 7 < x < 15\} \)
(ii) \( Y = \{y \mid y \in \mathbb{N}, y \text{ is a prime number from 1 to 20}\} \)
Answer:
The Universal set is \( U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\} \).

(i) \( X = \{x \mid x \in \mathbb{N}, \text{ and } 7 < x < 15\} \)
In roster form: \( X = \{8, 9, 10, 11, 12, 13, 14\} \)
The Venn diagram representation consists of:

• Universal Set (U): A rectangle containing all numbers from 1 to 20.
• Set X: A closed circle inside the rectangle containing the elements: 8, 9, 10, 11, 12, 13, 14.
• Outside Set X (but inside U): The remaining elements: 1, 2, 3, 4, 5, 6, 7, 15, 16, 17, 18, 19, 20.

(ii) \( Y = \{y \mid y \in \mathbb{N}, y \text{ is a prime number from 1 to 20}\} \)
In roster form: \( Y = \{2, 3, 5, 7, 11, 13, 17, 19\} \)
The Venn diagram representation consists of:

• Universal Set (U): A rectangle containing all numbers from 1 to 20.
• Set Y: A closed circle inside the rectangle containing the elements: 2, 3, 5, 7, 11, 13, 17, 19.
• Outside Set Y (but inside U): The remaining elements: 1, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20.

In simple words: A Venn diagram uses a big rectangle for the universal set and circles inside it for other sets to show which numbers belong inside the circle and which stay outside.

🎯 Exam Tip: Always list the elements of the universal set and the given sets in roster form first before drawing or describing the Venn diagram to avoid missing any numbers.

Question 2. Take the set of natural numbers from 1 to 20 as universal set and show set X and Y using Venn diagram.
(i) \( X = \{x \mid x \in N \text{ and } 7 < x < 15\} \)
(ii) \( Y = \{y \mid y \in N \text{ and } y \text{ is a prime number from 1 to 20}\} \)
Answer:
(i)
\( U = \{1, 2, 3, 4, \dots, 18, 19, 20\} \)
\( X = \{x \mid x \in N \text{ and } 7 < x < 15\} \)
\( \therefore X = \{8, 9, 10, 11, 12, 13, 14\} \)

Venn Diagram Representation:

• Universal Set (U): Represented by a rectangle containing all numbers from 1 to 20.
• Set X: Represented by a circle inside the rectangle containing elements: 8, 9, 10, 11, 12, 13, 14.
• Outside Set X (but inside U): Elements 1, 2, 3, 4, 5, 6, 7, 15, 16, 17, 18, 19, 20 are placed outside the circle.

(ii)
\( U = \{1, 2, 3, 4, \dots, 18, 19, 20\} \)
\( Y = \{y \mid y \in N, y \text{ is a prime number from 1 to 20}\} \)
\( \therefore Y = \{2, 3, 5, 7, 11, 13, 17, 19\} \)

Venn Diagram Representation:

• Universal Set (U): Represented by a rectangle containing all numbers from 1 to 20.
• Set Y: Represented by a circle inside the rectangle containing elements: 2, 3, 5, 7, 11, 13, 17, 19.
• Outside Set Y (but inside U): Elements 1, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20 are placed outside the circle.

In simple words: To show these sets, we list the numbers that fit the rules (like numbers between 7 and 15, or prime numbers) inside a circle, and put the rest of the numbers from 1 to 20 outside that circle but inside the big rectangle.

🎯 Exam Tip: Always remember to draw a rectangle for the Universal Set \( U \) and label it clearly in the corner, while using circles or ovals for subsets like \( X \) and \( Y \).

Question 3. U = {1, 2, 3, 7, 8, 9, 10, 11, 12}, P = {1, 3, 7, 10}, then
(i) show the sets U, P and P’ by Venn diagram.
(ii) Verify (P’)’ = P
Answer:
(i)
Here, \( U = \{1, 2, 3, 7, 8, 9, 10, 11, 12\} \)
\( P = \{1, 3, 7, 10\} \)
\( \therefore P' = \{2, 8, 9, 11, 12\} \)

Venn Diagram Representation:

• Universal Set (U): Represented by a rectangle containing all elements {1, 2, 3, 7, 8, 9, 10, 11, 12}.
• Set P: Represented by a circle inside the rectangle containing elements {1, 3, 7, 10}.
• Set P' (outside P but inside U): Contains elements {2, 8, 9, 11, 12} written outside the circle of set P.

(ii)
Here, \( U = \{1, 2, 3, 7, 8, 9, 10, 11, 12\} \)
\( P = \{1, 3, 7, 10\} \) ....(i)
\( \therefore P' = \{2, 8, 9, 11, 12\} \)
Now, \( (P')' \) is the complement of set \( P' \), which contains all elements of \( U \) that are not in \( P' \).
\( \therefore (P')' = \{1, 3, 7, 10\} \) ....(ii)
From (i) and (ii), we get:
\( (P')' = P \)
Hence verified.
In simple words: The complement of a set contains everything in the universal set that is not in that set. If you take the complement twice, you end up right back where you started with the original set.

🎯 Exam Tip: When verifying complement properties, clearly label your equations as (i) and (ii) to show the examiner that both sets contain the exact same elements.

Question 4. If \( A = \{1, 3, 2, 7\} \), then write any three subsets of \( A \).
Answer:
Any set containing elements only from set \( A \) is a subset of \( A \). Here are three possible subsets:
i. \( B = \{3\} \)
ii. \( C = \{2, 1\} \)
iii. \( D = \{1, 2, 7\} \)
[Note: The above problem has many solutions. Students may write other valid subsets such as the empty set or the set \( A \) itself.]
In simple words: A subset is just a smaller group made using some or all of the numbers from the main group. For example, we can make a subset using just the number 3, or the numbers 1 and 2.

🎯 Exam Tip: Remember that the empty set \( \emptyset \) and the set itself are always subsets of any given set. Writing these is a quick and easy way to get full marks.

Question 5.
i. Write the subset relation between the sets.
P is the set of all residents in Pune.
M is the set of all residents in Madhya Pradesh.
I is the set of all residents in Indore.
B is the set of all residents in India.
H is the set of all residents in Maharashtra.

ii. Which set can be the universal set for above sets ?
Answer:
i. Subset relations:
a. Since Pune is a city in India, all residents of Pune are also residents of India.
\( \therefore P \subseteq B \)
b. Since Pune is a city in Maharashtra, all residents of Pune are also residents of Maharashtra.
\( \dots P \subseteq H \)
c. Since Madhya Pradesh is a state in India, all residents of Madhya Pradesh are also residents of India.
\( \therefore M \subseteq B \)
d. Since Indore is a city in India, all residents of Indore are also residents of India.
\( \therefore I \subseteq B \)
e. Since Indore is a city in Madhya Pradesh, all residents of Indore are also residents of Madhya Pradesh.
\( \dots I \subseteq M \)
f. Since Maharashtra is a state in India, all residents of Maharashtra are also residents of India.
\( \therefore H \subseteq B \)

ii. Universal set:
Set \( B \) (the set of all residents in India) can be the universal set for all the above sets because all other sets \( P, M, I, \) and \( H \) are subsets of set \( B \). This is because India geographically encompasses both states and all their cities.
In simple words: A subset relation shows how one smaller group fits inside a bigger group, like how Pune is inside Maharashtra, and Maharashtra is inside India. Since India contains all these states and cities, the set of all residents in India acts as the big "universal" group that holds everyone.

🎯 Exam Tip: When identifying subset relations, think of geographical hierarchy (City \( \subseteq \) State \( \subseteq \) Country) to easily determine which set is contained within another.

Question 6. Which set of numbers could be the universal set for the sets given below?
(i) A = set of multiples of 5,
B = set of multiples of 7,
C = set of multiples of 12
(ii) P = set of integers which are multiples of 4.
T = set of all even square numbers.
Answer:
(i) A = set of multiples of 5
\( \therefore A = \{5, 10, 15, \dots\} \)
B = set of multiples of 7
\( \therefore B = \{7, 14, 21, \dots\} \)
C = set of multiples of 12
\( \dots C = \{12, 24, 36, \dots\} \)
Now, the set of natural numbers (\( N \)), whole numbers (\( W \)), integers (\( I \)), and rational numbers (\( Q \)) are as follows:
\( N = \{1, 2, 3, \dots\} \)
\( W = \{0, 1, 2, 3, \dots\} \)
\( I = \{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\} \)
\( Q = \left\{ \frac{p}{q} \;\middle|\; p, q \in I, q \neq 0 \right\} \)
Since sets A, B, and C are subsets of N, W, I, and Q, we can take any of these sets as the universal set.
\( \therefore \) For sets A, B, and C, we can take any one of the sets \( N \), \( W \), \( I \), or \( Q \) as the universal set.

(ii) P = set of integers which are multiples of 4.
\( P = \{4, 8, 12, \dots\} \)
T = set of all even square numbers.
\( T = \{2^2, 4^2, 6^2, \dots\} \)
Since sets P and T are subsets of N, W, I, and Q, we can take any of these sets as the universal set.
\( \dots \) For sets P and T, we can take any one of the sets \( N \), \( W \), \( I \), or \( Q \) as the universal set.
In simple words: A universal set is a larger set that contains all the elements of the smaller sets we are looking at. Since all the numbers in our lists are whole numbers, integers, or rational numbers, any of those big groups can act as the universal set.

🎯 Exam Tip: Remember that any set containing all elements of the given sets can serve as a universal set. Usually, the set of Natural numbers (\( N \)) or Integers (\( I \)) is the easiest choice for number sets.

Question 7. Let all the students of a class form a Universal set. Let set A be the students who secure 50% or more marks in Maths. Then write the complement of set A.
Answer: Let \( U \) be the universal set of all students in the class.
Set \( A \) = {students who secure 50% or more marks in Maths}
The complement of set \( A \) (denoted as \( A' \)) will consist of all the students in the universal set who are not in set \( A \). This means it includes all students who scored less than 50% marks.
\( \therefore A' \) = {students who secure less than 50% marks in Maths}
In simple words: The complement of a set contains everything that is NOT in that set. So, if set A is students with 50% or more marks, its complement is simply the students who got less than 50% marks.

🎯 Exam Tip: When writing the complement of a set, use the exact opposite condition of the original set (e.g., "less than 50%" is the complement of "50% or more").

Here, \( U \) = all the students of a class.
\( A \) = Students who secured 50% or more marks in Maths.
\( \therefore A' \) = Students who secured less than 50% marks in Maths.

Question 1. If \( A = \{1, 3, 4, 7, 8\} \), then write all possible subsets of \( A \).
i.e. \( P = \{1, 3\} \), \( T = \{4, 7, 8\} \), \( V = \{1, 4, 8\} \), \( S = \{1, 4, 7, 8\} \)
In this way many subsets can be written. Write five more subsets of set \( A \).
Answer: Five more subsets of set \( A \) are:
\( B = \{ \} \)
\( E = \{4\} \)
\( C = \{1, 4\} \)
\( D = \{3, 4, 7\} \)
\( F = \{3, 4, 7, 8\} \)
These subsets are formed by choosing different combinations of elements from the original set.
In simple words: A subset is just a smaller group made from the main group. We can make many different smaller groups, including an empty group or a group with just one number.

🎯 Exam Tip: Remember that the empty set \( \emptyset \) (or \( \{ \} \)) and the set itself are always subsets of any given set.

Question 2. Some sets are given below.
\( A = \{\dots, -4, -2, 0, 2, 4, 6, \dots\} \)
\( B = \{1, 2, 3, \dots\} \)
\( C = \{\dots, -12, -6, 0, 6, 12, 18, \dots\} \)
\( D = \{\dots, -8, -4, 0, 4, 8, \dots\} \)
\( I = \{\dots, -3, -2, -1, 0, 1, 2, 3, 4, \dots\} \)
Discuss and decide which of the following statements are true.
a. \( A \) is a subset of sets \( B \), \( C \) and \( D \).
b. \( B \) is a subset of all the sets which are given above.
Answer:
a. All elements of set \( A \) are not present in sets \( B \), \( C \), and \( D \).
\( \implies A \not\subseteq B \)
\( \implies A \not\subseteq C \)
\( \implies A \not\subseteq D \)
Therefore, statement (a) is false.

b. All elements of set \( B \) are not present in sets \( A \), \( C \), and \( D \).
\( \implies B \not\subseteq A \)
\( \implies B \not\subseteq C \)
\( \implies B \not\subseteq D \)
Therefore, statement (b) is false. Both statements are incorrect because the elements do not match up perfectly.
In simple words: For one set to be a subset of another, every single number in the first set must also be inside the second set. Since this is not the case here, both statements are false.

🎯 Exam Tip: To prove a subset statement is false, you only need to find a single element in the first set that is not present in the second set.

Question 3. Suppose \( U = \{1, 3, 9, 11, 13, 18, 19\} \), and \( B = \{3, 9, 11, 13\} \). Find \( (B')' \) and draw the inference. (Textbook pg. no. 10)
Answer:
Given:
\( U = \{1, 3, 9, 11, 13, 18, 19\} \)
\( B = \{3, 9, 11, 13\} \) ....(i)

The complement of set \( B \) (denoted as \( B' \)) contains elements of \( U \) that are not in \( B \):
\( \therefore B' = \{1, 18, 19\} \)

Now, the complement of \( B' \) (denoted as \( (B')' \)) contains elements of \( U \) that are not in \( B' \):
\( (B')' = \{3, 9, 11, 13\} \) ....(ii)

Comparing (i) and (ii):
\( \therefore (B')' = B \)

\( \therefore \) Complement of a complement is the given set itself. This fundamental property shows that double negation restores the original set.

Visual representation of the sets from the diagram:

• Universal Set \( U \) contains: 1, 3, 9, 11, 13, 18, 19
• Set \( B \) (inside the circle) contains: 3, 9, 11, 13
• Set \( B' \) (outside the circle \( B \)) contains: 1, 18, 19

In simple words: The complement of a set means everything outside of it. If you take the complement of that complement, you end up right back inside the original set, just like how a double negative makes a positive.

🎯 Exam Tip: Clearly label your equations as (i) and (ii) when proving set identities to make your logical steps easy for the examiner to follow.