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

Question 1. Choose the correct alternative answer for each of the following questions.
i. If \( M = \{1, 3, 5\} \), \( N = \{2, 4, 6\} \), then \( M \cap N = ? \)
(a) \( \{1, 2, 3, 4, 5, 6\} \)
(b) \( \{1, 3, 5\} \)
(c) \( \phi \)
(d) \( \{2, 4, 6\} \)
Answer: (c) \( \phi \)
In simple words: The intersection symbol \( \cap \) means we need to find the numbers that are in both set M and set N. Since set M has only odd numbers and set N has only even numbers, they have no numbers in common, so the answer is the empty set, represented by \( \phi \).

🎯 Exam Tip: When looking for the intersection (\( \cap \)) of two disjoint sets (sets with nothing in common), the result is always the empty set \( \phi \).

Question ii. P = \( \{x \mid x \text{ is an odd natural number}, 1 < x \le 5\} \). How to write this set in roster form?
(a) \( \{1, 3, 5\} \)
(b) \( \{1, 2, 3, 4, 5\} \)
(c) \( \{1, 3\} \)
(d) \( \{3, 5\} \)
Answer: (d) \( \{3, 5\} \)
In simple words: We need to find odd numbers that are strictly greater than 1 and less than or equal to 5. These numbers are 3 and 5, so the set is written as {3, 5}.

🎯 Exam Tip: Pay close attention to the inequality signs; "1 < x" means 1 is excluded, while "x ≤ 5" means 5 is included.

Question iii. P = \( \{1, 2, \dots, 10\} \). What type of set P is?
(a) Null set
(b) Infinite set
(c) Finite set
(d) None of the options
Answer: (c) Finite set
In simple words: Since we can count the exact number of elements in this set (there are exactly 10 elements), it is a finite set.

🎯 Exam Tip: If a set has a limited number of elements that you can count to the end, it is always a finite set.

Question iv. M ∪ N = \( \{1, 2, 3, 4, 5, 6\} \) and M = \( \{1, 2, 4\} \), then which of the following represent set N?
(a) \( \{1, 2, 3\} \)
(b) \( \{3, 4, 5, 6\} \)
(c) \( \{2, 5, 6\} \)
(d) \( \{4, 5, 6\} \)
Answer: (b) \( \{3, 4, 5, 6\} \)
In simple words: The union of M and N contains all elements from both sets. Since M has {1, 2, 4}, set N must contain the remaining elements {3, 5, 6} and can also contain some elements of M like 4 to form the union.

🎯 Exam Tip: Check each option by taking its union with M to see which one results exactly in the given union set.

Question v. If P ⊆ M, then which of the following set represent P ∩ (P ∪ M)?
(a) P
(b) M
(c) P ∪ M
(d) P’ ∩ M
Answer: (a) P
In simple words: Since P is a subset of M, their union is just M. Then, the intersection of P with M is simply P itself.

🎯 Exam Tip: Use Venn diagrams to quickly visualize subset relationships and operations to avoid algebraic mistakes.

Question vi. Which of the following sets are empty sets?
(a) Set of intersecting points of parallel lines.
(b) Set of even prime numbers.
(c) Month of an english calendar having less than 30 days.
Answer: (a) Set of intersecting points of parallel lines.
In simple words: Parallel lines run side-by-side and never cross each other, so they have zero points of intersection, making this an empty set.

🎯 Exam Tip: Remember that an empty set contains absolutely no elements. Always check if there is even a single element (like 2 for even primes) before deciding.

Question 2. Find the correct option for the given question.
(i) Which of the following collections is a set ?
(a) Colors of the rainbow
(b) Tall trees in the school campus.
(c) Rich people in the village
(d) Easy examples in the book

(ii) Which of the following set represent \( \text{N} \cap \text{W} \)?
(a) {1, 2, 3,....}
(b) {0, 1, 2, 3,....}
(c) {0}
(d) { }

(iii) P = {x | x is an odd natural number, \( 1 < x < 5 \)}. How to write this set in roster form?
(a) {1, 3, 5}
(b) {1, 2, 3, 4, 5}
(c) {1, 3}
(d) {3, 5}
Answer:
(i) (a) Colors of the rainbow
(ii) (a) {1, 2, 3,....}
(iii) (b) {1, 2, 3, 4, 5}
In simple words: A set is a collection of clearly defined objects, like rainbow colors. The intersection of natural and whole numbers gives all natural numbers, and roster form means listing the elements of the set.

🎯 Exam Tip: Always remember that a set must be well-defined, meaning its elements do not change from person to person. Double-check the definitions of natural numbers \( (\text{N}) \) and whole numbers \( (\text{W}) \) to avoid simple mistakes in set operations.

Question iv. If \( T = \{1, 2, 3, 4, 5\} \) and \( M = \{3, 4, 7, 8\} \), then \( T \cup M = ? \)
(a) \( \{1, 2, 3, 4, 5, 7\} \)
(b) \( \{1, 2, 3, 7, 8\} \)
(c) \( \{1, 2, 3, 4, 5, 7, 8\} \)
(d) \( \{3, 4\} \)
Answer: (c) \( \{1, 2, 3, 4, 5, 7, 8\} \)
In simple words: The union of two sets combines all the unique elements from both sets together without repeating any numbers.

🎯 Exam Tip: When finding the union, write down all elements of the first set, then add the elements of the second set that are not already listed.

Hints

• The elements of options B, C and D cannot be definitely and clearly decided.
• The common elements of \( N \) and \( W \) are 1, 2, 3, ...

Question 3. Out of 100 persons in a group, 72 persons speak English and 43 persons speak French. Each one out of 100 persons speak at least one language. Then how many speak only English? How many speak only French? How many of them speak English and French both?
Answer:
i. Let \( U \) be the set of all the persons,
\( E \) be the set of persons who speak English and
\( F \) be the set of persons who speak French.
\( \therefore n(E) = 72, n(F) = 43 \)
Since, each one out of 100 persons speak at least one language
\( \dots n(U) = n(E \cup F) = 100 \)

ii. \( n(E \cup F) = n(E) + n(F) - n(E \cap F) \)
\( 100 = 72 + 43 - n(E \cap F) \)
\( n(E \cap F) = 72 + 43 - 100 \)
\( \therefore n(E \cap F) = 15 \)
Number of people who speak English and French = 15

iii. Number of people who speak only English = \( n(E) - n(E \cap F) \)
\( = 72 - 15 = 57 \)

iv. Number of people who speak only French = \( n(F) - n(E \cap F) \)
\( = 43 - 15 = 28 \)
This step-by-step calculation clearly shows how the total group is divided among the language speakers.
In simple words: Out of 100 people, 15 speak both languages. By subtracting these bilingual people from the total English and French speakers, we find that 57 speak only English and 28 speak only French.

🎯 Exam Tip: Always start by writing down the given values in set notation, and remember the core formula \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \) to solve intersection problems easily.

Alternate Method:
Let U be the set of all the persons,
E be the set of persons who speak English,
F be the set of persons who speak French and x people speak both the languages.
Venn Diagram representation:

• Set E (English): 72 - x
• Intersection (E ∩ F): x
• Set F (French): 43 - x

Since, each one out of 100 persons speak at least one language.
\( \therefore n(U) = n(E \cup F) = 100 \)
\( \therefore 72 - x + x + 43 - x = 100 \)
\( \therefore 115 - x = 100 \)
\( \therefore x = 115 - 100 = 15 \)
Number of people who speak English and French = 15
Number of people who speak only English = \( 72 - x = 72 - 15 = 57 \)
Number of people who speak only French = \( 43 - x = 43 - 15 = 28 \). This helps us clearly distinguish between monolingual and bilingual speakers.
In simple words: By using a Venn diagram, we can easily find the number of people who speak both languages by subtracting the individual groups from the total.

🎯 Exam Tip: Venn diagrams are a great way to double-check your algebraic calculations in set theory problems.

Question 4. 70 trees were planted by Parth and 90 trees were planted by Pradnya on the occasion of Tree Plantation Week. Out of these 25 trees were planted by both of them together. How many trees were planted by Parth or Pradnya?
Answer:
i. Let P be the trees planted by Parth and Q be the trees planted by Pradnya.
\( \therefore n(P) = 70 \) and \( n(Q) = 90 \)
Total number of trees planted by Parth and Pradnya = \( n(P \cap Q) = 25 \)

ii. Number of trees planted by Parth or Pradnya = \( n(P \cup Q) \)
\( = n(P) + n(Q) - n(P \cap Q) \)
\( = 70 + 90 - 25 = 135 \)
\( \therefore \) A total of 135 trees were planted by Parth or Pradnya. This shows the collective effort of both individuals in greening their environment.

Alternate Method:
Let P be the trees planted by Parth and Q be the trees planted by Pradnya.
Venn Diagram representation:

• Set P (Trees planted by Parth only): 45 (i.e., 70 - 25)
• Intersection (P ∩ Q - Trees planted by both): 25
• Set Q (Trees planted by Pradnya only): 65 (i.e., 90 - 25)

Total trees planted = \( 45 + 25 + 65 = 135 \).
In simple words: To find the total number of unique trees planted, we add the trees planted by Parth and Pradnya, then subtract the trees they planted together so we do not count them twice.

🎯 Exam Tip: Remember that the word 'or' in set theory indicates the union (\( \cup \)) of sets, while 'and' indicates the intersection (\( \cap \)).

Question 5. If \( n(A) = 20 \), \( n(B) = 28 \) and \( n(A \cup B) = 36 \), then \( n(A \cap B) = ? \)
Answer:
Formula: \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
\( \therefore 36 = 20 + 28 - n(A \cap B) \)
\( \therefore n(A \cap B) = 20 + 28 - 36 \)
\( \therefore n(A \cap B) = 48 - 36 \)
\( \therefore n(A \cap B) = 12 \)
In simple words: We use a standard formula that connects the union and intersection of two sets. By putting the given numbers into this formula, we find that there are 12 elements shared between both sets.

🎯 Exam Tip: Always write down the standard formula \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \) clearly before substituting the values to secure step-wise marks.

Question 6. In a class, 8 students out of 28 have a dog as their pet animal at home, 6 students have a cat as their pet animal, 10 students have dog and cat both, then how many students do not have dog or cat as their pet animal at home?
Answer:
(i) Let \( U \) be the set of all the students, then \( n(U) = 28 \).
Let \( D \) be the set of students who have a dog as a pet and \( C \) be the set of students who have a cat as a pet.
Since 10 students have both a dog and a cat as their pet animal, \( n(D \cap C) = 10 \).

Venn Diagram Representation:

• Universal Set \( U \) = 28
• Students with only Dog \( D \) = 8
• Students with both Dog and Cat \( D \cap C \) = 10
• Students with only Cat \( C \) = 6

(ii) Number of students who have a cat or a dog as a pet:
\( n(D \cup C) = 8 + 10 + 6 = 24 \)

(iii) Number of students who do not have a dog or a cat as a pet:
\( n(U) - n(D \cup C) = 28 - 24 = 4 \)
Therefore, 4 students do not have a dog or a cat as their pet animal at home.
In simple words: Out of 28 total students, 24 students have at least one pet (either a dog, a cat, or both). Subtracting these 24 pet owners from the total 28 students leaves us with 4 students who do not have any dog or cat.

🎯 Exam Tip: When solving set word problems, always identify the intersection value first and place it in the overlapping region of the Venn diagram to avoid double-counting.

Question 7. Represent the union of two sets by Venn diagram for each of the following.
(i) \( A = \{3, 4, 5, 7\} \), \( B = \{1, 4, 8\} \)
(ii) \( P = \{a, b, c, e, f\} \), \( Q = \{l, m, n, e, b\} \)
(iii) \( X = \{x \mid x \text{ is a prime number between 80 and 100}\} \), \( Y = \{y \mid y \text{ is an odd number between 90 and 100}\} \)
Answer:
(i) \( A = \{3, 4, 5, 7\} \), \( B = \{1, 4, 8\} \)
The union of sets A and B is \( A \cup B = \{1, 3, 4, 5, 7, 8\} \).
Venn Diagram representation:

• Set A only: 3, 5, 7
• Intersection (both A and B): 4
• Set B only: 1, 8

(ii) \( P = \{a, b, c, e, f\} \), \( Q = \{l, m, n, e, b\} \)
The union of sets P and Q is \( P \cup Q = \{a, b, c, e, f, l, m, n\} \).
Venn Diagram representation:

• Set P only: a, c, f
• Intersection (both P and Q): b, e
• Set Q only: l, m, n

(iii) \( X = \{x \mid x \text{ is a prime number between 80 and 100}\} \)
\( \implies X = \{83, 89, 97\} \)
\( Y = \{y \mid y \text{ is an odd number between 90 and 100}\} \)
\( \implies Y = \{91, 93, 95, 97, 99\} \)
The union of sets X and Y is \( X \cup Y = \{83, 89, 91, 93, 95, 97, 99\} \).
Venn Diagram representation:

• Set X only: 83, 89
• Intersection (both X and Y): 97
• Set Y only: 91, 93, 95, 99

In simple words: To show the union of two sets in a Venn diagram, we list the common elements in the overlapping middle section and the unique elements in their respective circles. Combining all these elements gives us the union of the two sets.

🎯 Exam Tip: Always identify the common elements first and place them in the intersection area of the Venn diagram to avoid repeating elements in your sets.

Question 8. Write the subset relations between the following sets.
X = set of all quadrilaterals.
Y = set of all rhombuses.
S = set of all squares.
T = set of all parallelograms.
V = set of all rectangles.
Answer:
Based on the definitions of geometric shapes, we can establish the following subset relations:
1. Every square is a rhombus, a rectangle, a parallelogram, and a quadrilateral.
\( \implies S \subseteq Y \), \( S \subseteq V \), \( S \subseteq T \), \( S \subseteq X \)
2. Every rhombus is a parallelogram and a quadrilateral.
\( \implies Y \subseteq T \), \( Y \subseteq X \)
3. Every rectangle is a parallelogram and a quadrilateral.
\( \implies V \subseteq T \), \( V \subseteq X \)
4. Every parallelogram is a quadrilateral.
\( \implies T \subseteq X \)
Therefore, the subset relations are:
\( S \subseteq Y \subseteq T \subseteq X \)
\( S \subseteq V \subseteq T \subseteq X \)
In simple words: A subset means one group fits entirely inside another. For example, since every square is also a type of rectangle, the set of squares is a subset of the set of rectangles.

🎯 Exam Tip: Remember that a square is the most specific shape here, so it will be a subset of almost all other sets listed. Write down each relation clearly using the subset symbol \( \subseteq \).

Question 9. If M is any set, then write \( M \cup \Phi \) and \( M \cap \Phi \).
Answer: For any set \( M \), its union with the empty set \( \Phi \) is the set itself, while its intersection with the empty set is the empty set. Let \( M = \{2, 3, 4, 8\} \) and \( \Phi = \{ \} \)
\( \implies M \cup \Phi = \{2, 3, 4, 8\} = M \)
\( \implies M \cap \Phi = \{ \} = \Phi \)
In simple words: Combining any set with an empty set gives you the original set, but finding what they have in common with an empty set leaves you with nothing.

🎯 Exam Tip: Remember that the empty set \( \Phi \) contains no elements, so unioning with it changes nothing, and intersecting with it leaves nothing.

Question 10. Observe the Venn diagram and write the given sets U, A, B, A ∪ B and A ∩ B.
Answer: By observing the elements distributed within and outside the circles in the Venn diagram, we get:
\( U = \{1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13\} \)
\( A = \{1, 2, 3, 5, 7\} \)
\( B = \{1, 5, 8, 9, 10\} \)
\( A \cup B = \{1, 2, 3, 5, 7, 8, 9, 10\} \)
\( A \cap B = \{1, 5\} \)
In simple words: The universal set \( U \) lists every single number shown in the diagram, while \( A \) and \( B \) list numbers inside their respective circles, \( A \cup B \) combines them, and \( A \cap B \) shows only the overlapping numbers.

🎯 Exam Tip: Do not forget to include the numbers outside the circles (like 4, 11, and 13) when listing the elements of the universal set \( U \).

Question 11. If n(A) = 7, n(B) = 13, n(A ∩ B) = 4, then n(A ∪ B) = ?
Answer: We use the standard formula relating the cardinalities of sets:
\( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
\( \implies n(A \cup B) = 7 + 13 - 4 \)
\( \implies n(A \cup B) = 16 \) Therefore, the number of elements in the union of sets A and B is 16.
In simple words: To find the total number of unique items in both groups, add the counts of both groups together and subtract the items that were counted twice because they belong to both.

🎯 Exam Tip: Always write down the formula \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \) clearly before substituting the values to ensure you get full step-wise marks.

Question 1. Set of students in a class and set of students in the same class who can swim, are shown by the Venn diagram:

• Outer Set: Students in a class
• Inner Subset: Students who can swim

Observe the diagram and draw Venn diagrams for the following subsets.
i. a. Set of students in a class
b. Set of students who can ride bicycles in the same class
ii. A set of fruits is given as follows.
U = {guava, orange, mango, jackfruit, chickoo, jamun, custard apple, papaya, plum}
Show these subsets.
A = fruit with one seed
B = fruit with more than one seed. (Textbook pg. no. 8)
Answer:
i. The Venn diagram representing the subsets is as follows:

• Outer Set: Students in a class
• Inner Subset: Students who can ride bicycle

ii. The subsets based on the number of seeds are:
A = {mango, jamun, plum}
B = {guava, orange, jackfruit, chickoo, custard apple, papaya}
These can be represented as two separate sets within the universal set of fruits.
In simple words: A Venn diagram uses shapes like circles to show how different groups of things are related. Here, we grouped students who can ride bicycles inside the larger group of all students, and sorted fruits into two separate groups based on whether they have one seed or many seeds.

🎯 Exam Tip: When drawing Venn diagrams, always ensure the subset is fully enclosed within the larger set to represent proper containment. Clearly label each set with capital letters.

Question 2. Every student should take 9 triangular sheets of paper and one plate. Numbers from 1 to 9 should, be written on each triangle. Everyone should keep some numbered triangles in the plate. Now the triangles in each plate form a subset of the set of numbers from 1 to 9.
Answer: This is a hands-on activity to understand subsets. The universal set is \( U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \). Any collection of numbered triangles placed in a plate represents a subset of this universal set. For example, if a student puts triangles with numbers 2, 4, 6, and 8 in the plate, then the set \( A = \{2, 4, 6, 8\} \) is a subset of \( U \).
In simple words: This activity shows that any smaller group of numbers you choose from 1 to 9 is called a subset of the main group.

🎯 Exam Tip: Remember that an empty set (no triangles in the plate) and the set itself (all 9 triangles in the plate) are always subsets of the main set.

Question. Look at the plates of Sujata, Hameed, Mukta, Nandini, Joseph with the numbered triangles. Guess the thinking behind selecting these numbers. Hence write the subsets in set builder form.
Answer: By observing the numbers on each plate, we can identify the specific mathematical property that defines each set. The subsets written in set builder form are as follows:
Sujata: \( S = \{x \mid x = 2n - 1, n \in N, x < 9\} \)
Hameed: \( H = \{x \mid x = 2n, n \in N, x < 9\} \)
Mukta: \( M = \{x \mid x = n^2, n \in N, x \le 9\} \)
Nandini: \( N = \{x \mid x \in N, x \le 9\} \)
Joseph: \( J = \{x \mid x \text{ is a prime number between 1 and 9}\} \)
In simple words: Each person chose a different rule to pick their numbers: Sujata chose odd numbers, Hameed chose even numbers, Mukta chose perfect squares, Nandini chose all counting numbers up to 9, and Joseph chose prime numbers.

🎯 Exam Tip: When writing sets in set-builder form, always clearly define the variable, the rule it follows, and the set of numbers (like Natural numbers \( N \)) it belongs to.

Question 3. Collect the following information from 20 families nearby your house.
(i) Number of families subscribing for Marathi Newspaper.
(ii) Number of families subscribing for English Newspaper.
(iii) Number of families subscribing for both English as well as Marathi Newspaper.
Show the collected information using Venn diagram.
Answer: This is a practical survey-based activity where students need to collect real-world data from 20 neighboring households. Once the data is gathered, represent the counts using two overlapping circles within a universal set rectangle to show the intersection of families reading both newspapers.
In simple words: Go to 20 houses nearby and ask them which newspapers they read, then draw two overlapping circles to show how many read Marathi, how many read English, and how many read both.

🎯 Exam Tip: Ensure that the sum of all regions in your Venn diagram (only Marathi + only English + both + neither) equals the total number of families surveyed, which is 20.