Samacheer Kalvi Class 9 Maths Solutions Chapter 1 Set Language More Ques

Choose the Correct Answer

Question 1. If \( A = \{\text{set of odd natural numbers}\} \) and \( B = \{\text{set of even natural numbers}\} \), then \( A \) and \( B \) are..........
(a) equal set
(b) equivalent sets
(c) overlapping sets
(d) disjoint sets
Answer: (d) disjoint sets
In simple words: Odd numbers and even numbers have nothing in common. Sets that share no common elements are called disjoint sets.

🎯 Exam Tip: Remember that natural numbers start from 1. Disjoint sets are crucial in set theory for understanding relationships between collections.

Question 2. The number of subsets in set \( A = \{1, 2, 3\} \) is
(a) 3
(b) 6
(c) 8
(d) 9
Answer: (c) 8
In simple words: To find how many subsets a set has, you use the formula \( 2^n \), where \( n \) is the number of elements in the set. Here, there are 3 elements, so \( 2^3 = 8 \).

🎯 Exam Tip: Always remember that the empty set and the set itself are also considered subsets. The power set contains all possible subsets.

Question 3. The set that does not have a proper subset is
(a) Finite set
(b) Infinite set
(c) Null set
(d) Singleton set
Answer: (c) Null set
In simple words: A proper subset means a subset that is not the set itself. The null set, which is an empty set, has only one subset (itself), so it cannot have a proper subset.

🎯 Exam Tip: The null set \( \{\} \) or \( \emptyset \) is a subset of every set, and the only subset of itself. A proper subset must be smaller than the original set.

Question 4. Sets having the same number of elements are called
(a) overlapping sets
(b) disjoint sets
(c) equivalent sets
(d) equal sets
Answer: (c) equivalent sets
In simple words: When two sets have the exact same count of items, even if the items themselves are different, they are called equivalent sets. Equal sets mean they have the same items.

🎯 Exam Tip: Distinguish carefully between 'equivalent sets' (same cardinality or number of elements) and 'equal sets' (same elements). This is a common point of confusion.

Question 5. The set \( (A – B) \cup (B – A) \) is
(a) \( A\Delta B \)
(b) \( A \cup B \)
(c) \( A \cap B \)
(d) \( A' \cup B' \)
Answer: (a) \( A\Delta B \)
In simple words: The union of \( A \) minus \( B \) and \( B \) minus \( A \) represents elements that are in \( A \) or in \( B \), but not in both. This is precisely the definition of the symmetric difference of two sets.

🎯 Exam Tip: The expression \( (A – B) \cup (B – A) \) is the standard definition for the symmetric difference, often denoted as \( A\Delta B \). It can also be written as \( (A \cup B) – (A \cap B) \).

Question 6. The set \( (A \cup B) – (A \cap B) \) is
(a) \( (A \cup B)' \)
(b) \( A\Delta B \)
(c) \( (A \cap B)' \)
(d) \( A' \cup B' \)
Answer: (b) \( A\Delta B \)
In simple words: This expression means "all elements in A or B, but not in both A and B". This is another way to define the symmetric difference between two sets.

🎯 Exam Tip: Recognize this as an alternative, equivalent definition for the symmetric difference \( A\Delta B \). Being familiar with different forms helps solve problems faster.

Question 7. The set \( \{x: x \in A, x \in B, x \notin A \cap B\} \) is
(a) \( A \cap B \)
(b) \( A \cup B \)
(c) \( A – B \)
(d) \( A\Delta B \)
Answer: (d) \( A\Delta B \)
In simple words: This describes elements that are in set \( A \) or set \( B \), but not in both at the same time. This means the elements belong to the symmetric difference of \( A \) and \( B \).

🎯 Exam Tip: The notation \( x \in A, x \in B \) often implies elements shared by \( A \) and \( B \). However, when combined with \( x \notin A \cap B \), it indicates elements that are in either set but not in their common part, which defines the symmetric difference.

Question 8. The number of elements of the set \( \{x : x \in Z, x^2 = 1\} \) is
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (c) 2
In simple words: The equation \( x^2 = 1 \) has two solutions for integers: \( x = 1 \) and \( x = -1 \). Both are integers, so the set contains two elements.

🎯 Exam Tip: Remember that \( Z \) represents the set of all integers, which includes positive, negative, and zero. For \( x^2 = 1 \), both \( 1 \) and \( -1 \) are valid integer solutions.

Question 9. If \( A \) is a proper subset of \( B \), then \( A \cap B = \)
(a) \( A \)
(b) \( B \)
(c) \( \emptyset \)
(d) \( A \cup B \)
Answer: (a) \( A \)
In simple words: If set \( A \) is completely inside set \( B \) (a proper subset), then anything common to both \( A \) and \( B \) must just be all of \( A \).

🎯 Exam Tip: A proper subset means all elements of \( A \) are in \( B \), and \( B \) has at least one element not in \( A \). The intersection \( A \cap B \) will therefore always be \( A \).

Question 10. The shaded region in the adjoining diagram represents ..........

(a) \( A – B \)
(b) \( B – A \)
(c) \( A' \)
(d) \( B' \)
Answer: (a) \( A – B \)
In simple words: The diagram shows only the part of set \( A \) that does not overlap with set \( B \). This is exactly what \( A \) minus \( B \) means.

🎯 Exam Tip: Understanding how to visually represent set operations like \( A – B \), \( B – A \), \( A \cap B \), and \( A \cup B \) is key. \( A – B \) means "only A," excluding the shared portion.

Question 11. From the given Venn diagram \( (A \cup B)' \) is ...........

(a) \( \{5, 6\} \)
(b) \( \{1, 2, 3, 4, 7\} \)
(c) \( \{1, 2, 3, 4, 5, 6, 7\} \)
(d) \( \{8, 9\} \)
Answer: (d) \( \{8, 9\} \)
In simple words: \( (A \cup B)' \) means all the elements that are not in set \( A \) and not in set \( B \). Looking at the diagram, numbers 8 and 9 are outside both circles, so they are the complement of the union.

🎯 Exam Tip: The complement of a union \( (A \cup B)' \) includes all elements in the universal set \( U \) that are not part of either \( A \) or \( B \). Visually, this is the area outside both circles.

Question 12. If \( n(A \cup B \cup C) = 73 \), \( n(A) = 2x \), \( n(B) = 3x \), \( n(C) = 5x \), \( n(A \cap B) = 10 \), \( n(B \cap C) = 15 \), \( n(A \cap C) = 5 \) and \( n(A \cap B \cap C) = 3 \), then the value of \( x \) is .........
(a) 9
(b) 10
(c) 5
(d) 18
Answer: (b) 10
In simple words: We use a special formula for three sets to find the number of elements in their union. By plugging in all the given values into this formula, we can solve for \( x \).

🎯 Exam Tip: Remember the Principle of Inclusion-Exclusion for three sets: \( n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C) \). Substitute values and solve carefully for \( x \).

Question 13. For any three sets, \( n(A \cup B \cup C) = 60 \), \( n(A) = 25 \), \( n(B) = 20 \), \( n(C) = 15 \), \( n(A \cap B) = 10 \), \( n(B \cap C) = 7 \), \( n(A \cap C) = 3 \), then \( n(A \cap B \cap C) \) is..........
(a) 10
(b) 15
(c) 20
(d) 25
Answer: (c) 20
In simple words: Using the inclusion-exclusion principle formula for three sets, you can find the number of elements common to all three sets by rearranging the formula and plugging in the given values.

🎯 Exam Tip: This question tests your ability to apply the Principle of Inclusion-Exclusion for three sets. Be careful with the signs when rearranging the formula to solve for \( n(A \cap B \cap C) \).

Question 14. If \( n(U) = 70 \), \( n(A) = 25 \), \( n(B) = 30 \), \( n(A \cap B) = 5 \), then \( n(A \cup B)' \) is..........
(a) 5
(b) 10
(c) 15
(d) 20
Answer: (d) 20
In simple words: First, find the number of elements in \( A \cup B \) using its formula. Then, subtract this number from the total elements in the universal set \( U \) to find the complement.

🎯 Exam Tip: Remember the key formulas: \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \) and \( n(A \cup B)' = n(U) - n(A \cup B) \). Always calculate \( n(A \cup B) \) first.

Question 15. Which of the following is not correct?
(a) \( A – (B \cup C) = (A – B) \cap (A – C) \)
(b) \( A – (B \cap C) = (A – B) \cup (A – C) \)
(c) \( (A \cup B)' = A' \cap B' \)
(d) \( A' \cup B' = (A – B)' \)
Answer: (d) \( A' \cup B' = (A – B)' \)
In simple words: We are looking for the statement that is false. Statements (a), (b), and (c) are all true identities in set theory. Statement (d) is not a true identity. For example, \( A' \cup B' \) is the complement of \( A \cap B \), while \( (A – B)' \) is \( A' \cup B \).

🎯 Exam Tip: This question tests your knowledge of set identities, including De Morgan's Laws and properties of set difference. Thoroughly knowing these identities is essential for such problems.

Answer the Following Questions

Question 1. Write the following in “Roster" form?
(a) \( A = \text{set of the months having 31 days.} \)
(b) \( B = \{x : x \text{ is a natural number of 2 digits divisible by 13}\} \)
(c) \( C = \{\text{set of vowels in the word "father"}\} \)
(d) \( D = \{x : 5 < x \leq 10; x \in N\} \)
(e) \( E = \{x: x \text{ is a square natural number less than 16}\} \)
Answer:
(a) \( A = \{\text{Jan, March, May, July, Aug, Oct, Dec}\} \)
(b) \( B = \{13, 26, 39, 52, 65, 78, 91\} \)
(c) \( C = \{a, e\} \)
(d) \( D = \{6, 7, 8, 9, 10\} \)
(e) \( E = \{1, 4, 9\} \)
In simple words: Roster form lists all the elements of a set inside curly braces. For each part, we identify the elements that fit the description and write them down. For example, part (d) includes natural numbers greater than 5 but less than or equal to 10.

🎯 Exam Tip: When converting to roster form, ensure all elements are listed correctly and only once. Pay attention to inclusive and exclusive bounds in inequalities (like \( < \) vs \( \leq \)).

Question 2. Given that \( A = \{1, 3, 5, 7\} \) and \( B = \{1, 2, 4, 6, 8\} \). Find:
(i) \( A\Delta B \)
(ii) \( B\Delta A \)
Answer:
(i) To find \( A\Delta B \), we first calculate \( A – B \) and \( B – A \).
\( A – B = \{1, 3, 5, 7\} – \{1, 2, 4, 6, 8\} = \{3, 5, 7\} \)
\( B – A = \{1, 2, 4, 6, 8\} – \{1, 3, 5, 7\} = \{2, 4, 6, 8\} \)
Now, \( A\Delta B = (A – B) \cup (B – A) \)
\( A\Delta B = \{3, 5, 7\} \cup \{2, 4, 6, 8\} \)
\( A\Delta B = \{2, 3, 4, 5, 6, 7, 8\} \)
(ii) To find \( B\Delta A \), we use the definition \( B\Delta A = (B – A) \cup (A – B) \).
\( B\Delta A = \{2, 4, 6, 8\} \cup \{3, 5, 7\} \)
\( B\Delta A = \{2, 3, 4, 5, 6, 7, 8\} \)
The symmetric difference is the set of elements unique to each set. Observe that \( A\Delta B \) and \( B\Delta A \) yield the same result, showing symmetric difference is commutative.
In simple words: For \( A\Delta B \), we list items only in A, then items only in B, and combine those lists. For \( B\Delta A \), we do the same thing, starting with items only in B. Both will give the same answer because the order doesn't matter for symmetric difference.

🎯 Exam Tip: Remember that symmetric difference \( A\Delta B \) can be defined as \( (A – B) \cup (B – A) \) or \( (A \cup B) – (A \cap B) \). It's important to know that \( A\Delta B = B\Delta A \).

Question 3. From the Venn diagram, list the following:
(i) \( A \)
(ii) \( B \)
(iii) \( A \cap B \)
(iv) \( A \cup B \)
(v) \( A – B \)
(vi) \( B – A \)
(vii) \( (A – B) \cap (B – A) \)
Answer:
(i) \( A = \{1, 2, 5, 6, 7\} \)
(ii) \( B = \{3, 4, 5, 6\} \)
(iii) \( A \cap B = \{5, 6\} \)
(iv) \( A \cup B = \{1, 2, 3, 4, 5, 6, 7\} \)
(v) \( A – B = \{1, 2, 7\} \)
(vi) \( B – A = \{3, 4\} \)
(vii) \( (A – B) \cap (B – A) = \{\} \)
In simple words: We read the numbers from each section of the Venn diagram to form the sets. For example, set \( A \) includes all numbers within its circle. The intersection of \( (A-B) \) and \( (B-A) \) will always be an empty set because these two sets have no common elements.

🎯 Exam Tip: Carefully identify the elements belonging to each region of the Venn diagram. The intersection of disjoint sets (like \( A-B \) and \( B-A \)) is always the empty set \( \{\} \).

Question 4. In a class there are 40 students. 26 have opted for Mathematics and 24 have opted for Science. How many students have opted for Mathematics and Science?
Answer:
Let \( M \) be the set of students opting for Mathematics and \( S \) be the set of students opting for Science.
Given: \( n(M \cup S) = 40 \), \( n(M) = 26 \), \( n(S) = 24 \).
Using the formula for the union of two sets:
\( n(M \cup S) = n(M) + n(S) - n(M \cap S) \)
\( 40 = 26 + 24 - n(M \cap S) \)
\( 40 = 50 - n(M \cap S) \)
\( n(M \cap S) = 50 - 40 \)
\( n(M \cap S) = 10 \)
Thus, 10 students have opted for both Mathematics and Science.

Another Method (using Venn diagram approach):
Let \( x \) be the number of students who opted for both Mathematics and Science, i.e., \( n(M \cap S) = x \).
Students who opted only for Mathematics \( = n(M) - x = 26 - x \)
Students who opted only for Science \( = n(S) - x = 24 - x \)
The total number of students is the sum of those only in Math, only in Science, and in both:
\( n(M \cup S) = (26 - x) + x + (24 - x) \)
\( 40 = 26 + 24 - x \)
\( 40 = 50 - x \)
\( x = 50 - 40 \)
\( x = 10 \)
So, 10 students opted for both subjects. This method helps visualize the groups of students clearly.
In simple words: You can find the number of students who chose both subjects by either using the set formula (total students = Math only + Science only + both - intersection) or by drawing a Venn diagram and solving for the overlapping part. Both methods give the same answer.

🎯 Exam Tip: When solving problems involving groups, either apply the Principle of Inclusion-Exclusion formula directly or draw a Venn diagram to visualize the different regions and set up equations. Both methods are valid and can help check your answer.

Question 5. Given \( U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \), \( A = \{4, 5, 7, 9\} \), \( B = \{1, 3, 5, 7, 8\} \), verify De Morgan's Laws for complementation.
Answer:
De Morgan's Laws state:
(i) \( (A \cup B)' = A' \cap B' \)
(ii) \( (A \cap B)' = A' \cup B' \)

Let's verify the first law: \( (A \cup B)' = A' \cap B' \)
First, find \( A \cup B \):
\( A \cup B = \{4, 5, 7, 9\} \cup \{1, 3, 5, 7, 8\} \)
\( A \cup B = \{1, 3, 4, 5, 7, 8, 9\} \)
Next, find \( (A \cup B)' \):
\( (A \cup B)' = U - (A \cup B) \)
\( (A \cup B)' = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{1, 3, 4, 5, 7, 8, 9\} \)
\( (A \cup B)' = \{2, 6\} \quad \text{.........(1)} \)

Now, find \( A' \):
\( A' = U - A \)
\( A' = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{4, 5, 7, 9\} \)
\( A' = \{1, 2, 3, 6, 8\} \)
Next, find \( B' \):
\( B' = U - B \)
\( B' = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{1, 3, 5, 7, 8\} \)
\( B' = \{2, 4, 6, 9\} \)
Finally, find \( A' \cap B' \):
\( A' \cap B' = \{1, 2, 3, 6, 8\} \cap \{2, 4, 6, 9\} \)
\( A' \cap B' = \{2, 6\} \quad \text{.........(2)} \)
From (1) and (2), we can see that \( (A \cup B)' = A' \cap B' \). The first law is verified.

Now, let's verify the second law: \( (A \cap B)' = A' \cup B' \)
First, find \( A \cap B \):
\( A \cap B = \{4, 5, 7, 9\} \cap \{1, 3, 5, 7, 8\} \)
\( A \cap B = \{5, 7\} \)
Next, find \( (A \cap B)' \):
\( (A \cap B)' = U - (A \cap B) \)
\( (A \cap B)' = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{5, 7\} \)
\( (A \cap B)' = \{1, 2, 3, 4, 6, 8, 9\} \quad \text{.........(3)} \)

Using \( A' = \{1, 2, 3, 6, 8\} \) and \( B' = \{2, 4, 6, 9\} \) from the previous calculation, find \( A' \cup B' \):
\( A' \cup B' = \{1, 2, 3, 6, 8\} \cup \{2, 4, 6, 9\} \)
\( A' \cup B' = \{1, 2, 3, 4, 6, 8, 9\} \quad \text{.........(4)} \)
From (3) and (4), we can see that \( (A \cap B)' = A' \cup B' \). The second law is also verified.
De Morgan's Laws are fundamental in set theory and logic, providing a way to relate union and intersection with complements.
In simple words: We check if De Morgan's two rules are true using the given sets. For each rule, we calculate the left side and the right side separately. If both sides give the same final set, then the rule is proven true for these sets. We found that both rules hold true.

🎯 Exam Tip: When verifying De Morgan's Laws, calculate each part (union/intersection, then complement, and individual complements, then union/intersection) step-by-step. Clearly label your results and show that the final sets are equal.