Maharashtra Board Class 11 Maths Part 1 Chapter 1 Sets and Relations 1.1 Solutions

Question 1. Describe the following sets in Roster form:
(i) {x / x is a letter of the word ‘MARRIAGE’}
(ii) {x / x is an integer, \( -\frac{1}{2} < x < \frac{9}{2} \)}
(iii) {x / x = 2n, n \( \in \) N}
Answer: In roster form, we list all the elements of the set explicitly inside curly brackets, separated by commas, without repeating any elements.
(i) Let A = {x / x is a letter of the word ‘MARRIAGE’}
\( \implies \) A = {M, A, R, I, G, E}
(ii) Let B = {x / x is an integer, \( -\frac{1}{2} < x < \frac{9}{2} \)}
\( \implies \) B = {0, 1, 2, 3, 4}
(iii) Let C = {x / x = 2n, n \( \in \) N}
\( \implies \) C = {2, 4, 6, 8, ...}
In simple words: Roster form is just a way of writing a set by listing all its members inside curly brackets, making sure we don't repeat any letters or numbers.

🎯 Exam Tip: When writing a set in roster form, remember that the order of elements does not matter, but you must never repeat any element, such as the letters 'R' and 'A' in 'MARRIAGE'.

Question 2. Describe the following sets in Set-Builder form:
(i) {0}
(ii) {0, ±1, ±2, ±3}
(iii) \( \left\{ \frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \frac{4}{17}, \frac{5}{26}, \frac{6}{37}, \frac{7}{50} \right\} \)
Answer:
(i) Let \( A = \{0\} \). Here, 0 is a whole number but it is not a natural number.
\( \implies A = \{x \mid x \in W, x \notin N\} \)

(ii) Let \( B = \{0, \pm 1, \pm 2, \pm 3\} \). Here, B is the set of integers from -3 to 3.
\( \implies B = \{x \mid x \in Z, -3 \le x \le 3\} \)

(iii) Let \( C = \left\{ \frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \frac{4}{17}, \frac{5}{26}, \frac{6}{37}, \frac{7}{50} \right\} \). Here, the numerators are natural numbers \( n \le 7 \) and the denominators are of the form \( n^2 + 1 \).
\( \implies C = \left\{ x \;\middle|\; x = \frac{n}{n^2+1}, n \in N, n \le 7 \right\} \)
In simple words: Set-builder form is a way of writing a set by describing the rule or property that all its elements share, rather than listing each element individually.

🎯 Exam Tip: When writing in set-builder form, always clearly define the domain of the variable (like \( N \), \( W \), or \( Z \)) and the specific rule or formula that the elements follow.

Question 3. If \( A = \{x \mid 6x^2 + x - 15 = 0\} \), \( B = \{x \mid 2x^2 - 5x - 3 = 0\} \), \( C = \{x \mid 2x^2 - x - 3 = 0\} \), then find (i) \( (A \cup B \cup C) \) (ii) \( (A \cap B \cap C) \)
Answer:
First, let us find the elements of each set by solving the quadratic equations:

For Set A:
\( 6x^2 + x - 15 = 0 \)
\( \implies 6x^2 + 10x - 9x - 15 = 0 \)
\( \implies 2x(3x + 5) - 3(3x + 5) = 0 \)
\( \implies (3x + 5)(2x - 3) = 0 \)
\( \implies 3x + 5 = 0 \) or \( 2x - 3 = 0 \)
\( \implies x = -\frac{5}{3} \) or \( x = \frac{3}{2} \)
\( \implies A = \left\{-\frac{5}{3}, \frac{3}{2}\right\} \)

For Set B:
\( 2x^2 - 5x - 3 = 0 \)
\( \implies 2x^2 - 6x + x - 3 = 0 \)
\( \implies 2x(x - 3) + 1(x - 3) = 0 \)
\( \implies (2x + 1)(x - 3) = 0 \)
\( \implies 2x + 1 = 0 \) or \( x - 3 = 0 \)
\( \implies x = -\frac{1}{2} \) or \( x = 3 \)
\( \implies B = \left\{-\frac{1}{2}, 3\right\} \)

For Set C:
\( 2x^2 - x - 3 = 0 \)
\( \implies 2x^2 - 3x + 2x - 3 = 0 \)
\( \implies x(2x - 3) + 1(2x - 3) = 0 \)
\( \implies (x + 1)(2x - 3) = 0 \)
\( \implies x + 1 = 0 \) or \( 2x - 3 = 0 \)
\( \implies x = -1 \) or \( x = \frac{3}{2} \)
\( \implies C = \left\{-1, \frac{3}{2}\right\} \)

Now, we can find the required union and intersection:

(i) \( A \cup B \cup C \):
Combining all unique elements from sets A, B, and C, we get:
\( \implies A \cup B \cup C = \left\{-\frac{5}{3}, -1, -\frac{1}{2}, \frac{3}{2}, 3\right\} \)

(ii) \( A \cap B \cap C \):
Finding the elements that are common to all three sets A, B, and C:
Since there is no single element present in all three sets simultaneously, the intersection is empty.
\( \implies A \cap B \cap C = \emptyset \)
In simple words: To find the union, we combine all the elements from all three sets without repeating any. To find the intersection, we look for elements that are present in all three sets at the same time.

🎯 Exam Tip: Always solve the quadratic equations carefully by factoring, and remember that the intersection of three sets must contain only elements that are common to all three sets simultaneously.

Question 3. If \( A = \{-\frac{5}{3}, \frac{3}{2}\} \), \( B = \{x \mid 2x^2 - 5x - 3 = 0\} \) and \( C = \{x \mid 2x^2 - x - 3 = 0\} \), then find:
(i) \( A \cup B \cup C \)
(ii) \( A \cap B \cap C \)
Answer:
\( B = \{x \mid 2x^2 - 5x - 3 = 0\} \)
\( \implies 2x^2 - 5x - 3 = 0 \)
\( \implies 2x^2 - 6x + x - 3 = 0 \)
\( \implies 2x(x - 3) + 1(x - 3) = 0 \)
\( \implies (x - 3)(2x + 1) = 0 \)
\( \implies x - 3 = 0 \) or \( 2x + 1 = 0 \)
\( \implies x = 3 \) or \( x = -\frac{1}{2} \)
\( \implies B = \{-\frac{1}{2}, 3\} \)

\( C = \{x \mid 2x^2 - x - 3 = 0\} \)
\( \implies 2x^2 - x - 3 = 0 \)
\( \implies 2x^2 - 3x + 2x - 3 = 0 \)
\( \implies x(2x - 3) + 1(2x - 3) = 0 \)
\( \implies (2x - 3)(x + 1) = 0 \)
\( \implies 2x - 3 = 0 \) or \( x + 1 = 0 \)
\( \implies x = \frac{3}{2} \) or \( x = -1 \)
\( \implies C = \{-1, \frac{3}{2}\} \)

(i) \( A \cup B \cup C = \{-\frac{5}{3}, \frac{3}{2}\} \cup \{-\frac{1}{2}, 3\} \cup \{-1, \frac{3}{2}\} = \{-\frac{5}{3}, -1, -\frac{1}{2}, \frac{3}{2}, 3\} \)
(ii) \( A \cap B \cap C = \{\} \)
In simple words: First, we solve the quadratic equations to find the elements of sets B and C. Then, we combine all elements to find the union, and look for common elements across all three sets to find the intersection, which is empty here.

🎯 Exam Tip: When solving quadratic equations for sets, always write down the factorization steps clearly and verify the final values by substituting them back into the equation.

Question 4. If A, B, C are the sets for the letters in the words ‘college’, ‘marriage’ and ‘luggage’ respectively, then verify that [A – (B ∪ C)] = [(A – B) ∩ (A – C)].
Answer:
A = {c, o, l, g, e}
B = {m, a, r, i, g, e}
C = {l, u, g, a, e}
B ∪ C = {m, a, r, i, g, e, l, u}
A – (B ∪ C) = {c, o}
A – B = {c, o, l}
A – C = {c, o}

\( \implies [(A - B) \cap (A - C)] = \{c, o\} = A - (B \cup C) \)

\( \implies [A - (B \cup C)] = [(A - B) \cap (A - C)] \)
This verification shows that subtracting the union of two sets from a set is equivalent to intersecting the individual differences.
In simple words: We first list the unique letters of each word as sets. Then we find the union of B and C, subtract it from A, and compare it to the intersection of the differences (A - B) and (A - C) to show they are equal.

🎯 Exam Tip: Remember that sets do not contain duplicate letters, so write words like 'college' as {c, o, l, g, e} without repeating 'l' or 'e'.

Question 5. If \( A = \{1, 2, 3, 4\} \), \( B = \{3, 4, 5, 6\} \), \( C = \{4, 5, 6, 7, 8\} \) and universal set \( X = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \), then verify the following:
(i) \( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \)
(ii) \( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \)
(iii) \( (A \cup B)' = A' \cap B' \)
(iv) \( (A \cap B)' = A' \cup B' \)
(v) \( A = (A \cap B) \cup (A \cap B') \)
(vi) \( B = (A \cap B) \cup (A' \cap B) \)
(vii) \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
Answer: Given: \( A = \{1, 2, 3, 4\} \), \( B = \{3, 4, 5, 6\} \), \( C = \{4, 5, 6, 7, 8\} \), \( X = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \)

(i) To verify: \( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \)
\( B \cap C = \{4, 5, 6\} \)
\( \therefore A \cup (B \cap C) = \{1, 2, 3, 4, 5, 6\} \) ......(i)
\( A \cup B = \{1, 2, 3, 4, 5, 6\} \)
\( A \cup C = \{1, 2, 3, 4, 5, 6, 7, 8\} \)
\( \therefore (A \cup B) \cap (A \cup C) = \{1, 2, 3, 4, 5, 6\} \) ......(ii)
From (i) and (ii), we get:
\( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \)

(ii) To verify: \( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \)
\( B \cup C = \{3, 4, 5, 6, 7, 8\} \)
\( \therefore A \cap (B \cup C) = \{3, 4\} \) ......(i)
\( A \cap B = \{3, 4\} \)
\( A \cap C = \{4\} \)
\( \dots (A \cap B) \cup (A \cap C) = \{3, 4\} \) ......(ii)
From (i) and (ii), we get:
\( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \)

(iii) To verify: \( (A \cup B)' = A' \cap B' \)
\( A \cup B = \{1, 2, 3, 4, 5, 6\} \)
\( \therefore (A \cup B)' = \{7, 8, 9, 10\} \) ......(i)
\( A' = \{5, 6, 7, 8, 9, 10\} \), \( B' = \{1, 2, 7, 8, 9, 10\} \)
\( \therefore A' \cap B' = \{7, 8, 9, 10\} \) ......(ii)
From (i) and (ii), we get:
\( (A \cup B)' = A' \cap B' \)

(iv) To verify: \( (A \cap B)' = A' \cup B' \)
\( A \cap B = \{3, 4\} \)
\( \therefore (A \cap B)' = \{1, 2, 5, 6, 7, 8, 9, 10\} \) ......(i)
\( A' = \{5, 6, 7, 8, 9, 10\} \)
\( B' = \{1, 2, 7, 8, 9, 10\} \)
\( \therefore A' \cup B' = \{1, 2, 5, 6, 7, 8, 9, 10\} \) ......(ii)
From (i) and (ii), we get:
\( (A \cap B)' = A' \cup B' \)

(v) To verify: \( A = (A \cap B) \cup (A \cap B') \)
\( A \cap B = \{3, 4\} \)
\( B' = \{1, 2, 7, 8, 9, 10\} \)
\( A \cap B' = \{1, 2\} \)
\( \therefore (A \cap B) \cup (A \cap B') = \{3, 4\} \cup \{1, 2\} = \{1, 2, 3, 4\} \) ......(i)
But \( A = \{1, 2, 3, 4\} \) ......(ii)
From (i) and (ii), we get:
\( A = (A \cap B) \cup (A \cap B') \)

(vi) To verify: \( B = (A \cap B) \cup (A' \cap B) \)
\( A \cap B = \{3, 4\} \)
\( A' = \{5, 6, 7, 8, 9, 10\} \)
\( A' \cap B = \{5, 6\} \)
\( \therefore (A \cap B) \cup (A' \cap B) = \{3, 4\} \cup \{5, 6\} = \{3, 4, 5, 6\} \) ......(i)
But \( B = \{3, 4, 5, 6\} \) ......(ii)
From (i) and (ii), we get:
\( B = (A \cap B) \cup (A' \cap B) \)

(vii) To verify: \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
\( A = \{1, 2, 3, 4\} \)
\( \implies n(A) = 4 \)
\( B = \{3, 4, 5, 6\} \)
\( \implies n(B) = 4 \)
\( A \cap B = \{3, 4\} \)
\( \implies n(A \cap B) = 2 \)
L.H.S. = \( n(A \cup B) \)
Since \( A \cup B = \{1, 2, 3, 4, 5, 6\} \), \( n(A \cup B) = 6 \) ......(i)
R.H.S. = \( n(A) + n(B) - n(A \cap B) = 4 + 4 - 2 = 6 \) ......(ii)
From (i) and (ii), L.H.S. = R.H.S.
\( \therefore n(A \cup B) = n(A) + n(B) - n(A \cap B) \)

These set operations demonstrate fundamental laws of Boolean algebra and set theory.
In simple words: This problem helps us verify different rules of sets, like how they combine (union) or overlap (intersection), by using actual numbers. By calculating both sides of each equation step-by-step, we prove that they are equal.

🎯 Exam Tip: Always list the elements of each set clearly before performing operations like union or intersection, and number your equations (i) and (ii) to show LHS equals RHS clearly.

Question 5. Verify the set relations (Continued):
Answer:
(iv) (Continued)
\( B' = \{1, 2, 7, 8, 9, 10\} \)
\( \therefore A' \cup B' = \{1, 2, 5, 6, 7, 8, 9, 10\} \text{ ......(ii)} \)
From (i) and (ii), we get:
\( (A \cap B)' = A' \cup B' \). This confirms that the complement of the intersection is indeed equal to the union of the individual complements.

(v) \( A = \{1, 2, 3, 4\} \text{ ......(i)} \)
\( A \cap B = \{3, 4\} \)
\( B' = \{1, 2, 7, 8, 9, 10\} \)
\( A \cap B' = \{1, 2\} \)
\( \therefore (A \cap B) \cup (A \cap B') = \{1, 2, 3, 4\} \text{ ......(ii)} \)
From (i) and (ii), we get:
\( A = (A \cap B) \cup (A \cap B') \)

(vi) \( B = \{3, 4, 5, 6\} \text{ ......(i)} \)
\( A \cap B = \{3, 4\} \)
\( A' = \{5, 6, 7, 8, 9, 10\} \)
\( A' \cap B = \{5, 6\} \)
\( \therefore (A \cap B) \cup (A' \cap B) = \{3, 4, 5, 6\} \text{ ......(ii)} \)
From (i) and (ii), we get:
\( B = (A \cap B) \cup (A' \cap B) \)

(vii) \( A = \{1, 2, 3, 4\}, B = \{3, 4, 5, 6\} \),
\( A \cap B = \{3, 4\}, A \cup B = \{1, 2, 3, 4, 5, 6\} \)
\( \therefore n(A) = 4, n(B) = 4, \)
\( n(A \cap B) = 2, \)
\( n(A \cup B) = 6 \text{ ......(i)} \)
\( \therefore n(A) + n(B) - n(A \cap B) = 4 + 4 - 2 \)
\( \therefore n(A) + n(B) - n(A \cap B) = 6 \text{ ......(ii)} \)
From (i) and (ii), we get:
\( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
In simple words: These steps verify fundamental set properties and cardinalities by calculating the left-hand side and right-hand side of each identity using the given elements.

🎯 Exam Tip: When verifying set identities, clearly label your equations as (i) and (ii) to show that the left-hand side equals the right-hand side.

Question 6. If \( A \) and \( B \) are subsets of the universal set \( X \) and \( n(X) = 50 \), \( n(A) = 35 \), \( n(B) = 20 \), \( n(A' \cap B') = 5 \), find:
(i) \( n(A \cup B) \)
(ii) \( n(A \cap B) \)
(iii) \( n(A' \cap B) \)
(iv) \( n(A \cap B') \)
Answer:
Given:
\( n(X) = 50 \), \( n(A) = 35 \), \( n(B) = 20 \), \( n(A' \cap B') = 5 \)

(i) To find \( n(A \cup B) \):
By De Morgan's Law, \( A' \cap B' = (A \cup B)' \). This law is incredibly useful for converting intersections of complements into unions.
Therefore, \( n(A' \cap B') = n((A \cup B)') = n(X) - n(A \cup B) \)
\( 5 = 50 - n(A \cup B) \)

\( \implies n(A \cup B) = 50 - 5 = 45 \)

(ii) To find \( n(A \cap B) \):
We know that \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
Substituting the values:
\( 45 = 35 + 20 - n(A \cap B) \)
\( 45 = 55 - n(A \cap B) \)

\( \implies n(A \cap B) = 55 - 45 = 10 \)

(iii) To find \( n(A' \cap B) \):
We know that \( n(A' \cap B) = n(B) - n(A \cap B) \)
Substituting the values:
\( n(A' \cap B) = 20 - 10 = 10 \)

(iv) To find \( n(A \cap B') \):
We know that \( n(A \cap B') = n(A) - n(A \cap B) \)
Substituting the values:
\( n(A \cap B') = 35 - 10 = 25 \)
In simple words: We use the total number of elements in the universal set and De Morgan's laws to find the union of sets A and B. From there, we use standard set formulas to find their intersection and the elements belonging exclusively to one set.

🎯 Exam Tip: Always remember De Morgan's Law \( (A \cup B)' = A' \cap B' \) as it is frequently used to find the union of sets from their complements.

Question 7. Out of 200 students, 35 students failed in MHT-CET, 40 in AIEEE and 40 in IIT entrance, 20 failed in MHT-CET and AIEEE, 17 in AIEEE and IIT entrance, 15 in MHT-CET and IIT entrance, and 5 failed in all three examinations. Find how many students
(i) did not fail in any examination.
(ii) failed in AIEEE or IIT entrance.
Answer:
Let \( X \) be the set of all students, \( A \) be the set of students who failed in MHT-CET, \( B \) be the set of students who failed in AIEEE, and \( C \) be the set of students who failed in IIT entrance.
From the given data:
\( n(X) = 200 \)
\( n(A) = 35 \)
\( n(B) = 40 \)
\( n(C) = 40 \)
\( n(A \cap B) = 20 \)
\( n(B \cap C) = 17 \)
\( n(A \cap C) = 15 \)
\( n(A \cap B \cap C) = 5 \)

First, we calculate the number of students who failed in at least one of the three examinations using the union formula:
\( n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C) \)
\( n(A \cup B \cup C) = 35 + 40 + 40 - 20 - 17 - 15 + 5 \)
\( n(A \cup B \cup C) = 68 \)

(i) Students who did not fail in any examination:
The number of students who did not fail in any exam is the complement of the union of the three sets:
\( n(A' \cap B' \cap C') = n(X) - n(A \cup B \cup C) \)
\( = 200 - 68 \)
\( = 132 \)
Therefore, 132 students did not fail in any examination.

(ii) Students who failed in AIEEE or IIT entrance:
The number of students who failed in AIEEE or IIT entrance is represented by \( n(B \cup C) \):
\( n(B \cup C) = n(B) + n(C) - n(B \cap C) \)
\( = 40 + 40 - 17 \)
\( = 63 \)
Therefore, 63 students failed in AIEEE or IIT entrance.
In simple words: Out of 200 students, we find that 68 students failed in at least one exam. Subtracting this from the total gives us 132 students who passed everything, while adding the failures of AIEEE and IIT and subtracting their overlap gives us 63 students.

🎯 Exam Tip: Clearly define your sets at the beginning of the solution and write down the formula for the union of three sets to secure step-by-step marks.

Question 8. From amongst 2000 literate individuals of a town, 70% read Marathi newspapers, 50% read English newspapers and 32.5% read both Marathi and English newspapers. Find the number of individuals who read
(i) at least one of the newspapers.
(ii) neither Marathi nor English newspaper.
(iii) only one of the newspapers.
Answer:
Let \( M \) = set of individuals who read Marathi newspapers
Let \( E \) = set of individuals who read English newspapers
Let \( X \) = set of all literate individuals

Given:
\( n(X) = 2000 \)
\( n(M) = \frac{70}{100} \times 2000 = 1400 \)
\( n(E) = \frac{50}{100} \times 2000 = 1000 \)
\( n(M \cap E) = \frac{32.5}{100} \times 2000 = 650 \)

(i) At least one of the newspapers:
The number of individuals who read at least one newspaper is represented by \( n(M \cup E) \). This shows that a majority of the literate population in the town prefers reading at least one daily newspaper.
\( n(M \cup E) = n(M) + n(E) - n(M \cap E) \)
\( = 1400 + 1000 - 650 \)
\( = 1750 \)

(ii) Neither Marathi nor English newspaper:
The number of individuals who read neither newspaper is the complement of the union of both sets.
\( n(M' \cap E') = n(X) - n(M \cup E) \)
\( = 2000 - 1750 \)
\( = 250 \)

(iii) Only one of the newspapers:
The number of individuals who read only one newspaper is found by subtracting those who read both newspapers from those who read at least one.
\( = n(M \cup E) - n(M \cap E) \)
\( = 1750 - 650 \)
\( = 1100 \)
In simple words: We use set formulas to group the newspaper readers. By finding the total readers and subtracting those who read both, we can easily calculate who reads at least one paper, who reads none, and who reads only one.

🎯 Exam Tip: Clearly define each set with appropriate letters (like M for Marathi and E for English) and write down the standard union formula to secure step-wise marks.

Question 8. (Continued)
Answer:
Venn Diagram Representation:

• Universal Set \( X \): 2000
• Set \( M \) (Marathi): 750
• Set \( E \) (English): 350
• Intersection \( M \cap E \): 650

(i) No. of individuals who read at least one of the newspapers = \( n(M \cup E) = 1750 \).
(ii) No. of individuals who read neither Marathi nor English newspaper = \( n(M' \cap E') \)
\( = n(M \cup E)' \)
\( = n(X) - n(M \cup E) \)
\( = 2000 - 1750 \)
\( = 250 \)
(iii) No. of individuals who read only one of the newspapers = \( n(M \cap E') + n(M' \cap E) \)
\( = n(M \cup E) - n(M \cap E) \)
\( = 1750 - 650 \)
\( = 1100 \)
This calculation helps us understand the reading habits of the entire surveyed group.
In simple words: We use set formulas to find that 1750 people read at least one newspaper, 250 read neither, and 1100 read exactly one newspaper.

🎯 Exam Tip: Always double-check your calculations by ensuring the sum of individual regions in the Venn diagram equals the total universal set population.

Question 9. In a hostel, 25 students take tea, 20 students take coffee, 15 students take milk, 10 students take both tea and coffee, 8 students take both milk and coffee. None of them take tea and milk both and everyone takes atleast one beverage, find the number of students in the hostel.
Answer:
Let \( T \) = set of students who take tea
\( C \) = set of students who take coffee
\( M \) = set of students who take milk

\( \therefore n(T) = 25, n(C) = 20, n(M) = 15 \),
\( n(T \cap C) = 10, n(M \cap C) = 8, n(T \cap M) = 0, n(T \cap M \cap C) = 0 \)

Venn Diagram Representation:

• Set \( T \) (Tea only): 15
• Set \( C \) (Coffee only): 2
• Set \( M \) (Milk only): 7
• Intersection \( T \cap C \) (Tea and Coffee): 10
• Intersection \( M \cap C \) (Milk and Coffee): 8
• Intersection \( T \cap M \) (Tea and Milk): 0

\( \dots \)
\( \therefore \) Number of students in the hostel = \( n(T \cup C \cup M) \)
\( = n(T) + n(C) + n(M) - n(T \cap C) - n(M \cap C) - n(T \cap M) + n(T \cap M \cap C) \)
\( = 25 + 20 + 15 - 10 - 8 - 0 + 0 \)
\( = 42 \)
Thus, there are 42 students in the hostel in total.
In simple words: To find the total number of students, we add the students who drink each beverage and subtract those who were counted twice because they drink two different beverages.

🎯 Exam Tip: When three sets are involved, remember to subtract the double intersections and add back the triple intersection to avoid undercounting or overcounting.

Question 10. There are 260 persons with skin disorders. If 150 had been exposed to the chemical A, 74 to the chemical B, and 36 to both chemicals A and B, find the number of persons exposed to
(i) Chemical A but not Chemical B
(ii) Chemical B but not Chemical A
(iii) Chemical A or Chemical B.
Answer: Let \( A \) = set of persons exposed to chemical A
\( B \) = set of persons exposed to chemical B
\( X \) = set of all persons
\( \therefore n(X) = 260, n(A) = 150, n(B) = 74, n(A \cap B) = 36 \)

The Venn diagram representation of this data is as follows:
• Universal set \( X \) represents all 260 persons.
• Region of \( A \) only (Chemical A but not B) = 114
• Intersection region of \( A \) and \( B \) (Both chemicals) = 36
• Region of \( B \) only (Chemical B but not A) = 38

(i) No. of persons exposed to chemical A but not to chemical B:
\( n(A \cap B') = n(A) - n(A \cap B) \)
\( = 150 - 36 \)
\( = 114 \)

(ii) No. of persons exposed to chemical B but not to chemical A:
\( n(A' \cap B) = n(B) - n(A \cap B) \)
\( = 74 - 36 \)
\( = 38 \)

(iii) No. of persons exposed to chemical A or chemical B:
\( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
\( = 150 + 74 - 36 \)
\( = 188 \)
These calculations help dermatologists understand the exact exposure patterns among the patient group.
In simple words: To find people exposed to only one chemical, we subtract those exposed to both from the total of that chemical. To find those exposed to either chemical, we add both groups and subtract the overlap so we do not count them twice.

🎯 Exam Tip: Always draw a quick Venn diagram to visualize the regions; it helps prevent calculation errors when subtracting overlapping sets.

Question 11. If A = {1, 2, 3}, write the set of all possible subsets of A.
Answer: Given set \( A = \{1, 2, 3\} \).
The number of elements in set \( A \) is 3, which means there will be \( 2^3 = 8 \) subsets in total.
\( \therefore \{ \}, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{2, 3\}, \{1, 3\} \) and \( \{1, 2, 3\} \) are all the possible subsets of A.
In simple words: Subsets are all the smaller groups you can make from a main set, including the empty group and the full group itself. For three numbers, you can make exactly eight different combinations.

🎯 Exam Tip: Remember to always include the empty set \( \{ \} \) and the set itself as subsets, as these are frequently forgotten in exams.

Question 12. Write the following intervals in set-builder form:
(i) \( (-3, 0) \)
(ii) \( [6, 12] \)
(iii) \( (6, 12) \)
(iv) \( [-23, 5) \)
Answer:
(i) \( (-3, 0) = \{x \mid x \in \mathbb{R}, -3 < x < 0\} \)
(ii) \( [6, 12] = \{x \mid x \in \mathbb{R}, 6 \le x \le 12\} \)
(iii) \( (6, 12) = \{x \mid x \in \mathbb{R}, 6 < x < 12\} \)
(iv) \( [-23, 5) = \{x \mid x \in \mathbb{R}, -23 \le x < 5\} \)
These set-builder forms precisely define the boundaries of each real number interval.
In simple words: A round bracket means the end number is not included in the group, so we use a simple less-than sign. A square bracket means the end number is included, so we use a less-than-or-equal-to sign.

🎯 Exam Tip: Always pay close attention to the type of brackets used. Use \( < \) or \( > \) for open intervals (round brackets) and \( \le \) or \( \ge \) for closed intervals (square brackets) to avoid losing marks.