ML Aggarwal Class 11 Maths Solutions Chapter 01 Sets

Introduction

George Cantor was a mathematician from Russia who studied in Germany. He became the first person to recognize how important sets are. Set theory is helpful in nearly every area of mathematics. This chapter covers - the concept of a set, how to represent a set, different types of sets, set relations, subsets of real numbers called intervals, Venn diagrams, operations on sets, important results about the size of sets, and how to use sets to solve real-world problems.

1.1 Sets

Every day, we work with collections or groups of objects that are similar in some way. Think about these examples: (i) even natural numbers that are less than 15 (the numbers 2, 4, 6, 8, 10, 12, 14); (ii) vowels in the English alphabet (the letters a, e, i, o, u); (iii) all colors of the rainbow; (iv) all states of India; (v) all rivers of India; (vi) all prime factors of 330 (2, 3, 5, and 11); (vii) the roots of the equation x² - 2x - 3 = 0 (which are 3 and -1); (viii) all straight lines (drawn in a particular plane) that pass through a given point.

Each of these collections is a well-defined collection of objects. A "well-defined collection of objects" means that for any collection and any object, you can clearly decide whether that object belongs to the collection or not.

Set: A set is any well-defined collection of objects. The objects in the set are called its members or elements. The terms "objects," "members," or "elements" of a set all mean the same thing and are considered undefined terms.

Now think about the collection of all good books on mathematics. This is NOT a well-defined collection because what one person thinks is a good math book, another person might not. So this collection is NOT a set.

These collections are NOT well-defined: (i) all intelligent students of class XI of your school; (ii) all big cities of India; (iii) all beautiful girls of India; (iv) five most renowned scientists of the world. Because of this, none of these collections is a set.

Sets are usually shown using capital letters A, B, C, etc., and the members are shown using lowercase letters a, b, c, etc.

If x is a member of set A, we write x ∈ A (which means "x belongs to A"). If x is NOT a member of set A, we write x ∉ A (which means "x does not belong to A"). If both x and y belong to A, we write x, y ∈ A.

1.1.1 Representations of a Set

There are two main ways to show what a set contains.

1. Roster or Tabular Form: In this method, we list all members of the set inside braces (curly brackets) and separate them by commas.

For example:

(i) The set A of all even natural numbers less than 15 in roster form is written as A = {2, 4, 6, 8, 10, 12, 14}. Note that 2 ∈ A and 10 ∈ A while 5 ∉ A.

(ii) The set S of vowels in the English alphabet in tabular form is written as S = {a, e, i, o, u}.

(iii) The set M of months of a year having less than 31 days in roster form is written as M = {February, April, June, September, November}.

(iv) The set L of letters in the word 'JODHPUR' in tabular form is written as L = {J, O, D, H, P, U, R}.

Remarks:

1. The order in which you list elements can be changed. So {3, 7, 8, 12} can also be written as {7, 3, 8, 12} or {12, 7, 3, 8}.

2. If a set has repeated elements, the set stays the same. The set {a, b, c, b, b, a} is the same as {a, b, c}.

3. When listing elements of a set, list each member only once. The set X of letters in the word 'MATHEMATICS' in tabular form is written as X = {M, A, T, H, E, I, C, S}.

4. Roster form lets you see all members of a set at once. But if a set has very many elements, you can show it by writing a few elements that clearly show the pattern, then write three dots, and then write the last element (if there is one). The set A of odd natural numbers between 50 and 500 in tabular form can be written as A = {51, 53, 55, ..., 499}. The set P of even integers less than 10 in roster form can be written as P = {..., -4, -2, 0, 2, 4, 6, 8}.

2. Set Builder Form or Rule Method: In this method, we write a variable (such as x) that stands for any member of the set. This is followed by a colon ':' (or sometimes '|'). Then we write the property that each member has, and put the whole description inside braces. If A is a set with elements x that have property p, we write A = {x : x has property p}, which means "the set of elements x such that x has the property p."

The colon ':' means "such that." Sometimes '|' is used instead of ':'.

For example:

(i) The set A of all even natural numbers less than 15 in builder form is written as A = {x : x is an even natural number less than 15}.

(ii) The set S of vowels in the English alphabet in builder form is written as S = {x : x is a vowel in the English alphabet}.

(iii) The set S = {1, 4, 9, 16, 25, ...} in builder form can be written as S = {x : x is the square of a natural number}.

1.1.2 Kinds of Sets

1. Empty Set: A set that does not contain any elements is called the empty set, null set, or void set. There is only one such set. It is shown as φ or { }.

For example:

(i) The collection of natural numbers less than 1.

(ii) {x : 2x + 11 = 3 and x is a natural number}.

(iii) {x : x² = 9 and x is an even integer}.

(iv) {x : x is an even prime number greater than 2}.

Each of these is the empty set.

2. Singleton Set: A set that contains exactly one element is called a singleton set (or unit set).

For example:

(i) {0}.

(ii) {x : 3x - 1 = 8}.

(iii) {x : x is the capital of India}.

Each of these is a singleton set.

3. Finite Set: A set that contains a limited or fixed number of different elements is called a finite set.

For example:

(i) S = {a, e, i, o, u}.

(ii) A = {2, 4, 6, ..., 100}.

(iii) S = {x : x is the capital of India}.

(iv) M = {x : x is a month of a year}.

(v) P = {x : x ∈ N and x is a prime factor of 210} which is {2, 3, 5, 7}.

Each of these is a finite set.

Note: Since the empty set has no elements, φ is a finite set.

4. Infinite Set: A set that contains an unlimited number of different elements is called an infinite set. In other words, a set which is NOT finite is an infinite set.

For example:

(i) The set of even natural numbers, which is {2, 4, 6, ...}.

(ii) {x : x ∈ N and x is prime} which is {2, 3, 5, 7, 11, 13, ...}.

(iii) The set of all points on a line segment.

(iv) The set of all straight lines (drawn in a particular plane) that pass through a given point.

Each of these is an infinite set.

Note: Not all infinite sets can be written in roster form. For example, the set of real numbers cannot be written this way because the elements do not follow any clear pattern.

1.1.3 Cardinal Number (or Order) of a Finite Set

The number of different elements in a finite set A is called the cardinal number (or order) of A. It is shown as n(A) or O(A).

For example:

(i) Let A = {a, e, i, o, u}, then n(A) = 5.

(ii) Let A be the set of letters in the word SCHOOL, which is A = {S, C, H, O, L}, then n(A) = 5.

(iii) Let A = {x : x is a prime factor of 60} which is {2, 3, 5}, then n(A) = 3.

(iv) Let D = {x : x is a digit in our number system} which is {0, 1, 2, ..., 9}, then n(D) = 10.

Note: The cardinal number of the empty set is zero. The cardinal number of a singleton set is one. The cardinal number of an infinite set is never defined.

1.1.4 Some Standard Sets of Numbers

(i) Natural Numbers: The set of natural (or counting) numbers is shown by N. Thus N = {1, 2, 3, ...}.

(ii) Whole Numbers: The set of whole numbers is shown by W. Thus W = {0, 1, 2, 3, ...}.

(iii) Integers: The set of all integers is shown by I or Z. Thus I = {..., -2, -1, 0, 1, 2, ...}.

(iv) Rational Numbers: Any number that can be written as \( \frac{p}{q} \) where p, q ∈ I and q ≠ 0 is called a rational number. Examples include \( \frac{2}{3} \), \( \frac{7}{5} \), 6, \( \frac{-68}{1} \), and -7. The set of rational numbers is shown by Q.

(v) Real Numbers: All rational and irrational numbers are real numbers. Examples include -3, 0, 5, \( \frac{5}{3} \), \( \frac{7}{2} \), \( \sqrt{2} \), \( \sqrt{3} \), \( -2 + \sqrt{3} \), \( \sqrt[3]{2} \). The set of real numbers is shown by R.

(vi) Irrational Numbers: The set of irrational numbers is shown by T. Thus T = {x : x ∈ R and x ∉ Q}, which means T is the set of all real numbers that are NOT rational. Examples include \( \sqrt{2} \), \( \sqrt{3} \), \( -\sqrt[3]{5} \), and π.

(vii) Positive Rational Numbers: The set of positive rational numbers is shown by Q+.

(viii) Positive Real Numbers: The set of positive real numbers is shown by R+.

Illustrative Examples

Question 1. State whether the statement "collection of competent school teachers in Delhi is a set" is true or false. Justify your answer.
Answer: This statement is false. The reason is that the collection of competent school teachers in Delhi is not well-defined. Different people will have different opinions about which teachers are competent. One person might think a certain teacher is competent while another person disagrees.

Exam Tip: When asked if something is a set, always check if it is well-defined. A collection must let you decide without doubt whether any object belongs to it or not.

Question 2. Write the following sets in roster form:
(i) A = {x | x ∈ N and 4 < x ≤ 10}
(ii) H = {x | x is a letter in the word 'ARITHMETIC'}
(iii) B = {x | x ∈ N and 5 < x² < 50}
(iv) A = {x : x ∈ I and x² < 20}
(v) S = {x : x is a solution of the equation x² - x - 6 = 0}
(vi) B = {x : x = \( \frac{2n-1}{n+2} \), n ∈ W and n < 4}
(vii) A = {x : x is a two digit number such that the sum of its digits is 9}
(viii) P = {x | x is a positive integer less than 10 and 2x - 1 is an odd integer}
Answer:
(i) A = {5, 6, 7, 8, 9, 10}.
(ii) H = {A, R, I, T, H, M, E, C}.
(iii) We see that the squares of natural numbers 3, 4, 5, 6, 7 fall between 5 and 50. So A in roster form is A = {3, 4, 5, 6, 7}.
(iv) We see that the squares of integers 0, ±1, ±2, ±3, ±4 are less than 20. So A in roster form is A = {-4, -3, -2, -1, 0, 1, 2, 3, 4}.
(v) Starting with x² - x - 6 = 0:
\[ (x - 3)(x + 2) = 0 \]
\[ x - 3 = 0 \text{ or } x + 2 = 0 \implies x = 3 \text{ or } x = -2 \]
So S = {3, -2}.
(vi) Since n ∈ W and n < 4, we have n = 0, 1, 2, 3. Also x = \( \frac{2n-1}{n+2} \). Putting n = 0, 1, 2, 3, we get x = -\( \frac{1}{2} \), \( \frac{1}{3} \), \( \frac{3}{4} \), 1. So A = {-\( \frac{1}{2} \), \( \frac{1}{3} \), \( \frac{3}{4} \), 1}.
(vii) For x to be a two digit number with digit sum 9, the numbers are 18, 27, 36, 45, 54, 63, 72, 81, 90. So A = {18, 27, 36, 45, 54, 63, 72, 81, 90}.
(viii) The value 2x - 1 is always an odd positive number for all positive whole numbers x. Specifically, 2x - 1 is odd for x = 1, 2, 3, ..., 9. So P = {1, 2, 3, 4, 5, 6, 7, 8, 9}.

Exam Tip: When converting from builder to roster form, first find all values satisfying the condition, then list them inside braces without repeating any element.

Question 3. Write the following sets in builder form:
(i) The counting numbers which are multiples of 6 and less than 50
(ii) The fractions whose numerator is 1 and whose denominator is a counting number less than 10
(iii) The set of all positive integers whose cube is odd
(iv) A = {\( \frac{1}{2} \), \( \frac{2}{3} \), \( \frac{3}{4} \), ..., \( \frac{9}{10} \)}
(v) B = {1, \( \frac{1}{4} \), \( \frac{1}{9} \), \( \frac{1}{16} \), \( \frac{1}{25} \), ...}
Answer:
(i) {x : x is a multiple of 6 and 0 < x < 50}.
(ii) {\( \frac{1}{x} \) : x is a counting number and x < 10}.
(iii) Since the cube of an even positive whole number is even and the cube of an odd positive whole number is odd, all members of this set are positive odd whole numbers. So the set is {x : x is an odd positive integer} or {x : x = 2k + 1 and k ∈ W}.
(iv) Each member has numerator one less than the denominator. The numerator starts with 1 and ends with 9. So A = {x : x = \( \frac{n}{n+1} \), n ∈ N and 1 ≤ n ≤ 9}.
(v) B = {x : x = \( \frac{1}{n²} \), n ∈ N}.

Exam Tip: Look for the pattern in roster form. Write down what changes (the variable) and what stays the same (the conditions) to create the builder form.

Question 4. Match each set on the left (shown in roster form) with the same set on the right (shown in builder form):
(i) {3, 4, 5, 6, 7} <-> (a) {x : x is a solution of x² + x - 2 = 0}
(ii) {1, 3, 5, 7, 9} <-> (b) {x : x is a letter in the word TEACHER}
(iii) {A, C, H, R, T, E} <-> (c) {x : x is an odd natural number less than 10}
(iv) {1, -2} <-> (d) {x : x ∈ N and 2 < x ≤ 7}
Answer: For option (d), x ∈ N and 2 < x ≤ 7 means x can be 3, 4, 5, 6, 7. So (i) matches (d). For option (c), x is an odd natural number less than 10 means x can be 1, 3, 5, 7, 9. So (ii) matches (c). In option (b), the word TEACHER has 7 letters and E repeats, giving {T, E, A, C, H, R}, so (iii) matches (b). For option (a), solving x² + x - 2 = 0: (x - 1)(x + 2) = 0, so x = 1 or x = -2. So (iv) matches (a).

Exam Tip: Match each description carefully. Find the elements of each set, then check which roster form matches which builder form.

Question 5. State which of the following statements are true and which are false. Justify your answer.
(i) 31 ∉ {x : x has exactly two positive factors}
(ii) 77 ∈ {x : x has exactly four positive factors}
(iii) 28 ∈ {x : the sum of all positive factors of x is 2x}
Answer:
(i) False. The number 31 has exactly two positive factors: 1 and 31. So 31 belongs to this set, not outside it.
(ii) True. The number 77 has exactly four positive factors: 1, 7, 11, and 77. So 77 belongs to this set.
(iii) True. The sum of all positive factors of 28 is 1 + 2 + 4 + 7 + 14 + 28 = 56 = 2 × 28.

Exam Tip: Check membership by finding all factors and confirming whether the element satisfies the given condition.

Question 6. State which of the following sets are finite or infinite. For finite sets, mention the cardinal number:
(i) A = {x : x ∈ N and x² = 9}
(ii) B = {x : x ∈ W and 2x - 1 = 0}
(iii) C = {x : x ∈ N and x² - 3x + 2 = 0}
(iv) D = {x : x ∈ N and x is prime}
(v) E = {x : x ∈ N and x is odd}
(vi) F = {x : x is a month of a year having less than 31 days}
(vii) G = {x : x ∈ I and x > -3}
Answer:
(i) From x² = 9, we get x = 3 or x = -3. But x ∈ N (natural numbers), so A = {3}. This is a finite set. n(A) = 1.
(ii) From 2x - 1 = 0, we get x = \( \frac{1}{2} \). But x ∈ W (whole numbers), so B = φ. This is a finite set. n(φ) = 0.
(iii) From x² - 3x + 2 = 0, we get (x - 1)(x - 2) = 0, so x = 1 or x = 2. Both are in N, so C = {1, 2}. This is a finite set. n(C) = 2.
(iv) D = {x : x ∈ N and x is prime} = {2, 3, 5, 7, 11, 13, ...}. Since there are infinitely many primes, D is infinite.
(v) E = {x : x ∈ N and x is odd} = {1, 3, 5, 7, 9, 11, ...}. Since there are infinitely many odd numbers, E is infinite.
(vi) F = {February, April, June, September, November}. This is a finite set. n(F) = 5.
(vii) G = {x : x ∈ I and x > -3} = {-2, -1, 0, 1, 2, 3, ...}. This is an infinite set.

Exam Tip: First find all elements satisfying the condition. If you can list them all with a definite end, it is finite. If the list continues without stopping, it is infinite.

Exercise 1.1 - Very Short Answer Type Questions

Question 1. State which of the given collections of objects is a set:
(i) A collection of popular cinema actors of India
(ii) The collection of even natural numbers less than 51
(iii) The collection of counting numbers less than 1
(iv) Collection of interesting books written by Shakespeare
(v) The collection of novels written by Munshi Prem Chand
(vi) The collection of 10 most talented students of your school
(vii) Collection of all rivers flowing in India
(viii) Collection of 5 rivers flowing in India
(ix) Collection of all rational numbers which lie between -1 and 1
(x) A team of eleven best cricketers of the world
(xi) A collection of most dangerous animals of the world
Answer: (i) Not a set - "popular" is subjective. (ii) A set - we can clearly identify all even numbers less than 51. (iii) A set - there are no counting numbers less than 1, making it the empty set. (iv) Not a set - "interesting" depends on personal opinion. (v) A set - we can list all novels by this author. (vi) Not a set - "talented" is not well-defined. (vii) A set - India's rivers can be clearly identified. (viii) A set - we can name 5 specific rivers. (ix) A set - we can clearly identify all rational numbers in this range. (x) Not a set - "best" is subjective. (xi) Not a set - "dangerous" is not objective.

Exam Tip: A collection is a set only if you can always decide whether something belongs to it or not, without personal opinion.

Question 2. If A = {3, 5, 7, 9, 11}, then write which of the following statements are true. If a statement is not true, mention why.
(i) 3 ∈ A
(ii) 5, 9 ∈ A
(iii) 8 ∉ A
(iv) 9 ∉ A
(v) {3} ∈ A
(vi) {5, 7} ∈ A
Answer: (i) True - 3 is an element of A. (ii) True - both 5 and 9 are elements of A. (iii) True - 8 is not an element of A. (iv) False - 9 is an element of A, so 9 ∉ A is not true. (v) False - {3} is a set, not an element. The element is 3, not {3}. (vi) False - {5, 7} is a set containing two elements, not an element of A itself.

Exam Tip: Remember that an element (like 3) is different from a set containing that element (like {3}). Use ∈ for elements and ⊂ for subsets.

Question 3. Use the roster method to represent the following sets:
(i) The counting numbers which are multiples of 6 and less than 50
(ii) The fractions whose numerator is 1, and whose denominator is a counting number less than 10
(iii) {x : x ∈ N and x is a prime factor of 84}
(iv) The set of odd integers lying between -4 and 8
(v) The set of all natural numbers x for which x + 6 is less than 10
(vi) The set of all integers x for which x + 6 is greater than 10
(vii) The set of all integers x for which x + 6 is less than 10
(viii) The set of all integers x for which \( \frac{60}{x} \) is a natural number
(ix) {t : t³ = t, t ∈ R}
(x) {x : \( \frac{x-2}{x+3} \) = 3, x ∈ R}
(xi) {x : x⁴ - 5x² + 6 = 0, x ∈ R}
(xii) {x : x ∈ I, -\( \frac{1}{2} \) < x < \( \frac{9}{2} \)}
(xiii) {x : x ∈ N and 4x - 3 ≤ 15}
(xiv) {x : x ∈ N, x² < 40}
(xv) {x : x ∈ Z and x² < 16}
(xvi) The set of all digits in our decimal system
(xvii) The set of all letters in the word TRIGONOMETRY
(xviii) The set of all vowels in the English alphabet which precede q
(xix) {x : x is a consonant in the English alphabet which precedes k}
Answer: (i) {6, 12, 18, 24, 30, 36, 42, 48}. (ii) {\( \frac{1}{1} \), \( \frac{1}{2} \), \( \frac{1}{3} \), \( \frac{1}{4} \), \( \frac{1}{5} \), \( \frac{1}{6} \), \( \frac{1}{7} \), \( \frac{1}{8} \), \( \frac{1}{9} \)}. (iii) Prime factors of 84 are 2, 3, 7, so {2, 3, 7}. (iv) Odd integers between -4 and 8 are {-3, -1, 1, 3, 5, 7}. (v) x + 6 < 10 means x < 4. Natural numbers less than 4 are {1, 2, 3}. (vi) x + 6 > 10 means x > 4. Integers greater than 4 are {5, 6, 7, ...} - infinite, so {5, 6, 7, ...}. (vii) x + 6 < 10 means x < 4. Integers less than 4 are {..., -2, -1, 0, 1, 2, 3}. (viii) \( \frac{60}{x} \) is a natural number when x is a divisor of 60. Divisors of 60 are {1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60}. (ix) From t³ = t, we get t(t² - 1) = 0, so t(t - 1)(t + 1) = 0, giving {-1, 0, 1}. (x) From \( \frac{x-2}{x+3} \) = 3, we get x - 2 = 3(x + 3), so x - 2 = 3x + 9, giving -2x = 11, so x = -\( \frac{11}{2} \). The set is {-\( \frac{11}{2} \)}. (xi) Let y = x². Then y² - 5y + 6 = 0 gives (y - 2)(y - 3) = 0, so y = 2 or 3. Then x = ±\( \sqrt{2} \) or ±\( \sqrt{3} \), giving {-\( \sqrt{3} \), -\( \sqrt{2} \), \( \sqrt{2} \), \( \sqrt{3} \)}. (xii) -\( \frac{1}{2} \) < x < \( \frac{9}{2} \) means -0.5 < x < 4.5. Integers in this range are {0, 1, 2, 3, 4}. (xiii) 4x - 3 ≤ 15 means 4x ≤ 18, so x ≤ 4.5. Natural numbers ≤ 4.5 are {1, 2, 3, 4}. (xiv) x² < 40 with x ∈ N. Since 6² = 36 < 40 and 7² = 49 > 40, the set is {1, 2, 3, 4, 5, 6}. (xv) x² < 16 means |x| < 4. Integers satisfying this are {-3, -2, -1, 0, 1, 2, 3}. (xvi) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. (xvii) Distinct letters are {T, R, I, G, O, N, M, E, Y}. (xviii) Vowels before q are {a, e, i, o}. (xix) Consonants before k are {b, c, d, f, g, h, j}.

Exam Tip: When converting to roster form, find all elements satisfying the condition and list each only once.

Question 4. Write the following sets in builder form:
(i) {1, 3, 5, 7, 9, 11, 13}
(ii) {2, 4, 6, 8, ...}
(iii) {3, 6, 9, 12, 15}
(iv) {2, 4, 8, 16, 32, 64}
(v) {5, 25, 125, 625}
(vi) {1, 4, 9, 16, ..., 100}
(vii) {1, 4, 9, 16, 25, ...}
(viii) {\( \frac{1}{2} \), \( \frac{2}{3} \), \( \frac{3}{4} \), \( \frac{4}{5} \), ...}
Answer: (i) {x : x is an odd natural number and x ≤ 13}. (ii) {x : x is an even natural number}. (iii) {x : x is a multiple of 3 and x ≤ 15}. (iv) {x : x = 2^n, n ∈ N and 1 ≤ n ≤ 6}. (v) {x : x = 5^n, n ∈ N and 1 ≤ n ≤ 4}. (vi) {x : x = n², n ∈ N and 1 ≤ n ≤ 10}. (vii) {x : x = n², n ∈ N}. (viii) {x : x = \( \frac{n}{n+1} \), n ∈ N}.

Exam Tip: Look at the pattern and identify the relationship between elements. Express this relationship using a variable and conditions.

Question 5. Which of the following are examples of the null set?
(i) Set of even prime numbers
(ii) Set of odd natural numbers divisible by 2
(iii) Set of all Indian kids 5 metres tall
(iv) {x : x ∈ N, x < 5 and x > 8}
(v) {x : x is a point common to any two parallel straight lines}
(vi) {x : x is a student of your school presently studying in both classes XI and XII}
Answer: (i) Not null - 2 is an even prime number. (ii) Null - no odd number can be divisible by 2. (iii) Null - no child is 5 metres tall. (iv) Null - no number can be both less than 5 and greater than 8. (v) Null - parallel lines never meet. (vi) Null - a student cannot study in two different classes at the same time.

Exam Tip: The null set has no elements. Check whether any element can possibly satisfy all the conditions given.

Question 6. Which of the following sets are finite or infinite?
(i) The set of days of a week
(ii) The set of numbers which are multiples of 7
(iii) The set of animals living on Earth
(iv) The set of consonants in the English alphabet
(v) The set of circles drawn in a plane
(vi) The set of prime numbers which are less than one crore
Answer: (i) Finite - there are exactly 7 days. (ii) Infinite - multiples of 7 continue without end. (iii) Finite - though the count is huge, it is limited. (iv) Finite - there are 21 consonants in English. (v) Infinite - you can draw as many circles as you want. (vi) Finite - there is a limit (one crore) and a definite number of primes below it.

Exam Tip: A set is finite if all its elements can be counted and listed. It is infinite if the elements go on without stopping.

Question 7. Find the cardinal number of the following sets:
(i) { }
(ii) {0}
(iii) A = {1, 2, 2, 1, 3}
(iv) The set of all Indians having 8 legs
(v) The set of all letters in the word PRINCIPAL
(vi) The set of all vowels in the word PRINCIPAL
Answer: (i) n({ }) = 0 - this is the empty set. (ii) n({0}) = 1 - it contains one element: zero. (iii) A = {1, 2, 3}, so n(A) = 3. (iv) n = 0 - humans do not have 8 legs. (v) Letters are {P, R, I, N, C, A, L}, so n = 7. (vi) Vowels are {I, A}, so n = 2.

Exam Tip: Count each distinct element only once. Do not count repeats.

Question 8. (i) Write the cardinal number of the set A, where A = {x : x is a two digit number, sum of whose digits is 8}.
(ii) Write the cardinal number of the set of all integers x for which \( \frac{30}{x} \) is a natural number
(iii) What is the cardinal number of the set X, where X = {x : x is a letter in the word 'CHANDIGARH'}?
(iv) If S = {x : x is a positive multiple of 3 less than 100} and P = {x : x is a prime number less than 20}, then write n(S) + n(P).
Answer: (i) Two digit numbers with digit sum 8 are: 17, 26, 35, 44, 53, 62, 71, 80. So n(A) = 8. (ii) \( \frac{30}{x} \) is a natural number when x is a divisor of 30. Divisors of 30 are: ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30. So n = 16. (iii) Letters in CHANDIGARH are {C, H, A, N, D, I, G, R}. So n(X) = 8. (iv) Positive multiples of 3 less than 100 are 3, 6, 9, ..., 99. The count is 33, so n(S) = 33. Primes less than 20 are 2, 3, 5, 7, 11, 13, 17, 19. So n(P) = 8. Thus n(S) + n(P) = 33 + 8 = 41.

Exam Tip: For cardinal number, count all distinct elements. Use divisors carefully - remember to consider negative divisors for integers.

Question 9. Match each set on the left (in roster form) with the same set on the right (in builder form):
(i) {2, 3} <-> (a) {x : x ∈ N and is a divisor of 6}
(ii) {5, -5} <-> (b) {x : x ∈ N and is a prime divisor of 6}
(iii) {1, 3, 5} <-> (c) {x : x is a letter in the word LITTLE}
(iv) {1, 2, 3, 6} <-> (d) {x : x is an odd natural number less than 6}
(v) {T, E, L, I} <-> (e) {x : x is a root of the equation x² - 25 = 0}
Answer: For (a), divisors of 6 in N are {1, 2, 3, 6} - so (iv) matches (a). For (b), prime divisors of 6 are {2, 3} - so (i) matches (b). For (c), letters in LITTLE are {L, I, T, E} - so (v) matches (c). For (d), odd natural numbers less than 6 are {1, 3, 5} - so (iii) matches (d). For (e), roots of x² - 25 = 0 are x = ±5, giving {5, -5} - so (ii) matches (e).

Exam Tip: Work through each option to find all elements, then match carefully.

Question 10. State which of the following statements are true and which are false. Justify your answer.
(i) 37 ∉ {x : x has exactly two positive factors}
(ii) 35 ∈ {x : x has exactly four positive factors}
(iii) 128 ∈ {y : the sum of all positive factors of y is 2y}
(iv) 7747 ∈ {t : t is a multiple of 37}
Answer: (i) False - 37 is prime and has exactly two positive factors (1 and 37), so it belongs to the set. (ii) True - 35 = 5 × 7 has factors 1, 5, 7, 35 (four factors). (iii) False - 128 = 2^7 has factors 1, 2, 4, 8, 16, 32, 64, 128. Their sum is 255 ≠ 2 × 128. (iv) True - 7747 ÷ 37 = 209, so 7747 is a multiple of 37.

Exam Tip: For "exactly two factors," check if the number is prime. Count all factors carefully.

Question 11. Classify the following sets into finite set and infinite set. In case of finite sets, mention the cardinal number.
(i) A = {x : x ∈ I, x < 5}
(ii) A = {x : x ∈ W, x is divisible by 4 and 9}
(iii) P = {x : x is an even prime number > 2}
(iv) F = {x : x ∈ N and x is a factor of 84}
(v) B = {x : x is a two digit number, sum of whose digits is 12}
(vi) C = {x : x ∈ W, 3x - 7 ≤ 8}
(vii) {x : x = 5n, n ∈ N and x < 20}
(viii) {x : x = 5n, n ∈ I and x < 20}
(ix) {x : x = \( \frac{n}{n+1} \), n ∈ W and n ≤ 10}
(x) {x : x = \( \frac{2n}{n+3} \), n ∈ N and 5 < n < 20}
Answer: (i) Infinite - integers less than 5 go on forever in the negative direction. (ii) Finite - numbers divisible by both 4 and 9 are divisible by 36. They are 0, 36, 72, ..., but within W they are limited. Numbers are {0, 36, 72}, so n = 3. (iii) Finite - there is only one even prime number, which is 2. Since we need > 2, this set is empty, n = 0. (iv) Finite - factors of 84 are {1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84}, so n = 12. (v) Finite - two digit numbers with digit sum 12 are 39, 48, 57, 66, 75, 84, 93. So n = 7. (vi) From 3x - 7 ≤ 8, we get 3x ≤ 15, so x ≤ 5. Whole numbers ≤ 5 are {0, 1, 2, 3, 4, 5}, so n = 6. (vii) 5n < 20 means n < 4. Natural numbers < 4 are {1, 2, 3}. So x = {5, 10, 15}, n = 3. (viii) 5n < 20 means n < 4. Integers < 4 are {..., -2, -1, 0, 1, 2, 3}. So x includes ..., -10, -5, 0, 5, 10, 15. Infinite. (ix) n ∈ W and n ≤ 10 means n = {0, 1, 2, ..., 10}. So 11 values give 11 elements. Finite, n = 11. (x) 5 < n < 20 with n ∈ N means n = {6, 7, 8, ..., 19}. That is 14 values. Finite, n = 14.

Exam Tip: Carefully identify the domain (I, W, N, etc.) and apply constraints to determine if elements can be listed or continue infinitely.

1.2 Set Relations

1.2.1 Equivalent Sets

Two finite sets A and B are called equivalent if they have the same number of elements. We write this as A ↔ B (read as "A is equivalent to B") if n(A) = n(B).

For example:

(i) Let A = {a, b, c, d, e} and B = {2, 3, 5, 7, 9}. Then n(A) = 5 = n(B). So A ↔ B.

(ii) Let A = {x : x is a color of the rainbow} and B = {x : x ∈ W, x < 7}. Then n(A) = 7 = n(B). So A ↔ B.

(iii) Let P = {x : x is a letter in the word 'FLOWER'} and Q = {x : x is a letter in the word 'FOLLOWER'}. Both sets equal {F, L, O, W, E, R}, so n(P) = 6 = n(Q). So P ↔ Q.

1.2.2 Equal Sets

Two sets A and B are said to be equal if they have exactly the same elements. We write this as A = B. This means every member of A is also a member of B, and every member of B is also a member of A. If A and B are not equal, we write A ≠ B.

For example:

(i) A = {1, 2} and B = {2, 1, 1, 2, 1}. Then A = B (order and repetition do not matter).

(ii) Let P = {x : x is a vowel in the word 'EQUALITY'} and Q = {x : x is a vowel in the word 'QUATITATIVE'}. Both equal {E, U, A, I}, so P = Q.

(iii) Let A = {-3, -2, -1, 0, 1, 2, 3} and B = {x : x ∈ I and x² < 10}. Both equal the same set, so A = B.

(iv) Let P = {x : x ∈ N and x² - 2 = 0} and Q = {x : x is a triangle having 5 sides}. Both are empty sets (φ), so P = Q.

Remark: If A and B are finite sets and A = B, then n(A) = n(B), so A ↔ B. This means equal sets are always equivalent. However, the reverse is not always true. For example, A = {2, 3, 5} and B = {2, 3, 4} have n(A) = 3 = n(B), so A ↔ B, but A ≠ B. Thus, equal sets are always equivalent, but equivalent sets may not be equal.

1.2.3 Subset

Let A and B be any two sets. A is called a subset of B if every member of A is also a member of B. We write this as A ⊂ B (read as "A is a subset of B" or "A is contained in B"). This means: if x ∈ A, then x ∈ B.

If A ⊂ B (A is contained in B), we may also say that B contains A or that B is a superset of A. We write this as B ⊃ A (read as "B contains A" or "B is a superset of A").

If there exists at least one element in A that is not a member of B, then A is not a subset of B. We write this as A ⊄ B.

For example:

(i) Let A = {2, 3, 5} and B = {1, 2, 3, 5, 6}. Since every member of A is also in B, we have A ⊂ B. Note that 1 ∈ B but 1 ∉ A, so B ⊄ A.

(ii) Let A = {a, e, i, o, u} and B = {a, b, c, d, e}. We see that i ∈ A but i ∉ B, so A ⊄ B. Also, b ∈ B but b ∉ A, so B ⊄ A.

(iii) Let P be the set of letters in 'SCHOOL' and Q be the set of letters in 'SCHOLAR'. In roster form, P = {S, C, H, O, L} and Q = {S, C, H, O, L, A, R}. Clearly P ⊂ Q while Q ⊄ P.

(iv) Let A = {x : x is a divisor of 56} and B = {x : x is a prime divisor of 56}. Then A = {1, 2, 4, 7, 8, 14, 28, 56} and B = {2, 7}. We see that B ⊂ A while A ⊄ B.

(v) Let A = {1, 3, 5, 3, 1} and B = {x : x is an odd natural number less than 6}. We see that A ⊂ B and B ⊂ A. In fact, A = B.

Proper Subset: Let A be any set and B be a non-empty set. Then A is called a proper subset of B if every member of A is also a member of B AND there exists at least one element in B that is not a member of A. If A is a proper subset of B, we write A ⊂ B and A ≠ B.

In the examples above, (i) A is a proper subset of B, (iii) P is a proper subset of Q, and (iv) B is a proper subset of A.

Remark: If A = B, then A ⊂ B and B ⊂ A. Conversely, if A ⊂ B and B ⊂ A, then A = B. Thus A = B if and only if every a ∈ A is also in B and every b ∈ B is also in A.

Note: For any set A:

(1) A ⊂ A - every set is a subset of itself, but not a proper subset. A subset that is not proper is called an improper subset.

(2) Every set has exactly one improper subset (itself).

(3) Since the empty set has no elements, φ ⊂ A - the empty set is a subset of every set.

(4) The empty set is a proper subset of every set except itself.

Subsets of a Set:

(i) Let A = {a}. The subsets of A are φ and A. Note n(A) = 1 and the number of subsets = 2 = 2¹.

(ii) Let A = {1, 2}. The subsets of A are φ, {1}, {2}, and A. Note n(A) = 2 and the number of subsets = 4 = 2².

(iii) Let A = {1, 2, 3}. The subsets of A are φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, and A. Note n(A) = 3 and the number of subsets = 8 = 2³.

Remark: If A has n(A) = m, then the number of subsets of A is 2^m and the number of proper subsets is 2^m - 1.

1.2.4 Power Set

The power set of a given set A is the set formed by all the subsets of A. It is shown as P(A).

For example, if A = {1, 2}, then P(A) = {φ, {1}, {2}, A}.

Remark: If A has n(A) = m, then n(P(A)) = 2^m.

1.2.5 Universal Set

A universal set is a set that contains all the elements under consideration in a given problem. It is shown as ξ or U. It is a kind of "parent set." Every set being discussed is a subset of the universal set.

Note that the choice of universal set is not fixed. It can vary from one problem to another. Therefore, we always specify the universal set in a given problem.

For example:

(i) For A = {b, c, g, m, u}, the universal set might be {x : x is a letter in the English alphabet}.

(ii) For A = {x : x ∈ N, 3 ≤ x < 12}, the universal set might be {1, 2, 3, ..., 20} or N.

(iii) For A = {Earth, Mars}, the universal set might be {x : x is a planet of our solar system}.

1.2.6 Subsets of Real Numbers

We know several standard sets of numbers. These are all subsets of the set of real numbers. It is easy to see that:

N ⊂ W ⊂ I ⊂ Q ⊂ R, and T ⊂ R, and N ⊄ T.

Also Q+ ⊂ Q ⊂ R and R+ ⊂ R.

Intervals as Subsets of R

Intervals are special types of subsets of the set of real numbers. Let a and b be two distinct real numbers where a < b.

Open Interval: The set of all real numbers lying between a and b is called an open interval. It is shown as (a, b). More precisely, (a, b) = {x : x ∈ R, a < x < b}. The number a is the left end point and b is the right end point. Note that the open interval (a, b) does NOT include the endpoints a and b.

On a number line, if A and B represent a and b, the open interval is shown as:

Closed Interval: The set of all real numbers lying between a and b, including a and b, is called a closed interval. It is shown as [a, b]. More precisely, [a, b] = {x : x ∈ R, a ≤ x ≤ b}.

On a number line, it is shown as:

Open-Closed Interval: The set of all real numbers between a and b, including b but not a, is called an open-closed interval. It is shown as (a, b]. More precisely, (a, b] = {x : x ∈ R, a < x ≤ b}. This interval is open on the left and closed on the right.

On a number line, it is shown as:

Closed-Open Interval: The set of all real numbers between a and b, including a but not b, is called a closed-open interval. It is shown as [a, b). More precisely, [a, b) = {x : x ∈ R, a ≤ x < b}. This interval is closed on the left and open on the right.

On a number line, it is shown as:

The length of any of these intervals is (b - a). All the intervals above are finite intervals. There are also infinite intervals.

Infinite Intervals:

The set of all real numbers x such that x > a is shown as (a, ∞). More precisely, (a, ∞) = {x : x ∈ R, x > a}.

The set of all real numbers x such that x ≥ a is shown as [a, ∞). More precisely, [a, ∞) = {x : x ∈ R, x ≥ a}.

The set of all real numbers x such that x < a is shown as (-∞, a). More precisely, (-∞, a) = {x : x ∈ R, x < a}.

The set of all real numbers x such that x ≤ a is shown as (-∞, a]. More precisely, (-∞, a] = {x : x ∈ R, x ≤ a}.

The set of all real numbers is shown as (-∞, ∞). More precisely, (-∞, ∞) = {x : x ∈ R}. This is the entire real number line.

Remark: Note that '∞' (read as infinity) is NOT a number - it cannot be treated as one. It is a symbol showing largeness without any limit - meaning greater than any positive real number, no matter how large. Similarly, '-∞' shows smallness without limit - meaning smaller than any negative real number, no matter how small.

Illustrative Examples

Question 1. Find all pairs of equal sets (if any):
A = {0}, B = {x : x < 5 and x > 15}, C = {x : x - 5 = 0}, D = {x : x² = 25}, E = {x : x is a positive integral root of the equation x² - 2x - 15 = 0}
Answer: Let me find each set. A = {0}. For B, no number can be both less than 5 and greater than 15, so B = φ. For C, x - 5 = 0 gives x = 5, so C = {5}. For D, x² = 25 gives x = 5 or x = -5, so D = {5, -5}. For E, solving x² - 2x - 15 = 0: (x - 5)(x + 3) = 0, so x = 5 or x = -3. Since we need positive integral roots, E = {5}. The only pair of equal sets is C and E, both equal {5}.

Exam Tip: Carefully solve equations to find the exact elements of each set. Two sets are equal only if they have exactly the same elements.

Question 2. Consider the following sets: φ, A = {1, 3}, B = {1, 5, 9} and C = {1, 2, 3, 5, 7, 9}. Insert the correct symbol ⊂ or ⊄ between the following pairs of sets:
(i) φ ... B
(ii) A ... B
(iii) A ... C
(iv) B ... C
Answer: (i) φ ⊂ B - the empty set is a subset of every set. (ii) A ⊄ B - we see that 3 ∈ A but 3 ∉ B. (iii) A ⊂ C - both 1 and 3 are members of C. (iv) B ⊂ C - all members 1, 5, 9 of B are in C.

Exam Tip: To check if A ⊂ B, verify that every member of A is also in B. If even one member of A is missing from B, then A ⊄ B.

Question 3. State whether each of the following statements is true or false for the sets A, B and C where A = {x : x is a letter of the word 'BOWL'}, B = {x : x is a letter of the word 'ELBOW'}, C = {x : x is a letter of the word 'BELLOW'}:
(i) A ⊂ B
(ii) B ⊃ C
(iii) B = C
(iv) B ↔ C
(v) A is a proper subset of B
(vi) B is a proper subset of C
Answer: First, let me write the sets in roster form. A = {B, O, W, L}, B = {E, L, B, O, W}, C = {B, E, L, O, W}. (i) True - every member of A {B, O, W, L} is in B. (ii) True - B ⊃ C means C ⊂ B. Every member of C is in B. (iii) True - B and C have the same elements {B, E, L, O, W}. (iv) True - n(B) = 5 = n(C). (v) True - A ⊂ B and A ≠ B (since L ∈ A but L ∉ B is false; actually L ∈ B, so let me recheck: A = {B, O, W, L} and B = {E, L, B, O, W}. All of A are in B, and E ∈ B but E ∉ A, so A is a proper subset of B). (vi) False - B = C, so B cannot be a proper subset of C.

Exam Tip: Write sets in roster form carefully, counting each distinct letter only once.

Question 4. Let ξ = {1, 2, 3, ..., 50}, A = {x : x is divisible by 2 and 3}, B = {x : x = n², n ∈ N} and C = {x : x is a factor of 42}, then:
(i) Write the sets A, B and C in roster form
(ii) State n(A), n(B) and n(C)
(iii) State whether A ↔ B
(iv) State whether A ↔ C
Answer: (i) We work within ξ = {1, 2, 3, ..., 50}. A consists of numbers divisible by both 2 and 3, meaning divisible by 6: A = {6, 12, 18, 24, 30, 36, 42, 48}. B consists of perfect squares: B = {1, 4, 9, 16, 25, 36, 49}. C consists of factors of 42: C = {1, 2, 3, 6, 7, 14, 21, 42}. (ii) n(A) = 8, n(B) = 7, n(C) = 8. (iii) No; since n(A) = 8 ≠ 7 = n(B). (iv) Yes; since n(A) = 8 = n(C). Note that A ≠ C even though they are equivalent.

Exam Tip: Equivalence means same number of elements; equality means same exact elements. They are different concepts.

Question 5. Let A, B and C be three sets. (i) If A ∈ B and B ⊂ C, is it true that A ⊂ C? If not, give an example.
(ii) If A ⊂ B and B ∈ C, is it true that A ∈ C? If not, give an example.
Answer: (i) No, this is not necessarily true. Let A = {1}, B = {{1}, 2} and C = {{1}, 2, 3}. Here A ∈ B (the element A is in set B) and B ⊂ C (every element of B is in C). However, A ⊄ C because 1 ∈ A but 1 ∉ C. (ii) No, this is not necessarily true. Let A = {1}, B = {1, 2} and C = {{1, 2}, 3}. Here A ⊂ B (A is a subset of B) and B ∈ C (B is an element of C). However, A ∉ C because A is not listed as an element of C.

Exam Tip: Be careful to distinguish between ∈ (element of) and ⊂ (subset of). Using one where the other is meant gives wrong conclusions.

Question 6. Let A = {x : x is a letter in the word 'GEORGE CANTOR'} and B = {x : x is a vowel in the word 'GEORGE CANTOR'}, then:
(i) Write the sets A, B in tabular form
(ii) State n(A) and n(B)
(iii) Write the number of proper subsets of A
(iv) Write the power set of B
Answer: (i) The word GEORGE CANTOR has letters G, E, O, R, G, E, C, A, N, T, O, R. Removing repeats: A = {G, E, O, R, C, A, N, T}. The vowels in GEORGE CANTOR are E, O, A: B = {E, O, A}. (ii) n(A) = 8 and n(B) = 3. (iii) The number of proper subsets of A = 2^8 - 1 = 256 - 1 = 255. (iv) P(B) = {φ, {E}, {O}, {A}, {E, O}, {E, A}, {O, A}, {E, O, A}}.

Exam Tip: List distinct elements only - do not repeat. The power set includes the empty set and the set itself.

Question 7. Let A = {1, 2, 3}, B = {2, 4} and C = {1, 2, 3, 4}. Find all sets X such that (i) X ⊂ A and X ⊂ B, (ii) X ⊂ C but X ⊄ A
Answer: (i) X must be a subset of both A and B. So X can only contain elements that are in both A and B. The common elements are {2}. All subsets of {2} are {φ, {2}}. So X = φ or X = {2}. (ii) X must be a subset of C but not a subset of A. This means X contains at least one element from C that is not in A. The elements in C but not in A are {4}. So X must contain 4. Also, X can contain any subset of {1, 2, 3}. The subsets of C containing 4 are {4}, {1, 4}, {2, 4}, {3, 4}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}. All of these are subsets of C but not of A.

Exam Tip: For X ⊂ A and X ⊂ B, find X ⊂ (A ∩ B). For X ⊂ C but X ⊄ A, X must have at least one element in C - A.

Question 14. There are 200 individuals with a skin disorder; 120 had been exposed to the chemical C₁, 50 to the chemical C₂ and 30 to both the chemicals C₁ and C₂. Find the number of individuals exposed to
(i) chemical C₁ but not chemical C₂
(ii) chemical C₂ but not chemical C₁
(iii) chemical C₁ or chemical C₂.
Answer: Let A be the set of individuals exposed to C₁ and B be the set of individuals exposed to C₂. We are given: |A| = 120, |B| = 50, |A ∩ B| = 30.
(i) Individuals exposed to C₁ but not C₂ = |A - B| = |A| - |A ∩ B| = 120 - 30 = 90
(ii) Individuals exposed to C₂ but not C₁ = |B - A| = |B| - |A ∩ B| = 50 - 30 = 20
(iii) Individuals exposed to C₁ or C₂ = |A ∪ B| = |A| + |B| - |A ∩ B| = 120 + 50 - 30 = 140
In simple words: Use the formula: people in A only = A total minus those in both; people in B only = B total minus those in both; people in A or B = add both groups, then subtract the overlap so you don't count them twice.

Exam Tip: Always draw a Venn diagram with two overlapping circles to organize the given numbers and check your answers - this visual helps catch mistakes in identifying "only," "both," and "or" regions.

Question 15. In an examination, 56 per cent of the candidates failed in English and 48 per cent failed in Science. If 18 per cent failed in both English and Science, find the percentage of those who passed in both the subjects.
Answer: Let E be the set of candidates who failed in English and S be the set of candidates who failed in Science. We have: |E| = 56%, |S| = 48%, |E ∩ S| = 18%.
The percentage who failed in English or Science = |E ∪ S| = |E| + |S| - |E ∩ S| = 56 + 48 - 18 = 86%
Therefore, the percentage who passed in both subjects = 100% - 86% = 14%
In simple words: If 86 percent of students failed in at least one subject, then 14 percent passed in both subjects (since 100 - 86 = 14).

Exam Tip: The phrase "passed in both" means they did NOT fail in either subject - use the complement rule: percentage passing both = 100 - (percentage failing in at least one).

Question 16. From amongst the 6000 literate individuals of a city, 50% read newspaper A, 45% read newspaper B and 25% read neither A nor B. How many individuals read both the newspapers A as well as B?
Answer: Let A and B be the sets of individuals reading newspapers A and B respectively. Total = 6000; |A| = 50% of 6000 = 3000; |B| = 45% of 6000 = 2700. Those reading neither = 25% of 6000 = 1500.
Those reading at least one newspaper = 6000 - 1500 = 4500
Using |A ∪ B| = |A| + |B| - |A ∩ B|:
4500 = 3000 + 2700 - |A ∩ B|
|A ∩ B| = 5700 - 4500 = 1200
Therefore, 1200 individuals read both newspapers A and B.
In simple words: First find how many read at least one newspaper (total minus those reading neither). Then use the overlap formula to find how many read both.

Exam Tip: Always convert percentages to actual numbers first, then apply set formulas - this reduces calculation errors and makes the answer clearer to the examiner.

Question 17. In a beauty contest, half the number of judges voted for Miss A, 2/3 of them voted for Miss B, 10 voted for both and 6 did not vote for either Miss A or Miss B. Find how many judges, in all, were present there.
Answer: Let the total number of judges be n. Let A be the set of judges voting for Miss A and B be the set of judges voting for Miss B.
Given: |A| = n/2, |B| = 2n/3, |A ∩ B| = 10, and those voting for neither = 6
Using the principle: n = |A ∪ B| + (those voting for neither)
|A ∪ B| = |A| + |B| - |A ∩ B| = n/2 + 2n/3 - 10
n = (n/2 + 2n/3 - 10) + 6
n = 7n/6 - 4
6n = 7n - 24
n = 24
Therefore, there were 24 judges in total.
In simple words: Set up an equation where total judges equals those voting for at least one person plus those voting for neither, then solve for n.

Exam Tip: When using fractional parts of an unknown total, express everything in terms of that unknown (n) before solving - this approach keeps algebra manageable and reduces errors.

Question 18. In a group of 50 students, the number of students studying French, English and Sanskrit were found to be as follows: French = 17, English = 13, Sanskrit = 15; French and English = 9, English and Sanskrit = 4, French and Sanskrit = 5; English, French and Sanskrit = 3. Find the number of students who study:
(i) French only
(ii) French and Sanskrit but not English
(iii) English only
(iv) French and English but not Sanskrit
(v) Sanskrit only
(vi) English and Sanskrit but not French
(vii) atleast one of the three languages
(viii) none of the three languages.
Answer: Let F, E, and S denote the sets of students studying French, English, and Sanskrit respectively.
Given: |F| = 17, |E| = 13, |S| = 15; |F ∩ E| = 9, |E ∩ S| = 4, |F ∩ S| = 5; |F ∩ E ∩ S| = 3

(i) French only = |F| - |F ∩ E| - |F ∩ S| + |F ∩ E ∩ S| = 17 - 9 - 5 + 3 = 6

(ii) French and Sanskrit but not English = |F ∩ S| - |F ∩ E ∩ S| = 5 - 3 = 2

(iii) English only = |E| - |F ∩ E| - |E ∩ S| + |F ∩ E ∩ S| = 13 - 9 - 4 + 3 = 3

(iv) French and English but not Sanskrit = |F ∩ E| - |F ∩ E ∩ S| = 9 - 3 = 6

(v) Sanskrit only = |S| - |F ∩ S| - |E ∩ S| + |F ∩ E ∩ S| = 15 - 5 - 4 + 3 = 9

(vi) English and Sanskrit but not French = |E ∩ S| - |F ∩ E ∩ S| = 4 - 3 = 1

(vii) Atleast one of the three languages = 6 + 2 + 3 + 6 + 9 + 1 + 3 = 30

(viii) None of the three languages = 50 - 30 = 20
In simple words: For each region in a three-circle Venn diagram, subtract overlaps from individual totals. The center (all three) is counted 3 times, so add it back twice. Sum all regions for "at least one," then subtract from total for "none."

Exam Tip: Use the inclusion-exclusion principle and organize your work by region (only one, exactly two, all three, none) - drawing a labeled Venn diagram with all given values entered first helps prevent arithmetic errors on multi-set problems.