Sets

Which of the following collections are sets ?

(i) All books in your school library.

(ii) All red flowers in a park.

(iii) All good players in your school.

(iv) All fiction movies.

(v) All easy problems in your book on mathematics.

(vi) All poor people in Mumbai.

(vii) All boys in your class weighing less than 50 kg.

(viii) All persons of repute in your colony.

(ix) All even numbers greater than 100.

(x) All integers less than -5.

Answer

(i) All books in your school library.

The collection is well-defined because one can objectively check if a book belongs to the library's catalog.

∴ It is a set.

(ii) All red flowers in a park.

Colour is a physical property that can be objectively identified.

∴ It is a set.

(iii) All good players in your school.

The term "good" is subjective; a player considered good by one person might not be considered good by another.

∴ It is not a set.

(iv) All fiction movies.

The genre of a movie is a defined category.

∴ It is a set.

(v) All easy problems in your book on mathematics.

The term "easy" is relative; a problem that is easy for one student may be difficult for another.

∴ It is not a set.

(vi) All poor people in Mumbai.

"Poor" is a relative term unless a specific economic threshold (like a poverty line) is provided.

∴ It is not a set.

(vii) All boys in your class weighing less than 50 kg.

Weight is a measurable, objective quantity.

∴ It is a set.

(viii) All persons of repute in your colony.

"Repute" (reputation) is based on personal opinion and varies from person to person.

∴ It is not a set.

(ix) All even numbers greater than 100.

The criteria for being even and being greater than 100 are mathematically definite.

∴ It is a set.

(x) All integers less than -5.

Integers less than -5 (such as -6, -7, .....) are clearly defined.

∴ It is a set.

Write each of the following sets in Roster form :

(i) A = set of all prime numbers between 70 and 100.

(ii) B = set of all whole numbers less than 8.

(iii) C = set of all integers lying between -7 and 2.

(iv) D = set of all composite numbers between 23 and 33.

(v) E = set of letters in the word, 'MATHEMATICS'.

(vi) F = set of consonants in the word, 'SECONDARY'.

(vii) G = set of vowels in the word, 'INTERMEDIATE'.

Answer

(i) A = set of all prime numbers between 70 and 100.

Prime numbers are numbers that have only two factors: 1 and the number itself.

∴ A = {71, 73, 79, 83, 89, 97}

(ii) B = set of all whole numbers less than 8.

Whole numbers start from 0.

∴ B = {0, 1, 2, 3, 4, 5, 6, 7}

(iii) C = set of all integers lying between -7 and 2.

"Between" means we do not include -7 or 2.

∴ C = {-6, -5, -4, -3, -2, -1, 0, 1}

(iv) D = set of all composite numbers between 23 and 33.

Composite numbers have more than two factors.

∴ D = {24, 25, 26, 27, 28, 30, 32}

(v) E = set of letters in the word, 'MATHEMATICS'.

We list each unique letter once, even if it appears multiple times in the word.

∴ E = {M, A, T, H, E, I, C, S}

(vi) F = set of consonants in the word, 'SECONDARY'.

Consonants are letters that are not vowels (a, e, i, o, u).

∴ F = {S, C, N, D, R, Y}

(vii) G = set of vowels in the word, 'INTERMEDIATE'.

Vowels are a, e, i, o, u. We list each unique vowel once.

∴ G = {I, E, A}

Write each of the following sets in Roster form and write the cardinal number of each :

(i) A = {x : x is an integer, -3 < x ≤ 4}.

(ii) B = {x : x ∈ N, 3x - 6 < 9}.

(iii) C = {x : x = n², n ∈ N, 10 < n < 16}.

(iv) D = {x : x ∈ W, x - 3 < 2}.

(v) E = {x : x = 2n - 1, n ∈ N and n < 6}.

(vi) F = {x : x is a letter in the word 'COMMON'}.

(vii) G = {x : x is a primary colour}.

(viii) H = {x : x is a digit in the numeral 2362}.

(ix) J = $\Big\lbrace x : x = \dfrac{1}{n}, n ∈ N, 4 \lt n \lt 10\Big\rbrace$.

Answer

(i) A = {x : x is an integer, -3 < x ≤ 4}.

Integers greater than -3 and less than or equal to 4 are -2, -1, 0, 1, 2, 3, 4.

A = {-2, -1, 0, 1, 2, 3, 4}, n(A) = 7

(ii) B = {x : x ∈ N, 3x - 6 < 9}.

3x - 6 < 9

⇒ 3x < 9 + 6

⇒ 3x < 15

⇒ x < $\dfrac{15}{3}$

⇒ x < 5.

Since x is a natural number (N), x can be 1, 2, 3, 4.

B = {1, 2, 3, 4}, n(B) = 4

(iii) C = {x : x = n², n ∈ N, 10 < n < 16}.

n can be 11, 12, 13, 14, 15.

Calculating x(n²) = (11)², (12)², (13)², (14)², (15)²

= (11 x 11), (12 x 12), (13 x 13), (14 x 14), (15 x 15)

= 121, 144, 169, 196, 225.

C = {121, 144, 169, 196, 225}, n(C) = 5

(iv) D = {x : x ∈ W, x - 3 < 2}.

x - 3 < 2

⇒ x < 2 + 3

⇒ x < 5

Since x is a whole number (W), x can be 0, 1, 2, 3, 4.

D = {0, 1, 2, 3, 4}, n(D) = 5

(v) E = {x : x = 2n - 1, n ∈ N and n < 6}.

n < 6, so n = 1, 2, 3, 4, 5.

Calculating x = (2n - 1):

For n = 1, x = 2(1) - 1 = 1

For n = 2, x = 2(2) - 1 = 3

For n = 3, x = 2(3) - 1 = 5

For n = 4, x = 2(4) - 1 = 7

For n = 5, x = 2(5) - 1 = 9

E = {1, 3, 5, 7, 9}, n(E) = 5

(vi) F = {x : x is a letter in the word 'COMMON'}.

The letters in 'COMMON' are C, O, M, M, O, N.

Removing repeated letters, we get C, O, M, N.

F = {C, O, M, N}, n(F) = 4

(vii) G = {x : x is a primary colour}.

The primary colours are Red, Blue, and Yellow.

G = {Red, Blue, Yellow}, n(G) = 3

(viii) H = {x : x is a digit in the numeral 2362}.

The digits in 2362 are 2, 3, 6, 2.

Removing the repeated digit '2', we get 2, 3, 6.

H = {2, 3, 6}, n(H) = 3

(ix) J = $\Big\lbrace x : x = \dfrac{1}{n}, n ∈ N, 4 \lt n \lt 10\Big\rbrace$.

n can be 5, 6, 7, 8, 9.

Calculating x = $\dfrac{1}{n}$:

For n = 5, $x = \dfrac{1}{5}$

For n = 6, $x = \dfrac{1}{6}$

For n = 7, $x = \dfrac{1}{7}$

For n = 8, $x = \dfrac{1}{8}$

For n = 9, $x = \dfrac{1}{9}$

$J = \Big\lbrace\dfrac{1}{5}, \dfrac{1}{6}, \dfrac{1}{7}, \dfrac{1}{8}, \dfrac{1}{9}\Big\rbrace$, n(J) = 5

Write each of the following sets in set-builder form :

(i) A = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18}.

(ii) B = {1, 2, 3, 5, 6, 10, 15, 30}.

(iii) C = {-9, -6, -3, 0, 3, 6, 9, 12, 15}.

(iv) D = $\Big\lbrace\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},...............,\dfrac{8}{9}\Big\rbrace$.

(v) E = $\Big\lbrace\dfrac{1}{3},\dfrac{1}{5},\dfrac{1}{7},\dfrac{1}{11},\dfrac{1}{13},\dfrac{1}{17},\dfrac{1}{19}\Big\rbrace$.

(vi) F = {April, June, September, November}.

(vii) G = {0}.

(viii) H = { }.

Answer

(i) A = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18}.

These are composite numbers (numbers with more than two factors) less than 20.

∴ A = {x : x is a composite number, 1 < x < 20}

(ii) B = {1, 2, 3, 5, 6, 10, 15, 30}.

These are all the natural numbers that divide 30 without a remainder.

∴ B = {x : x is a factor of 30}

(iii) C = {-9, -6, -3, 0, 3, 6, 9, 12, 15}.

These are multiples of 3 ranging from -9 to 15.

∴ C = {x : x = 3n, n ∈ I, -3 ≤ n ≤ 5}

(iv) D = $\Big\lbrace\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},...............,\dfrac{8}{9}\Big\rbrace$.

Each element is a fraction where the denominator is 1 more than the numerator $\Big(\dfrac{n}{n+1}\Big)$, and the numerator n is a natural number from 1 to 8.

∴ $D = {x : x = \dfrac{n}{n+1}, n \in N \text{ and } 1 \le n \le 8}$

(v) E = $\Big\lbrace\dfrac{1}{3},\dfrac{1}{5},\dfrac{1}{7},\dfrac{1}{11},\dfrac{1}{13},\dfrac{1}{17},\dfrac{1}{19}\Big\rbrace$.

The denominators are prime numbers between 2 and 20.

∴ $E = {x : x = \dfrac{1}{n}, n \text{ is prime, } 2 \lt n \lt 20}$

(vi) F = {April, June, September, November}.

These are the specific months of the year that contain exactly 30 days.

∴ F = {x : x is a month of the year having 30 days}

(vii) G = {0}.

Zero is the only whole number that is not a natural number.

∴ G = {x : x + 1 = 1, x ∈ W}

(viii) H = { }.

An empty set contains no elements. We can describe it using a property that is impossible to satisfy.

∴ H = {x : x is a number, x ≠ x}

State whether the given set is finite or infinite :

(i) Set of all even natural numbers.

(ii) Set of all odd integers.

(iii) Set of all rivers in India.

(iv) Set of all points on a line segment 1 cm long.

(v) Set of all factors of 1200.

(vi) Set of all multiples of 6.

(vii) Set of all drops of water in a bucket.

Answer

(i) Set of all even natural numbers.

Natural numbers go on forever (2, 4, 6, 8, .....), so the counting process never ends.

∴ It is an infinite set.

(ii) Set of all odd integers.

Integers extend infinitely in both positive and negative directions (...., -3, -1, 1, 3, ....).

∴ It is an infinite set.

(iii) Set of all rivers in India.

Although there are many rivers, the total number is a specific, countable figure that can be listed.

∴ It is a finite set.

(iv) Set of all points on a line segment 1 cm long.

Mathematically, a line segment consists of an uncountable infinity of points, regardless of its length.

∴ It is an infinite set.

(v) Set of all factors of 1200.

Every number has a limited (finite) number of factors.

∴ It is a finite set.

(vi) Set of all multiples of 6.

Multiples are generated by multiplying 6 by natural numbers (6, 12, 18, ....), which are endless.

∴ It is an infinite set.

(vii) Set of all drops of water in a bucket.

The number is extremely large and difficult to count.

∴ It is an infinite set.

Identify the null sets among the following :

(i) A = {x : x is a whole number, x < 1}.

(ii) B = {x : x is a number, x > x}.

(iii) C = {x : x is an even prime number}.

(iv) D = {x : x ∈ I, x² = -4}.

(v) E = {x : x is a perfect square number, 40 < x < 50}.

(vi) F = {x : x ∈ N, 5 < x < 6}.

Answer

(i) A = {x : x is a whole number, x < 1}.

Whole numbers start from 0. The only whole number less than 1 is 0.

A = {0}, it contains one element.

∴ It is not a null set.

(ii) B = {x : x is a number, x > x}.

Mathematically, no number can be strictly greater than itself.

∴ It is a null set.

(iii) C = {x : x is an even prime number}.

The number 2 is the only even prime number.

C = {2}

∴ It is not a null set.

(iv) D = {x : x ∈ I, x² = -4}.

The square of any integer (I) is always non-negative (e.g., 2² = 4 and (-2)² = 4). There is no integer whose square is -4.

D = { }

∴ It is a null set.

(v) E = {x : x is a perfect square number, 40 < x < 50}.

Perfect squares near this range are 6² = 36 and 7² = 49. Since 49 falls between 40 and 50, the set contains an element.

E = {49}

∴ It is not a null set.

(vi) F = {x : x ∈ N, 5 < x < 6}.

Natural numbers (N) are counting numbers (1, 2, 3, .....). There is no natural number between 5 and 6.

F = { }

∴ It is a null set.

Identify whether the given pair consists of equal or equivalent but not equal sets or none :

(i) A = set of letters of the word 'FLOWER'.
    B = set of letters of the word 'FOLLOWER'.

(ii) C = {x : x ∈ N, x + 5 = 6} and D = {x : x ∈ W, x < 1}.

(iii) E = set of first five whole numbers.
      F = set of first five natural numbers.

(iv) G = {a, b, c} and H = {x, y, z}.

(v) J = {x : x ∈ N, x ≠ x} and K = {x : x ∈ N, 6 < x < 7}.

Answer

(i)

A = Set of letters of the word 'FLOWER'.
B = Set of letters of the word 'FOLLOWER'.

A = {F, L, O, W, E, R} and B = {F, O, L, W, E, R}

In roster form, repeated letters in 'FOLLOWER' are listed only once.

Both sets contain the same elements {F, L, O, W, E, R}. These are equal sets.

∴ A = B

(ii) C = {x : x ∈ N, x + 5 = 6} and D = {x : x ∈ W, x < 1}.

For Set C:

x + 5 = 6

⇒ x = 6 - 5

⇒ x = 1

Since 1 is a natural number, x = {1}.

C = {1}, n(C) = 1

For Set D:

x < 1

The only whole number less than 1 is 0.

D = {0}, n(D) = 1

n(C) = 1 and n(D) = 1. They have the same number of elements but different elements.

They are equivalent but not equal sets

∴ C ↔ D

(iii)

E = set of first five whole numbers.
F = set of first five natural numbers.

E = {0, 1, 2, 3, 4} and F = {1, 2, 3, 4, 5}

E contains 0 to 4 (first five whole numbers). F contains 1 to 5 (first five natural numbers).

n(E) = 5 and n(F) = 5. The cardinal numbers are the same, but the elements are different.

They are equivalent but not equal sets

∴ E ↔ F

(iv) G = {a, b, c} and H = {x, y, z}.

Both sets have three distinct elements.

n(G) = 3 and n(H) = 3. The elements are entirely different.

They are equivalent but not equal sets

∴ G ↔ H

(v) J = {x : x ∈ N, x ≠ x} and K = {x : x ∈ N, 6 < x < 7}.

No natural number is unequal to itself. J = { } (Null set).

There is no natural number between 6 and 7. K = { } (Null set).

Both are empty sets and therefore contain exactly the same (zero) elements. These are equal sets.

∴ J = K

For each of the following pairs of sets, identify the disjoint and overlapping sets :

(i)

A = {x : x is a prime number, x < 8}.
B = {x : x is an even natural number, x < 8}.

(ii) C = {x : x ∈ N, x < 10} and D = {x : x ∈ N, x is a multiple of 5}.

(iii) E = {x : x = 4n, n ∈ N} and F = {x : x = 9n, n ∈ N}.

(iv) G = {x : x = 8n, n ∈ N and n < 7} and H = {x : x = 9n, n ∈ N and n < 7}.

Answer

(i)

A = {x : x is a prime number, x < 8}.
B = {x : x is an even natural number, x < 8}.

Roster Form of A = {2, 3, 5, 7}

Roster Form of B = {2, 4, 6}

Both sets share the element {2}.

∴ These are overlapping sets

(ii) C = {x : x ∈ N, x < 10} and D = {x : x ∈ N, x is a multiple of 5}.

Roster Form of C = {1, 2, 3, 4, 5, 6, 7, 8, 9}

Roster Form of D = {5, 10, 15, 20, $\dots$}

The number 5 is present in both sets.

∴ These are overlapping sets

(iii) E = {x : x = 4n, n ∈ N} and F = {x : x = 9n, n ∈ N}.

Roster Form of E = {4, 8, 12, 16, 20, 24, 28, 32, 36, ....}

Roster Form of F = {9, 18, 27, 36, 45, ....}

The first common multiple is 36. Since these are infinite sets of multiples, they will continue to have common elements (like 72, 108, etc.).

∴ These are overlapping sets

(iv) G = {x : x = 8n, n ∈ N and n < 7} and H = {x : x = 9n, n ∈ N and n < 7}.

Roster Form of G = {8, 16, 24, 32, 40, 48} (Multiples of 8 up to n=6)

Roster Form of H = {9, 18, 27, 36, 45, 54} (Multiples of 9 up to n=6)

There are no common elements in these finite lists. The first shared multiple of 8 and 9 is 72, which is outside the range of both sets.

∴ These are disjoint sets

State in each case, whether the given statement is true or false :

(i) If A is the set of all non-negative integers, then 0 ∈ A.

(ii) If B is the set of all consonants, then c ∈ B.

(iii) If C is the set of all prime numbers less than 80, then 57 ∈ C.

(iv) {x : x ∈ W, x + 5 = 5} is a singleton set.

(v) If D = {x : x ∈ W, x < 4}, then n(D) = 4.

(vi) {a, b, c, 1, 2, 3} is not a set.

(vii) {1, 2, 3, 1, 2, 3, 1, 2, 3,...............} is an infinite set.

(viii) 0 ∈ Φ.

(ix) {3, 5} ∈ (1, 3, 5, 7, 9).

Answer

(i) True
Reason — Non-negative integers include all natural numbers and zero (0, 1, 2, 3, ....). Therefore, 0 is an element of set A.

(ii) True
Reason — Consonants are all letters of the alphabet except for vowels (a, e, i, o, u). Since 'c' is not a vowel, it belongs to the set of consonants B.

(iii) False
Reason — A prime number has only two factors: 1 and itself. Since 57 is divisible by 3 (3 x 19 = 57), it is a composite number, not a prime number.

(iv) True
Reason — Solving x + 5 = 5 gives x = 0. Since 0 is a whole number (W), the set contains exactly one element: {0}. A set with one element is called a singleton set.

(v) True
Reason — The whole numbers (W) less than 4 are {0, 1, 2, 3}. Counting these elements gives a cardinal number n(D) = 4.

(vi) False
Reason — A set can contain any well-defined collection of distinct objects, including a mix of letters and numbers. Therefore, {a, b, c, 1, 2, 3} is a valid set.

(vii) False
Reason — In set theory, repeated elements are counted only once. The set {1, 2, 3, 1, 2, 3, ....} contains only the distinct elements {1, 2, 3}. Since it has a limited number of distinct elements, it is a finite set, not an infinite one.

(viii) False
Reason — The symbol Φ represents a null set, which by definition contains no elements at all. Therefore, 0 cannot be an element of Φ.

(ix) False
Reason — The notation {3, 5} represents a subset, not an element. The correct notation would be {3, 5} ⊂ {1, 3, 5, 7, 9} or 3, 5 ∈ {1, 3, 5, 7, 9}.

Indicate whether the given statement is true or false :

(i) {Triangles} ⊆ {Quadrilaterals}

(ii) {Squares} ⊆ {Rectangles}

(iii) {Rhombuses} ⊆ {Parallelograms}

(iv) {Natural numbers} ⊆ {Whole numbers}

(v) {Integers} ⊆ {Whole numbers}

(vi) {Composite numbers} ⊆ {Odd numbers}

Answer

(i) False
Reason — A triangle is a polygon with 3 sides, whereas a quadrilateral is a polygon with 4 sides. Since no triangle can be a quadrilateral, the set of triangles is not a subset of the set of quadrilaterals.

(ii) True
Reason — A square is defined as a special type of rectangle where all four sides are equal. Since every square satisfies the properties of a rectangle, the set of squares is a subset of the set of rectangles.

(iii) True
Reason — A rhombus is a quadrilateral with both pairs of opposite sides parallel and all sides equal. Since it satisfies the definition of a parallelogram (a quadrilateral with two pairs of parallel sides), the set of rhombuses is a subset of the set of parallelograms.

(iv) True
Reason — Natural numbers (N) are {1, 2, 3, ...} and whole numbers (W) are {0, 1, 2, 3, ....}. Since every natural number is also a whole number, {Natural numbers} ⊆ {Whole numbers}.

(v) False
Reason — Integers include both, negative and positive numbers (...., -2, -1, 0, 1, 2, ....), whereas whole numbers consist only of zero and positive counting numbers. Since negative integers are not whole numbers, the set of integers is not a subset of whole numbers.

(vi) False
Reason — Composite numbers are numbers with more than two factors, such as 4, 6, 8, 9, 10, ... . Many composite numbers (like 4, 6, and 8) are even, so the set of composite numbers is not a subset of the set of odd numbers.

Write down all possible subsets of each of the sets given below :

(i) {1}

(ii) {3, 4}

(iii) {2, 3, 5}

(iv) Φ

(v) {c, d, e}

(vi) {a, b, c, d}

Answer

(i) {1}

Subsets are: Φ, {1}

(ii) {3, 4}

Subsets are: Φ, {3}, {4}, {3, 4}

(iii) {2, 3, 5}

Subsets are: Φ, {2}, {3}, {5}, {2, 3}, {3, 5}, {2, 5}, {2, 3, 5}

(iv) Φ

The empty set has only itself as a subset.

Subsets are: Φ

(v) {c, d, e}

Subsets are: Φ, {c}, {d}, {e}, {c, d}, {d, e}, {c, e}, {c, d, e}

(vi) {a, b, c, d}

Subsets are: Φ, {a}, {b}, {c}, {d}, {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d}, {a, b, c}, {a, b, d}, {a, c, d}, {b, c, d}, {a, b, c, d}

Write down all possible proper subsets of each of the sets given below :

(i) {x}

(ii) {p, q}

(iii) {m, n, p}

(iv) {1, 2, 3, 4}

Answer

(i) {x}

Proper Subset is Φ

(ii) {p, q}

Proper Subsets are: Φ, {p}, {q}

(iii) {m, n, p}

Proper Subsets are: Φ, {m}, {n}, {p}, {m, n}, {n, p}, {m, p}

(iv) {1, 2, 3, 4}

Proper Subsets are: Φ, {1}, {2}, {3}, {4} {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4} {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}

Write down :

(i) The set C of letters of the word 'PAPAYA'.

(ii) All subsets of C.

(iii) All proper subsets of C.

Answer

(i) The set C of letters of the word 'PAPAYA'.

In roster form, every repeated element in a set is taken only once.

The distinct letters in the word 'PAPAYA' are P, A, and Y.

C = {P, A, Y}

(ii) All subsets of C.

The empty set (Φ) is a subset of every set.

Every set is a subset of itself.

Since set C has 3 elements, it will have 2³ = 8 subsets.

Subsets are: Φ, {P}, {A}, {Y}, {P, A}, {A, Y}, {P, Y}, {P, A, Y}

(iii) All proper subsets of C.

A proper subset includes all subsets of the set except for the set itself.

There are 2ⁿ - 1 proper subsets, which means 8 - 1 = 7 for this set.

Proper subsets are: Φ, {P}, {A}, {Y}, {P, A}, {A, Y}, {P, Y}

How many subsets in all are there of a set containing 4 elements?

Answer

Number of elements (n) = 4.

Formula for total number of subsets = 2ⁿ.

By replacing 'n' with 4, we get:

2⁴ = 2 x 2 x 2 x 2 = 16.

Hence, there are 16 subsets in all.

How many subsets in all are there of a set with cardinal number 6?

Answer

Cardinal number (n) = 6.

Formula for total number of subsets = 2ⁿ.

By replacing 'n' with 6, we get:

2⁶ = 2 x 2 x 2 x 2 x 2 x 2 = 64.

Hence, there are 64 subsets in all.

How many proper subsets in all are there of a set containing 3 elements?

Answer

Number of elements (n) = 3.

Formula for number of proper subsets = 2ⁿ - 1

By replacing 'n' with 3, we get:

2³ - 1 ⇒ 8 - 1 ⇒ 7

Hence, there are 7 proper subsets in all.

How many proper subsets in all are there of a set with cardinal number 5?

Answer

Cardinal number (n) = 5.

Formula for number of proper subsets = 2ⁿ - 1

By replacing 'n' with 5, we get:

2⁵ - 1 ⇒ 32 - 1 ⇒ 31

Hence, there are 31 proper subsets in all.

Which of the following statements are true ?

(i) {a} ⊂ {a, b, c}

(ii) {a} ⊂ {b, c, d, e}

(iii) Φ ⊂ {a, b, c}

(iv) Φ ∈ {a, b, c}

(v) 0 ∉ Φ

(vi) {1} ⊂ {0, 1}

(vii) Every subset of a finite set is finite.

(viii) Every subset of an infinite set is infinite.

Answer

(i) True
Reason — The element 'a' is present in the set {a, b, c}. Since {a} is a set containing an element from the second set, it is a proper subset.

(ii) False
Reason — For {a} to be a subset of {b, c, d, e}, the element 'a' must be present in the second set. Since it is not, the statement is false.

(iii) True
Reason — By definition, the empty set (Φ) is a subset of every set.

(iv) False
Reason — The symbol ∈ means "is an element of." Φ is a subset of {a, b, c}, not an element of it.

(v) True
Reason — The empty set (Φ) contains no elements at all. Therefore, it is correct to say that 0 is not an element of Φ.

(vi) True
Reason — The element 1 is present in the set {0, 1}, making {1} a proper subset.

(vii) True
Reason — A finite set has a specific number of elements. Any collection of elements taken from it will also have a specific, countable number of elements.

(viii) False
Reason — While an infinite set has endless elements, you can still pick a limited number of elements from it to form a subset. For example, {1, 2} is a finite subset of the infinite set of natural numbers {1, 2, 3, ....}.

Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.

Write the subset of A containing :

(i) all odd numbers

(ii) all prime numbers

(iii) all multiples of 4.

Answer

(i) all odd numbers

Given A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.

Odd numbers are: {1, 3, 5, 7, 9, 11}

(ii) all prime numbers

Prime numbers are natural numbers greater than 1 that have exactly two factors: 1 and the number itself. From set A, these are 2, 3, 5, 7, and 11.

All prime numbers are: {2, 3, 5, 7, 11}

(iii) all multiples of 4.

Multiples of 4 are numbers that can be divided by 4 without a remainder. Within the range of set A, these are 4 x 1, 4 x 2, and 4 x 3.

All multiples of 4 are: {4, 8, 12}

Let U = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16} be the universal set and let A = {5, 7, 11, 13}, B = {6, 8, 10, 12, 14, 16} and C = {5, 6, 8, 10, 11, 12} be its subsets.

Find:

(i) A'

(ii) B'

(iii) C'

Answer

(i) A'

Universal Set (U) = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}

Set A = {5, 7, 11, 13}

Remove 5, 7, 11, and 13 from U.

A' = {6, 8, 9, 10, 12, 14, 15, 16}

(ii) B'

Universal Set (U) = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}

B = {6, 8, 10, 12, 14, 16}

Remove 6, 8, 10, 12, 14, and 16 from U.

B' = {5, 7, 9, 11, 13, 15}

(iii) C'

Universal Set (U) = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}

C = {5, 6, 8, 10, 11, 12}

Remove 5, 6, 8, 10, 11, and 12 from U.

C' = {7, 9, 13, 14, 15, 16}

Let the set I of all integers be the universal set and let A = {x : x is a negative integer} be its subset. Find A'.

Answer

Universal Set (I) = The set of all integers, which includes negative integers, zero, and positive integers. I = {..., -2, -1, 0, 1, 2, ....}.

A = The set of all negative integers = {..., -3, -2, -1}.

Complement (A') = This set consists of all integers that are not negative. This includes zero and all positive integers (0, 1, 2, 3, ....).

∴ A' = {x : x is a non-negative integer}

Suggest a universal set for the sets given below :

(i) {5, 7, 9}, {3, 5, 7}, {1, 3, 9} and {2, 4, 8}.

(ii) {odd numbers less than 8}, {prime numbers less than 8} and {even numbers between 3 and 8}.

(iii) {vowels in English alphabet}, {consonants in English alphabet}.

Answer

(i) {5, 7, 9}, {3, 5, 7}, {1, 3, 9} and {2, 4, 8}.

Subsets = {5, 7, 9}, {3, 5, 7}, {1, 3, 9} and {2, 4, 8}.

The elements present across all sets are 1, 2, 3, 4, 5, 7, 8, and 9.

∴ Universal Set U = {1, 2, 3, 4, 5, 6, 7, 8, 9}

(ii) {odd numbers less than 8}, {prime numbers less than 8} and {even numbers between 3 and 8}.

Subsets = {odd numbers less than 8}, {prime numbers less than 8} and {even numbers between 3 and 8}

Odd numbers less than 8 = {1, 3, 5, 7}

Prime numbers less than 8 = {2, 3, 5, 7}

Even numbers between 3 and 8 = {4, 6}

The elements present across all sets are 1, 2, 3, 4, 5, 6, 7.

∴ Universal Set U = {1, 2, 3, 4, 5, 6, 7}

(iii) {vowels in English alphabet}, {consonants in English alphabet}.

Subsets = {vowels in English alphabet}, {consonants in English alphabet}.

The first set contains {a, e, i, o, u} and the second contains all other letters of the alphabet.

∴ Universal Set U = {x : x is a letter in English alphabet}.

Let A = {2, 4, 6, 8}, B = {6, 8, 10, 12} and C = {7, 8, 9, 10}. Find :

(i) A ∪ B

(ii) A ∪ C

(iii) B ∪ C

(iv) A ∩ B

(v) A ∩ C

(vi) B ∩ C

Answer

(i) A ∪ B

List all elements from both A and B, avoiding duplicates.

We have:

A = {2, 4, 6, 8}

B = {6, 8, 10, 12}

A ∪ B = {2, 4, 6, 8} ∪ {6, 8, 10, 12}

∴ A ∪ B = {2, 4, 6, 8, 10, 12}

(ii) A ∪ C

List all elements from both A and C, avoiding duplicates.

We have:

A = {2, 4, 6, 8}

C = {7, 8, 9, 10}

A ∪ C = {2, 4, 6, 8} ∪ {7, 8, 9, 10}

∴ A ∪ C = {2, 4, 6, 7, 8, 9, 10}

(iii) B ∪ C

List all elements from both B and C, avoiding duplicates.

We have:

B = {6, 8, 10, 12}

C = {7, 8, 9, 10}

B ∪ C = {6, 8, 10, 12} ∪ {7, 8, 9, 10}

∴ B ∪ C = {6, 7, 8, 9, 10, 12}

(iv) A ∩ B

List all elements common to both A and B.

We have:

A = {2, 4, 6, 8}

B = {6, 8, 10, 12}

A ∩ B = {2, 4, 6, 8} ∩ {6, 8, 10, 12}

∴ A ∩ B = {6, 8}

(v) A ∩ C

List all elements common to both A and C.

We have:

A = {2, 4, 6, 8}

C = {7, 8, 9, 10}

A ∩ C = {2, 4, 6, 8} ∩ {7, 8, 9, 10}

∴ A ∩ C = {8}

(vi) B ∩ C

List all elements common to both B and C.

We have:

B = {6, 8, 10, 12}

C = {7, 8, 9, 10}

B ∩ C = {6, 8, 10, 12} ∩ {7, 8, 9, 10}

∴ B ∩ C = {8, 10}

Let P = {x : x is a factor of 18} and Q = {x : x is a factor of 24}.

(i) Write each one of P and Q in Roster form.

(ii) Find:

(a) P ∪ Q

(b) P ∩ Q

Answer

(i) Write each one of P and Q in Roster form.

P = {x : x is a factor of 18}

Factors of 18 are numbers that divide 18 exactly: 1, 2, 3, 6, 9, 18.

P = {1, 2, 3, 6, 9, 18}

And

Q = {x : x is a factor of 24}

Factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24.

Q = {1, 2, 3, 4, 6, 8, 12, 24}

∴ P = {1, 2, 3, 6, 9, 18}, Q = {1, 2, 3, 4, 6, 8, 12, 24}

(ii) Find:

(a) P ∪ Q

P = {1, 2, 3, 6, 9, 18}

Q = {1, 2, 3, 4, 6, 8, 12, 24}

P ∪ Q = {1, 2, 3, 6, 9, 18} ∪ {1, 2, 3, 4, 6, 8, 12, 24}

∴ P ∪ Q = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24}

(b) P ∩ Q

P = {1, 2, 3, 6, 9, 18}

Q = {1, 2, 3, 4, 6, 8, 12, 24}

P ∩ Q = {1, 2, 3, 6, 9, 18} ∩ {1, 2, 3, 4, 6, 8, 12, 24}

∴ P ∩ Q = {1, 2, 3, 6}

Let A = {a, b, c}, B = {b, d, e} and C = {e, f, g}, verify that :

(i) A ∪ B = B ∪ A

(ii) (A ∪ B) ∪ C = A ∪ (B ∪ C)

(iii) A ∩ B = B ∩ A

(iv) (A ∩ B) ∩ C = A ∩ (B ∩ C)

Answer

(i) We have:

A = {a, b, c}

B = {b, d, e}

A ∪ B = {a, b, c} ∪ {b, d, e} = {a, b, c, d, e} $\quad$.....(1)

B ∪ A = {b, d, e} ∪ {a, b, c} = {a, b, c, d, e} $\quad$......(2)

Since, (1) and (2) are equal,

∴ A ∪ B = B ∪ A

(ii) (A ∪ B) ∪ C = A ∪ (B ∪ C)

We have:

A = {a, b, c}

B = {b, d, e}

C = {e, f, g}

A ∪ B = {a, b, c} ∪ {b, d, e} = {a, b, c, d, e}

(A ∪ B) ∪ C = {a, b, c, d, e} ∪ {e, f, g} = {a, b, c, d, e, f, g} $\quad$.....(1)

Again, B ∪ C = {b, d, e} ∪ {e, f, g} = {b, d, e, f, g}

A ∪ (B ∪ C) = {a, b, c} ∪ {b, d, e, f, g} = {a, b, c, d, e, f, g} $\quad$......(2)

Since, (1) and (2) are equal,

∴ (A ∪ B) ∪ C = A ∪ (B ∪ C)

(iii) A ∩ B = B ∩ A

We have:

A = {a, b, c}

B = {b, d, e}

A ∩ B = {a, b, c} ∩ {b, d, e} = {b} $\quad$.....(1)

B ∩ A = {b, d, e} ∩ {a, b, c} = {b} $\quad$......(2)

Since, (1) and (2) are equal,

∴ A ∩ B = B ∩ A

(iv) (A ∩ B) ∩ C = A ∩ (B ∩ C)

We have:

A = {a, b, c}

B = {b, d, e}

C = {e, f, g}

A ∩ B = {a, b, c} ∩ {b, d, e} = {b}

(A ∩ B) ∩ C = {b} ∩ {e, f, g} = { } or ϕ $\quad$.....(1)

B ∩ C = {b, d, e} ∩ {e, f, g} = {e}

A ∩ (B ∩ C) = {a, b, c} ∩ {e} = { } or ϕ $\quad$......(2)

Since, (1) and (2) are equal,

∴ (A ∩ B) ∩ C = A ∩ (B ∩ C)

Let A = {x : x is a multiple of 2, x < 15}, B = {x : x is a multiple of 3, x < 20}, C = {x : x is a prime, x < 20}.

(i) Write each one of the sets A, B, C in Roster form.

(ii) Find:

(a) A ∪ B

(b) A ∪ C

(c) B ∪ C

(d) A ∩ B

(e) A ∩ C

(f) B ∩ C

Answer

(i) Write each one of the sets A, B, C in Roster form.

Given:

A = {x : x is a multiple of 2, x < 15}

Multiples of 2 less than 15 are: 2, 4, 6, 8, 10, 12, 14.

A = {2, 4, 6, 8, 10, 12, 14}

B = {x : x is a multiple of 3, x < 20}

Multiples of 3 less than 20 are: 3, 6, 9, 12, 15, 18.

B = {3, 6, 9, 12, 15, 18}

C = {x : x is a prime, x < 20}

Prime numbers between 1 and 20 are: 2, 3, 5, 7, 11, 13, 17, 19.

C = {2, 3, 5, 7, 11, 13, 17, 19}

∴ A = {2, 4, 6, 8, 10, 12, 14}, B = {3, 6, 9, 12, 15, 18} and C = {2, 3, 5, 7, 11, 13, 17, 19}

(ii) Find:

(a) A ∪ B

Given:

A = {2, 4, 6, 8, 10, 12, 14}

B = {3, 6, 9, 12, 15, 18}

A ∪ B = {2, 4, 6, 8, 10, 12, 14} ∪ {3, 6, 9, 12, 15, 18} = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 18}

∴ A ∪ B = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 18}

(b) A ∪ C

Given:

A = {2, 4, 6, 8, 10, 12, 14}

C = {2, 3, 5, 7, 11, 13, 17, 19}

A ∪ C = {2, 4, 6, 8, 10, 12, 14} ∪ {2, 3, 5, 7, 11, 13, 17, 19} = {2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 17, 19}

∴ A ∪ C = {2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 17, 19}

(c) B ∪ C

Given:

B = {3, 6, 9, 12, 15, 18}

C = {2, 3, 5, 7, 11, 13, 17, 19}

B ∪ C = {3, 6, 9, 12, 15, 18} ∪ {2, 3, 5, 7, 11, 13, 17, 19} = {2, 3, 5, 6, 7, 9, 11, 12, 13, 15, 17, 18, 19}

∴ B ∪ C = {2, 3, 5, 6, 7, 9, 11, 12, 13, 15, 17, 18, 19}

(d) A ∩ B

Given:

A = {2, 4, 6, 8, 10, 12, 14}

B = {3, 6, 9, 12, 15, 18}

A ∩ B = {2, 4, 6, 8, 10, 12, 14} ∩ {3, 6, 9, 12, 15, 18} = {6, 12}

∴ A ∩ B = {6, 12}

(e) A ∩ C

Given:

A = {2, 4, 6, 8, 10, 12, 14}

C = {2, 3, 5, 7, 11, 13, 17, 19}

A ∩ C = {2, 4, 6, 8, 10, 12, 14} ∩ {2, 3, 5, 7, 11, 13, 17, 19} = {2}

∴ A ∩ C = {2}

(f) B ∩ C

Given:

B = {3, 6, 9, 12, 15, 18}

C = {2, 3, 5, 7, 11, 13, 17, 19}

B ∩ C = {3, 6, 9, 12, 15, 18} ∩ {2, 3, 5, 7, 11, 13, 17, 19} = {3}

∴ B ∩ C = {3}

Let A = {b, d, e, f}, B = {c, d, g, h} and C = {e, f, g, h}. Find :

(i) A - B

(ii) B - C

(iii) C - A

(iv) (A - B) ∪ (B - A)

(v) (B - C) ∪ (C - B)

Answer

(i) A - B

Given:

A = {b, d, e, f}

B = {c, d, g, h}

A - B = Elements of A which are not in B

A - B = {b, d, e, f} - {c, d, g, h} = {b, e, f}

∴ A - B = {b, e, f}

(ii) B - C

Given:

B = {c, d, g, h}

C = {e, f, g, h}

B - C = Elements of B which are not in C.

B - C = {c, d, g, h} - {e, f, g, h} = {c, d}

∴ B - C = {c, d}

(iii) C - A

Given:

C = {e, f, g, h}

A = {b, d, e, f}

C - A = Elements of C which are not in A.

C - A = {e, f, g, h} - {b, d, e, f} = {g, h}

∴ C - A = {g, h}

(iv) (A - B) ∪ (B - A)

Given:

A = {b, d, e, f}

B = {c, d, g, h}

(A - B) = {b, d, e, f} - {c, d, g, h} = {b, e, f}

(B - A) = {c, d, g, h} - {b, d, e, f} = {c, g, h}

(A - B) ∪ (B - A) = {b, e, f} ∪ {c, g, h} = {b, c, e, f, g, h}

∴ (A - B) ∪ (B - A) = {b, c, e, f, g, h}

(v) (B - C) ∪ (C - B)

Given:

B = {c, d, g, h}

C = {e, f, g, h}

B - C = {c, d, g, h} - {e, f, g, h} = {c, d}

(C - B) = {e, f, g, h} - {c, d, g, h} = {e, f}

(B - C) ∪ (C - B) = {c, d} ∪ {e, f} = {c, d, e, f}

∴ (B - C) ∪ (C - B) = {c, d, e, f}

Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} be the universal set and let A = {2, 3, 4, 5, 6} and B = {3, 5, 7, 8} be its subsets.

Find:

(i) A'

(ii) B'

(iii) A' ∩ B'

(iv) A' ∪ B'

Verify that:

(v) (A ∪ B)' = (A' ∩ B')

(vi) (A ∩ B)' = (A' ∪ B')

Answer

Given:

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {2, 3, 4, 5, 6}

B = {3, 5, 7, 8}

(i) A'

A' = Elements in U which are not in A.

A' = U - A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {2, 3, 4, 5, 6} = {1, 7, 8, 9, 10}

∴ A' = {1, 7, 8, 9, 10}

(ii) B'

B' = Elements in U which are not in B.

B' = U - B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {3, 5, 7, 8} = {1, 2, 4, 6, 9, 10}

∴ B' = {1, 2, 4, 6, 9, 10}

(iii) A' ∩ B'

We have:

A' = {1, 7, 8, 9, 10}

B' = {1, 2, 4, 6, 9, 10}

A' ∩ B' = {1, 7, 8, 9, 10} ∩ {1, 2, 4, 6, 9, 10} = {1, 9, 10}

∴ A' ∩ B' = {1, 9, 10}

(iv) A' ∪ B'

We have:

A' = {1, 7, 8, 9, 10}

B' = {1, 2, 4, 6, 9, 10}

A' ∪ B' = {1, 7, 8, 9, 10} ∪ {1, 2, 4, 6, 9, 10} = {1, 2, 4, 6, 7, 8, 9, 10}

∴ A' ∪ B' = {1, 2, 4, 6, 7, 8, 9, 10}

Verify

(v) (A ∪ B)' = (A' ∩ B')

We have:

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {2, 3, 4, 5, 6}

B = {3, 5, 7, 8}

A' = {1, 7, 8, 9, 10}

B' = {1, 2, 4, 6, 9, 10}

First let us find A ∪ B:

A ∪ B = {2, 3, 4, 5, 6} ∪ {3, 5, 7, 8} = {2, 3, 4, 5, 6, 7, 8}

(A ∪ B)' = Elements in U which are not in (A ∪ B).

LHS = (A ∪ B)' = U - (A ∪ B)

LHS = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {2, 3, 4, 5, 6, 7, 8}

LHS = {1, 9, 10}

RHS = (A' ∩ B') = {1, 7, 8, 9, 10} ∩ {1, 2, 4, 6, 9, 10}

RHS = {1, 9, 10}

Since LHS = RHS,

∴ The statement (A ∪ B)' = (A' ∩ B') is verified.

(vi) (A ∩ B)' = (A' ∪ B')

Given:

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {2, 3, 4, 5, 6}

B = {3, 5, 7, 8}

A' = {1, 7, 8, 9, 10}

B' = {1, 2, 4, 6, 9, 10}

First let us find A ∩ B:

A ∩ B = {2, 3, 4, 5, 6} ∩ {3, 5, 7, 8} = {3, 5}

(A ∩ B)' = Elements in U which are not in (A ∩ B).

LHS = (A ∩ B)' = U - (A ∩ B)

LHS = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {3, 5}

LHS = {1, 2, 4, 6, 7, 8, 9, 10}

RHS = (A' ∪ B') = {1, 7, 8, 9, 10} ∪ {1, 2, 4, 6, 9, 10}

RHS = {1, 2, 4, 6, 7, 8, 9, 10}

Since LHS = RHS,

∴ The statement (A ∩ B)' = (A' ∪ B') is verified.

Let U = {a, b, c, d, e, f, g} be the universal set and let its subsets be A = {a, b, d, e} and B = {b, e, g}.

Verify that:

(i) (A ∪ B)' = (A' ∩ B')

(ii) (A ∩ B)' = (A' ∪ B')

Answer

Given:

Universal set U = {a, b, c, d, e, f, g}

Subset A = {a, b, d, e}

Subset B = {b, e, g}

(i) (A ∪ B)' = (A' ∩ B')

First let us find A ∪ B:

A ∪ B = {a, b, d, e} ∪ {b, e, g} = {a, b, d, e, g}

(A ∪ B)' = Elements in U which are not in (A ∪ B).

LHS = (A ∪ B)' = U - (A ∪ B)

LHS = {a, b, c, d, e, f, g} - {a, b, d, e, g}

LHS = {c, f}

Now, find A' and B':

A' = U - A

A' = {a, b, c, d, e, f, g} - {a, b, d, e} = {c, f, g}

B' = U - B

B' = {a, b, c, d, e, f, g} - {b, e, g} = {a, c, d, f}

RHS = (A' ∩ B') = {c, f, g} ∩ {a, c, d, f}

RHS = {c, f}

Since LHS = RHS,

∴ The statement (A ∪ B)' = (A' ∩ B') is verified.

(ii) (A ∩ B)' = (A' ∪ B')

First let us find A ∩ B:

A ∩ B = {a, b, d, e} ∩ {b, e, g} = {b, e}

LHS = (A ∩ B)' = U - (A ∩ B).

LHS = (A ∩ B)' = {a, b, c, d, e, f, g} - {b, e}

LHS = {a, c, d, f, g}

Now, find A' and B':

A' = U - A

A' = {a, b, c, d, e, f, g} - {a, b, d, e} = {c, f, g}

B' = U - B

B' = {a, b, c, d, e, f, g} - {b, e, g} = {a, c, d, f}

RHS = (A' ∪ B') = {c, f, g} ∪ {a, c, d, f}

RHS = {a, c, d, f, g}

Since LHS = RHS,

∴ The statement (A ∩ B)' = (A' ∪ B') is verified.

Let U = {3, 6, 9, 12, 15, 18, 21, 24} be the universal set and let A = {6, 12, 18, 24} be its subset.

Verify that:

(i) A ∪ A = A

(ii) A ∩ A = A

(iii) A ∩ A' = Φ

(iv) A ∪ A' = U

(v) (A')' = A

Answer

Given:

Universal set U = {3, 6, 9, 12, 15, 18, 21, 24}

Subset A = {6, 12, 18, 24}

(i) A ∪ A = A

LHS = A ∪ A = {6, 12, 18, 24} ∪ {6, 12, 18, 24}

LHS = {6, 12, 18, 24} $\quad$[since repeated elements are written once]

RHS = A = {6, 12, 18, 24}

Since LHS = RHS,

∴ The statement A ∪ A = A is verified.

(ii) A ∩ A = A

LHS = A ∩ A = {6, 12, 18, 24} ∩ {6, 12, 18, 24}

LHS = {6, 12, 18, 24}

RHS = A = {6, 12, 18, 24}

Since LHS = RHS,

∴ The statement A ∩ A = A is verified.

(iii) A ∩ A' = Φ

A' = U - A = {3, 6, 9, 12, 15, 18, 21, 24} - {6, 12, 18, 24} = {3, 9, 15, 21}

LHS = A ∩ A' = {6, 12, 18, 24} ∩ {3, 9, 15, 21}

LHS = Φ

RHS = Φ

Since LHS = RHS,

∴ The statement A ∩ A' = Φ is verified.

(iv) A ∪ A' = U

A' = {3, 9, 15, 21} $\quad$[From previous step]

LHS = A ∪ A' = {6, 12, 18, 24} ∪ {3, 9, 15, 21}

LHS = {3, 6, 9, 12, 15, 18, 21, 24}

RHS = U = {3, 6, 9, 12, 15, 18, 21, 24}

Since LHS = RHS,

∴ The statement A ∪ A' = U is verified.

(v) (A')' = A

We know A' = {3, 9, 15, 21}

LHS = (A')' = U - A' = {3, 6, 9, 12, 15, 18, 21, 24} - {3, 9, 15, 21}

LHS = {6, 12, 18, 24}

RHS = A = {6, 12, 18, 24}

Since LHS = RHS,

∴ The statement (A')' = A is verified.

Let A and B be two sets such that n(A) = 52, n(B) = 60 and n(A ∩ B) = 16. Draw a Venn diagram and find :

(i) n(A ∪ B)

(ii) n(A - B)

(iii) n(B - A)

Answer

Given:

n(A) = 52

n(B) = 60

n(A ∩ B) = 16

(i) n(A ∪ B)

The number of elements in the union of two sets is found using the formula:

n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

Substituting the values in above, we get:

n(A ∪ B) = 52 + 60 - 16

n(A ∪ B) = 112 - 16

∴ n(A ∪ B) = 96

(ii) n(A - B)

The number of elements that belong to A but not to B is found by subtracting the intersection from n(A):

n(A - B) = n(A) - n(A ∩ B)

Substituting the values in above, we get:

n(A - B) = 52 - 16

∴ n(A - B) = 36

(iii) n(B - A)

The number of elements that belong to B but not to A is found by subtracting the intersection from n(B):

n(B - A) = n(B) - n(A ∩ B)

Substituting the values in above, we get:

n(B - A) = 60 - 16

∴ n(B - A) = 44

Let P and Q be two sets such that n(P ∪ Q) = 70, n(P) = 45 and n(Q) = 38. Draw a Venn diagram and find :

(i) n(P ∩ Q)

(ii) n(P - Q)

(iii) n(Q - P)

Answer

Given:

n(P ∪ Q) = 70

n(P) = 45

n(Q) = 38

(i) n(P ∩ Q)

To find the number of elements in the intersection, we use the formula:

n(P ∪ Q) = n(P) + n(Q) - n(P ∩ Q)

Rearranging to solve for the intersection:

n(P ∩ Q) = n(P) + n(Q) - n(P ∪ Q)

Substituting the values in above, we get:

n(P ∩ Q) = 45 + 38 - 70

n(P ∩ Q) = 83 - 70

∴ n(P ∩ Q) = 13

(ii) n(P - Q)

This represents elements that are in set P but not in set Q.

We use the formula:

n(P - Q) = n(P) - n(P ∩ Q)

Substituting the values in above, we get:

n(P - Q) = 45 - 13

∴ n(P - Q) = 32

(iii) n(Q - P)

This represents elements that are in set Q but not in set P.

We use the formula:

n(Q - P) = n(Q) - n(P ∩ Q)

Substituting the values in above, we get:

n(Q - P) = 38 - 13

∴ n(Q - P) = 25

In a city, there are 25 Hindi medium schools, 18 English medium schools and 7 schools have both the mediums. Find

(i) how many schools are there in all in the city ;

(ii) how many schools have Hindi medium only ;

(iii) how many schools have English medium only.

Answer

Given:

Total Hindi medium schools: n(H) = 25

Total English medium schools: n(E) = 18

Schools with both mediums: n(H ∩ E) = 7

(i) how many schools are there in all in the city

This represents the union of the two sets, n(H ∪ E).

We know the formula:

n(H ∪ E) = n(H) + n(E) - n(H ∩ E)

Substituting the values in above, we get:

n(H ∪ E) = 25 + 18 - 7

n(H ∪ E) = 43 - 7

n(H ∪ E) = 36

∴ There are 36 schools in all in the city.

(ii) how many schools have Hindi medium only

This represents the set H - E, consisting of schools that are Hindi medium but not English medium.

We use the formula:

n(H - E) = n(H) - n(H ∩ E)

Substituting the values in above, we get:

n(H - E) = 25 - 7

n(H - E) = 18

∴ 18 schools have Hindi medium only.

(iii) how many schools have English medium only.

This represents the set E - H, consisting of schools that are English medium but not Hindi medium.

We use the formula:

n(E - H) = n(E) - n(H ∩ E)

Substituting the values in above, we get:

n(E - H) = 18 - 7

n(E - H) = 11

∴ 11 schools have English medium only.

There is a group of 50 persons who can speak English or Tamil or both. Out of these persons, 37 can speak English and 30 can speak Tamil.

(i) How many can speak both English and Tamil?

(ii) How many can speak English only?

(iii) How many can speak Tamil only?

Answer

Given:

Total persons in the group: n(E ∪ T) = 50

Persons who can speak English: n(E) = 37

Persons who can speak Tamil: n(T) = 30

(i) How many can speak both English and Tamil?

This represents the intersection of the two sets, n(E ∩ T).

We use the formula:

n(E ∩ T) = n(E) + n(T) - n(E ∪ T)

Substituting the values in above, we get:

n(E ∩ T) = 37 + 30 - 50

n(E ∩ T) = 67 - 50

n(E ∩ T) = 17

∴ 17 persons can speak both English and Tamil.

(ii) How many can speak English only?

This represents the set E - T, consisting of people who speak English but not Tamil.

We use the formula:

n(E - T) = n(E) - n(E ∩ T)

Substituting the values in above, we get:

n(E - T) = 37 - 17

n(E - T) = 20

∴ 20 persons can speak English only.

(iii) How many can speak Tamil only?

This represents the set T - E, consisting of people who speak Tamil but not English.

We use the formula:

n(T - E) = n(T) - n(E ∩ T)

Substituting the values in above, we get:

n(T - E) = 30 - 17

n(T - E) = 13

∴ 13 persons can speak Tamil only.

In a class of 40 students, each one plays either Tennis or Badminton or both. If 28 play Tennis and 26 play Badminton, find

(i) how many play both the games;

(ii) how many play Tennis only;

(iii) how many play Badminton only.

Answer

Given:

Total students in the class: n(T ∪ B) = 40

Students who play Tennis: n(T) = 28

Students who play Badminton: n(B) = 26

(i) how many play both the games

This represents the intersection of the two sets, n(T ∩ B).

We use the formula:

n(T ∩ B) = n(T) + n(B) - n(T ∪ B)

Substituting the values in above, we get:

n(T ∩ B) = 28 + 26 - 40

n(T ∩ B) = 54 - 40

n(T ∩ B) = 14

∴ 14 students play both the games.

(ii) how many play Tennis only

This represents the set T - B, consisting of students who play Tennis but do not play Badminton.

We use the formula:

n(T - B) = n(T) - n(T ∩ B)

Substituting the values in above, we get:

n(T - B) = 28 - 14

n(T - B) = 14

∴ 14 students play Tennis only.

(iii) how many play Badminton only

This represents the set B - T, consisting of students who play Badminton but do not play Tennis.

We use the formula:

n(B - T) = n(B) - n(T ∩ B)

Substituting the values in above, we get:

n(B - T) = 26 - 14

n(B - T) = 12

∴ 12 students play Badminton only.

The Venn diagram is shown below:

In a class of 45 pupils, 21 play chess, 23 play cards and 5 play both the games. Find

(i) how many do not play any of the games;

(ii) how many play chess only;

(iii) how many play cards only.

Answer

Total pupils in the class: n(U) = 45

Pupils who play chess: n(C) = 21

Pupils who play cards: n(D) = 23

Pupils who play both games: n(C ∩ D) = 5

(i) how many do not play any of the games

First find n(C ∪ D):

n(C ∪ D) = n(C) + n (D) - n(C ∩ D)

Substituting the values in above, we get:

n(C ∪ D) = 21 + 23 - 5

n(C ∪ D) = 44 - 5

n(C ∪ D) = 39

∴ 39 pupils play both the games.

Pupils who do not play any of the games = n(U) - n(C ∪ D)

Substituting the values in above, we get:

Pupils who do not play any of the games = 45 - 39 = 6

∴ Number of pupils who do not play any of the games = 6.

(ii) how many play chess only

This represents the set C - D, consisting of pupils who play chess but not cards.

We use the formula:

n(C - D) = n(C) - n(C ∩ D)

Substituting the values in above, we get:

n(C - D) = 21 - 5

n(C - D) = 16

∴ 16 pupils play chess only.

(iii) how many play cards only

This represents the set D - C, consisting of pupils who play cards but not chess.

We use the formula:

n(D - C) = n(D) - n(C ∩ D)

Substituting the values in above, we get:

n(D - C) = 23 - 5

n(D - C) = 18

∴ 18 pupils play cards only.

In a group of 36 girls, each one can either stitch or weave or can do both. If 25 girls can stitch and 17 can stitch only, how many can weave only?

Answer

Given:

Total number of girls: n(S ∪ W) = 36

Girls who can stitch: n(S) = 25

Girls who can stitch only: n(S - W) = 17

Girls who can weave only: n(W - S) = ?

First find how many can do both, we use the formula:

n(S - W) = n(S) - n(S ∩ W)

Substituting the values in above, we get:

17 = 25 - n(S ∩ W)

⇒ n(S ∩ W) = 25 - 17

⇒ n(S ∩ W) = 8

So, 8 girls can do both.

Since every girl in the group of 36 does at least one activity, the total is the sum of "stitch only," "weave only," and "both."

n(S ∪ W) = n(S - W) + n(W - S) + n(S ∩ W)

n(W - S) = n(S ∪ W) - n(S - W) - n(S ∩ W) $\quad$[Solving for n(W - S)]

Substituting the values in above, we get:

n(W - S) = 36 - 17 - 8

n(W - S) = 36 - 25

n(W - S) = 11

∴ 11 girls can weave only.

The Venn diagram is shown below:

In a group of 24 children, each one plays cricket or hockey or both. If 16 play cricket and 12 play cricket only, find how many play hockey only.

Answer

Given:

Total number of children: n(C ∪ H) = 24

Children who play cricket : n(C) = 16

Children who play cricket only: n(C - H) = 12

Children who play hockey only: n(W - C) = ?

First find how many children play both using the formula:

n(C ∩ H) = n(C) - n(C - H)

Substituting the values in above, we get:

n(C ∩ H) = 16 - 12

n(C ∩ H) = 4

So, 4 children play both cricket and hockey.

Since every child in the group of 24 plays at least one game, the total is the sum of “cricket only,” “hockey only,” and “both.”

n(C ∪ H) = n(C - H) + n(H - C) + n(C ∩ H)

n(H - C) = n(C ∪ H) - n(C - H) - n(C ∩ H) $\quad$[Solving for n(H - C)]

Substituting the values in above, we get:

n(H - C) = 24 - 12 - 4

n(H - C) = 24 - 16

n(H - C) = 8

∴ Number of children who play hockey only = 8.

The Venn diagram is shown below:

In a group of 40 persons, 10 drink tea but not coffee and 26 drink tea. How many drink coffee but not tea?

Answer

Given:

Total number of persons: n(T ∪ C) = 40

Persons who drink tea: n(T) = 26

Persons who drink tea but not coffee: n(T - C) = 10

Persons who drink coffee only = n(C - T) = ?

First find number of persons who drink both tea and coffee by using the formula:

n(T ∩ C) = n(T) - n(T - C)

Substituting the values in above, we get:

n(T ∩ C) = 26 - 10

n(T ∩ C) = 16

So, 16 persons drink both tea and coffee.

Since every person in the group of 40 drinks at least one of the two beverages, the total is the sum of “tea only,” “coffee only,” and “both.”

n(T ∪ C) = n(T - C) + n(C - T) + n(T ∩ C)

n(C - T) = n(T ∪ C) - n(T - C) - n(T ∩ C) $\quad$[Solving for n(C - T)]

Substituting the values in above, we get:

n(C - T) = 40 - 10 - 16

n(C - T) = 40 - 26

n(C - T) = 14

∴ Number of persons who drink coffee but not tea = 14.

The Venn diagram is shown below:

All the people in a locality read the daily newspaper Indian Express or Hindustan Times or both. If 120 read Indian Express and 150 read Hindustan Times and 36 read both, find :

(i) how many people are there in the locality;

(ii) how many people read only Indian Express.

Answer

Given:

People who read Indian Express: n(I) = 120

People who read Hindustan Times: n(H) = 150

People who read both: n(I ∩ H) = 36

(i) how many people are there in the locality

Since every person in the locality reads at least one of the two papers, the total population is equal to the union of the two sets, n(I ∪ H).

n(I ∪ H) = n(I) + n(H) − n(I ∩ H)

Substituting the values in above, we get:

n(I ∪ H) = 120 + 150 - 36

n(I ∪ H) = 270 - 36

n(I ∪ H) = 234

∴ Total people in the locality = 234.

(ii) how many people read only Indian Express

This represents the set I - H, consisting of people who read Indian Express but do not read Hindustan Times.

n(I - H) = n(I) - n(I ∩ H)

Substituting the values in above, we get:

n(I - H) = 120 - 36

n(I - H) = 84

∴ People who read only Indian Express = 84.

The Venn diagram is shown below:

Which of the following is a finite set?

1. Set of all natural numbers greater than 500.
2. Set of all integers less than 5.
3. Set of all even prime numbers.
4. Set of all multiples of 3.

Answer

The only even prime number is 2. Therefore, the set is {2}, which has a countable (finite) number of elements. All other options describe infinite sets (natural numbers > 500, integers < 5, and multiples of 3).

Hence, option 3 is the correct option.

Which of the following is a pair of disjoint sets?

1. A = {even natural numbers} and B = {prime numbers}
2. C = {multiples of 2} and D = {multiples of 3}
3. E = {factors of 24} and F = {factors of 9}
4. G = {letters in the word 'RHYTHM'} and H = {Vowels in English Alphabet}

Answer

Disjoint sets have no common elements. The set G = {R, H, Y, T, M} and H = {a, e, i, o, u} share nothing. In all other options, there is at least one common element (e.g., 2 is both even and prime).

Hence, option 4 is the correct option.

The cardinal number of the set A, defined as A = Set of all prime numbers less than 100, is

1. -20
2. 22
3. 24
4. 25

Answer

The prime numbers less than 100 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. Counting them gives a total of 25.

Hence, option 4 is the correct option.

The number of proper subsets of the set {α, β, Φ, θ, η} is

1. 24
2. 25
3. 31
4. 32

Answer

The number of proper subsets for a set with n elements is 2^n-1.
Here, n=5,
so 2^5-1 = 2⁴ = 32 - 1 = 31

Hence, option 3 is the correct option.

The sets {a, b, c} and {x, y, z} are

1. overlapping sets
2. equal sets
3. singleton sets
4. equivalent sets

Answer

Equivalent sets have the same number of elements (the same cardinal number). Both sets have exactly 3 elements (n=3). They are not "equal" because the elements themselves are different.

Hence, option 4 is the correct option.

A = {x : x is a prime number} and B = {x : x is an even natural number}. Then A ∩ B is

1. an empty set
2. a singleton set
3. an infinite set
4. equal to A

Answer

A ∩ B represents numbers that are both prime and even natural numbers. The only number that fits this is 2. Since the resulting set is {2}, it contains exactly one element, making it a singleton set.

Hence, option 2 is the correct option.

If A = {x : x ∈ N, 5 ≤ x < 10} and B = {1, 2, 3, 4, 5}, then n(A - B) is equal to

1. 1
2. 2
3. 4
4. 5

Answer

First, write the sets in roster form:

A = {5, 6, 7, 8, 9}

B = {1, 2, 3, 4, 5}

A - B = {5, 6, 7, 8, 9} - {1, 2, 3, 4, 5} = {6, 7, 8, 9} $\quad$(Elements in A not in B).

The count n(A - B) = 4.

Hence, option 3 is the correct option.

If X = {x : x is a multiple of 5} and Y = {x : x is a multiple of 7}, then X ∩ Y is

1. Φ
2. {x : x is a multiple of 2}
3. {5, 7}
4. {x : x is a multiple of 35}

Answer

The intersection of multiples of two numbers is the set of multiples of their Least Common Multiple (LCM). Since 5 and 7 are co-prime, their LCM is 35.

Hence, option 4 is the correct option.

{Φ} is a / an

1. empty set
2. infinite set
3. singleton set
4. universal set

Answer

An empty set is denoted by { } or Φ. However, {Φ} is a set that contains the empty set as an element. Since it contains one element, it is a singleton set.

Hence, option 3 is the correct option.

If A is any set and U is the universal set, then the complement of A is

1. Φ
2. A
3. U - A
4. U

Answer

By definition, the complement of set A (A') consists of all elements in the Universal set U that are not in A. This is mathematically written as U - A.

Hence, option 3 is the correct option.

n(A - B) is equal to

1. n(A) - n(B)
2. n(A) - n(A ∪ B)
3. n(A ∪ B) - n(A ∩ B)
4. n(A) - n(A ∩ B)

Answer

To find the number of elements only in A, you take the total number of elements in A and subtract the elements it shares with B (the intersection).

Hence, option 4 is the correct option.

Fill in the blanks :

(i) In a ............... form, we list the properties satisfied by each element of the set.

(ii) A void set is usually denoted by ............... .

(iii) If A ⊆ X, then X is called a ............... of A.

(iv) Two finite sets having the same number of elements are said to be ............... .

(v) n(A ∪ B) + n(A ∩ B) = ............... .

Answer

(i) In a set-builder form, we list the properties satisfied by each element of the set.

(ii) A void set is usually denoted by ϕ.

(iii) If A ⊆ X, then X is called a superset of A.

(iv) Two finite sets having the same number of elements are said to be equivalent.

(v) n(A ∪ B) + n(A ∩ B) = n(A) + n(B).

Write true (T) or false (F):

(i) The number of proper subsets of a set containing n elements is 2ⁿ.

(ii) Any set A and its complement are equivalent sets.

(iii) The complement of a set is a subset of U.

(iv) If n(A ∩ B) = Φ, then n(B - A) = n(B)

(v) If two sets A and B are disjoint, then n(A ∪ B) = n(A) + n(B)

Answer

(i) False
Reason — The total number of subsets of a set containing n elements is 2ⁿ. However, a proper subset must be smaller than the set itself (it cannot be the set itself). Therefore, the number of proper subsets is 2ⁿ-1.

(ii) False
Reason — For two sets to be equivalent, they must have the same number of elements (n(A) = n(A')). This is only true if the set A contains exactly half the elements of the Universal set U. In most cases, the number of elements in a set and its complement are different.

(iii) True
Reason — By definition, the complement of a set A (A') consists of all elements that are in the Universal set (U) but not in A. Since every element of A' is an element of U, A' is a subset of U (A' ⊆ U).

(iv) True
Reason — The term n(B - A) represents the elements in B that are not in A. The formula is n(B) - n(A ∩ B). If n(A ∩ B) = 0 (meaning the sets are disjoint), then n(B) - 0 = n(B).

(v) True
Reason — For any two sets, n(A ∪ B) = n(A) + n(B) - n(A ∩ B). If sets A and B are disjoint, their intersection is empty (n(A ∩ B) = 0), which simplifies the formula to n(A ∪ B) = n(A) + n(B).

Sid is given a few sets. These are represented as A, B, C, D, E, F and G on a plain sheet of paper as a Venn Diagram.

(1) How many empty sets are there in the given Venn Diagram ?

1. 3
2. 2
3. 1
4. 0

(2) How many singleton sets are there in the given Venn Diagram ?

1. 0
2. 1
3. 2
4. 3

(3) How many pairs of equivalent sets are there in the given Venn Diagram ?

1. 1
2. 2
3. 3
4. 4

(4) How many pairs of equal sets are there in the given Venn Diagram ?

1. 0
2. 1
3. 2
4. 3

Answer

Sets in the diagram are:

A = {α, x, a, 1}
B = {1, 4}
C = {4}
D = {8, 7, 2, α, x}
E = {α, x, a, 1}
F = {7, 2, 4}
G = {8}

(1) An empty set has no elements.
Looking at the list above all sets A, B, C, D, E, F, and G contain at least one element. Therefore there are no empty sets.

Hence, option 4 is the correct option.

(2) A singleton set contains exactly one element.

C = {4}

G = {8}

Total singleton sets = 2

Hence, option 3 is the correct option.

(3) Equivalent sets have the same number of elements (n(A) = n(B)).

Pairs with same number of elements:

A and E (4 elements)

C and G (1 element)

Total pairs = 2

Hence, option 2 is the correct option.

(4) Equal sets must have the exact same elements.

Pairs with same elements:

A and E = {α, x, a, 1}

Hence, option 2 is the correct option.

In a class there are 27 students. Out of these 14 study Psychology and 19 study Geography. There are 11 students who study both Psychology and Geography.

(1) How many students study Psychology but not Geography ?

1. 3
2. 4
3. 5
4. 6

(2) How many students study Geography but not Psychology ?

1. 7
2. 8
3. 9
4. 11

(3) How many students study neither Psychology nor Geography ?

1. 8
2. 7
3. 6
4. 5

(4) What is the difference between the number of students who study Psychology only and those who study Geography only ?

1. 4
2. 5
3. 6
4. 7

Answer

Given:

Total students: n(U) = 27

Students studying Psychology: n(P) = 14

Students studying Geography: n(G) = 19

Students studying both: n(P ∩ G) = 11

(1) This represents the number of students who study Psychology only.

We use the formula:

n(P - G) = n(P) - n(P ∩ G)

Substituting the values in above, we get:

n(P - G) = 14 - 11

n(P - G) = 3

Number of students who study Psychology but not Geography = 3.

Hence, option 1 is the correct option.

(2) This represents the number of students who study Geography only.

We use the formula:

n(G - P) = n(G) - n(P ∩ G)

Substituting the values in above, we get:

n(G - P) = 19 - 11

n(G - P) = 8

Number of students who study Geography but not Psychology = 8.

Hence, option 2 is the correct option.

(3) First, find the total students who study at least one subject (n(P ∪ G)):

n(P ∪ G) = n(P) + n(G) - n(P ∩ G)

Substituting the values in above, we get:

n(P ∪ G) = 14 + 19 - 11

n(P ∪ G) = 33 - 11

n(P ∪ G) = 22

So, 22 students study either of two subjects.

To find students who study neither Psychology nor Geography, we use formula:

Students who study neither Psychology nor Geography = Total students - Students who study either of two subjects

Students who study neither Psychology nor Geography = 27 - 22

Students who study neither Psychology nor Geography = 5

Hence, option 4 is the correct option.

(4) Difference between the number of students who study Psychology only and those who study Geography only = ?

Number of students who study Psychology only = 3.

Number of students who study Geography only = 8.

Difference = 8 - 3 = 5

Difference between the number of students who study Psychology only and those who study Geography only = 5.

Hence, option 2 is the correct option.

Assertion: The collection of all tall girls of your class is not a set.

Reason: A well defined collection of objects is called a set.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
3. Assertion (A) is true but Reason (R) is false.
4. Assertion (A) is false but Reason (R) is true.

Answer

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Explanation

Assertion is true. In set theory, a collection must be "well-defined." The term "tall" is relative and subjective (someone 5'6" might be tall to one person but short to another). Because there is no fixed criteria, it is not a set.

The reason correctly states the definition of a set (a well-defined collection of objects), which explains why the assertion is true.

Hence, option 1 is the correct option.

Assertion: Let P = {x : x is a factor of 24} and Q = {x : x is a factor of 30}.
P ∩ Q = {1, 2, 3, 6}.

Reason: The intersection of two sets A and B is the set of all those elements of A which are not in B.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
3. Assertion (A) is true but Reason (R) is false.
4. Assertion (A) is false but Reason (R) is true.

Answer

Assertion (A) is true but Reason (R) is false.

Explanation

P = Factors of 24: {1, 2, 3, 4, 6, 8, 12, 24}

Q = Factors of 30: {1, 2, 3, 5, 6, 10, 15, 30}

Common elements (P ∩ Q ) = {1, 2, 3, 6}

So the assertion is true.

Reason is false. The definition provided in the reason actually describes the Difference of Sets (A - B). The intersection (A ∩ B) is the set of elements that are common to both A and B.

Hence, option 3 is the correct option.