Sets

We know that in roster method, we write all the elements of a set in curly bracket. Each of the element is written only once and separated by commas. The order of an element is not important but it is necessary to write all the elements of the set.

i. A = {2, 4, 6, 8, 10, 12, 14, … }

ii. B = {2}

iii. C = { … -7, -6, -5, -4, -3, -2, -1}

iv. D = {Sa, Re, Ga, Ma, Pa, Dha, Ni}

i. We know that Q is denoted as set of rational numbers.

We write 4/3 ∈ Q as ‘4/3 is an element of set of rational numbers’.

ii. We know that N is denoted as set of natural numbers.

We write -2 N as ‘-2 is not an element of set of natural numbers.’

iii. We write P = {p| p is an odd number} as ‘Set P contains all the odd numbers’.

We know that in listing method, we write all the elements of a set in curly bracket. Each of the elements is written only once and separated by commas. The order of an element is not important but it is necessary to write all the elements of the set and In rule method, we do not write the list of elements but write the general element using variable followed by a vertical line or colon and write the property of the variable.

i. We know that in listing method, we write all the elements of a set in curly bracket. Each of the elements is written only once and separated by commas. The order of an element is not important but it is necessary to write all the elements of the set.

A = {Mesa, Vrsabha, Mithuna, Karkata, Simha, Kanya, Tula, Vrscika, Dhanu, Makara, Kumbha, Mina}

ii. B = {C, O, M, P, L, E, N, T}

iii. C = {Eyes, Ears, Nose, Mouth, Skin}

iv. D = {2, 3, 5, 7, 11, 13, 17, 19}

v. E = {Asia, Africa, North America, South America, Antarctica, Europe, Australia}

We know that in rule method, we do not write the list of elements but write the general element using variable followed by a vertical line or colon and write the property of the variable.

i. A = {x| x is a perfect square number from 1 to 101}

ii. B = {y| y is a multiple of 6 and is less than 50}

iii. C = {z| z is a letter of the world ‘SMILE’.}

iv. D = {p | p is a day of the week}

We know that two sets A and B are said to be equal, if every element of set A is in set B and every element of set B is in set A. It is symbolically written as A = B.

A = {x|3x – 1 = 2}

⇒ 3x – 1 = 2

⇒ 3x = 2 + 1 = 3

⇒ x = 3/3 = 1

∴ A = {1}

B = {x |x is a natural number but x is neither prime nor composite}

We know that 1 is neither prime nor composite and is a natural number.

∴ B = {1}

C = {x |x ∈ N, x < 2}

We know that natural number, N = {0, 1, 2, 3, …}

∴ C = {0, 1}

A = B, B ≠ C, A ≠ C

Ans. A and B are equal sets but B and C (and) A and C are not equal sets.

We know that two sets A and B are said to be equal, if every element of set A is in set B and every element of set B is in set A. It is symbolically written as A = B.

A = Even Prime Numbers = {2}

B = {x| 7x – 1 = 13}

⇒ 7x – 1 = 13

⇒ 7x = 13 + 1 = 14

⇒ x = 14/7

∴ x = 2

∴ A = B

Ans. Set A and set B are equal sets.

i. We know that If there is not a single element in the set which satisfies the given condition then it is called a Null set or an Empty set. Empty set is represented by { }.

We know that Natural number, N = {0, 1, 2, … }.

∴ A = {a |a is a natural number smaller than zero} = {}

Ans. A is an empty set.

ii. B = {x|x² = 0}

⇒ x² = 0

∴ x = 0

B = {x|x² = 0} = {0} ≠ {}

Ans. B is not an empty set.

iii. C = {x|5x – 2 = 0, x ∈ N}

⇒ 5x – 2 = 0

⇒ 5x = 2

⇒ x = 5/2

We know that natural number, N = {0, 1, 2,… }

∴ x N

∴ C = {x|5x – 2 = 0, x ∈ N} = {}

Ans. C is an empty set.

i. A = {x|x < 10, x is a natural number}

∴ A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Ans. A is a finite set.

ii. B = {y |y< -1, y is an integer}

∴ B = {… -5, -4, -3, -2}

Ans. B is an infinite set.

iii. C = {a, b, c, d, e, …}

Ans. C is an infinite set.

iv. D = {Sarpanch, villager1, villager2, …}

Ans. D is an infinite set.

v. E = {Beakers, funnels, chemicals, thermometers, …}

Ans. E is an infinite set.

vi. W = {1, 2, 3, 4, 5, …}

Ans. W is an infinite set.

vii. Q = {(p/q) |p, q ∈ I, q ≠ 0}

Q = {1/2, 1/3, 1/4, …}

Ans. Q is an infinite set.

We consider the following Venn diagram to check whether the above statements are true or not.

Now, we have A = {a, b, c, d, e}, B = {c, d, e, f}, C = {b, d} and D = {a, e}.

So, it is clear from above that every element of set C and set D is also an element of set A.

∴ C ⊆ A and D ⊆ A

Hence, (iv), (vi) are true and (ii) is false.

Also, it is clear from above that every element of set B is not an element of set A.

∴ (v) is false.

Similarly, it is clear that every element of set C is not an element of set B.

∴ (i) is false.

Also, every element of set D is not an element of set B.

∴ (iii) is false.

Consider the universal set:

U = {u | u ∈ ℕ, 1 ≤ x ≤ 20}

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

X = {x | x ∈ ℕ, 7 < x < 15}

= {8, 9, 10, 11, 12, 13, 14}

Y = {y | y ∈ ℕ, y is prime number from 1 to 20}

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

U = {1, 2, 3, 7, 8, 9, 10, 11, 12}

P = {1, 3, 7, 10}

P’ = {x | x ∈ U and x ∉ P} = {2, 8, 9, 11, 12}

(i) The Venn diagram showing U, P and P’ is:

(ii) Now, we consider (P’)’

(P’)’ = {x | x ∈ U and x ∉ P’}

= {1, 3, 7, 10}

= P

Hence, (P’)’ = P.

We know that any set S such that every element of S is an element of the set A is said to be a subset of A.

∴ Consider B = {1, 2, 3}, C = {1, 3} and D = {2, 3, 7}

It is clear from above that every element of the sets B, C and D is an element of set A, so, B, C and D are subsets of A.

Also, note that

Every set is a subset of itself, i. e. A ⊆ A and

Empty set is a subset of every set, i. e. ϕ ⊆ A.

So, A and ϕ are also subsets of A.

(i) Consider the following Venn diagram for the above sets:

(ii) From the above Venn diagram, it is clear that set B can serve as the universal set for above sets.

A universal set is the set which can accommodate all other sets in it, i.e. every element of all the sets should be an element of the universal set.

A = set of multiples of 5,

B = set of multiples of 7.

C = set of multiples of 12

Consider the set of all integers, I = {………, -2, -1, 0, 1, 2,………}

We see that every element of the sets A, B and C is an element of the set I.

So, set I can serve as the universal set for the sets A, B and C.

A universal set is the set which can accommodate all other sets in it, i.e. every element of all the sets should be an element of the universal set.

P = set of integers which are multiples of 4

= {x | x ∈ ℤ and x is a multiple of 4}

T = {y | y is an even square number}

Consider the set of all integers, I = {………, -2, -1, 0, 1, 2,………}

We see that every element of both the sets P and T is an element of the set I.

So, set I can serve as the universal set for the sets P and T.

Let U be the universal set.

Then, U = Set of all students of a class

A = Set of students who secure 50% or more marks in Maths

We know that a set A’ is said to be the complement of the set A if it contains all the elements of the universal set U, which are not in the set A.

∴The complement of set A is:

A’ = Set of all students who do not secure 50% or more marks in Maths

Or A’ = Set of all students who secured less than 50% marks in Maths
∴ A^'= U - A

Given: n(A ∪ B) = 29, n(A) = 15 and n (A ∩ B) = 7

As we know, n (A ∪ B) = n(A) + n(B) - n (A ∩ B)

⇒ 7 = 15 + n(B) – 29

⇒ n(B) = -14 - 7

⇒ n(B) = -21

Given: n(A) = 80, n(B) = 60 and n (A ∩ B) = x = 20

Total number of students = 125

n(A) = number of students who drink tea

n(B) = number of students who drink coffee

n (A ∩ B) = number of students who drink both tea and coffee

n (A ∪ B) = number of students who drink either tea or coffee

As we know, n (A ∪ B) = n(A) + n(B) - n (A ∩ B)

⇒ n (A ∪ B) = 80 + 60 - 20

⇒ n (A ∪ B) = 140 - 20

⇒ n (A ∪ B) = 120

Then, the number of students who do not drink tea or coffee = 125 – 120

= 5

Given: n(A) = 50, n(B) = 60 and n (A ∩ B) = x = 40

n(A) = number of students who passed in English

n(B) = number of students who passed in mathematics

n (A ∩ B) = number of students who passed both subjects

n (A ∪ B) = number of students who passed in at least one of the subject

As we know, n (A ∪ B) = n(A) + n(B) - n (A ∩ B)

⇒ n (A ∪ B) = 50 + 60 - 40

⇒ n (A ∪ B) = 110 - 40

⇒ n (A ∪ B) = 70

Hence, the number of students who passed in at least one of the subject = 70

Given: n(A) = 130, n(B) = 180 and n (A ∩ B) = x = 110

Total number of students = 220

n(A) = number of students whose hobby is rock climbing

n(B) = number of students whose hobby is sky watching

n (A ∩ B) = number of students whose hobby is both

n (A ∪ B) = number of students whose hobby is either rock climbing or sky watching

As we know, n (A ∪ B) = n(A) + n(B) - n (A ∩ B)

⇒ n (A ∪ B) = 130 + 180 - 110

⇒ n (A ∪ B) = 310 - 110

⇒ n (A ∪ B) = 200

Then,

(i) The number of students who do not have any of the two hobbies

= 220– 200

= 20

(ii) The number of students whose hobby is rock climbing

= 130– 110

= 20

(iii) The number of students whose hobby is sky watching

= 180– 110

= 70

(i) A = {x, y, z, m, n}

(ii) B = {p, q, r, m, n}

(iii) (A ∪ B) = { x, y, z, m, n, p, q, r}

(iv) U = { x, y, z, m, n, p, q, r, s, t}

(v) A’ = {p, q, r, s, t}

(vi) B’ = {x, y, z, s, t}

(vii) (A ∪ B)’ = { s, t}

Given: M = {1, 3, 5}, N = {2, 4, 6}

(M ∩ N) = set of all common elements of M and N

Hence, (M ∩ N) = ϕ

Option (C) is correct.

Given : P = {x | x is an odd natural number,

set in roster form:

⇒ P = {3, 5}

Option (D) is correct.

Given : P = {1, 2, ........., 10}

As the number of elements given are finite.

⇒ given set is finite set.

Option (C) is correct.

Given : (M ∪ N) = {1, 2, 3, 4, 5, 6}, M = {1, 2, 4}

(M ∪ N) = set of all elements of set M and N

Hence, N = {3, 4, 5, 6}

Option (B) is correct.

Given :

_As,

_{Then using property,}

(P ∪ M) = M

_As,

_{Then using property,}

(P ∩ M) = P

A. set of intersecting points of parallel lines = ϕ

Because no two parallel lines will intersect.

Hence, Option (A) is correct.

SET: If we can definitely and clearly decide the objects of a given collection

then that collection is called set.

Hence, option A is correct answer.

Because we can clearly say that colours of rainbow are seven in number as

(violet, indigo, blue, green, yellow, orange and red).

Hence, Option (A) is correct.

Given : N = {1, 2, .........∞} and W = {0, 1, 2, .........∞}

(N ∩ W) = set of all common elements of N and W

Hence, (N ∩ W) = {1, 2, .........∞}

Option (A) is correct.

Given : P = {x | x is a letter of the word ' indian'}

listing form = each of the element is written only once and separated by commas in curly brackets.

Hence, P = {i, n, d, a}

Option (B) is correct.

Given : T = {1, 2, 3, 4, 5} and M = {3, 4, 7, 8}

(T ∪ M) = set of all elements of set T and M

Hence, (T ∪ M) = {1, 2, 3, 4, 5, 7, 8}

Option (C) is correct.

Given: n(A) = 72, n(B) = 43 and n (A ∪ B) = 100

Total number of persons who speak at least one language = n (A ∪ B) = 100

n(A) = number of students who speaks English

n(B) = number of students who speaks French

n (A ∩ B) = number of students who speak both

As we know, n (A ∪ B) = n(A) + n(B) - n (A ∩ B)

⇒ n (A ∩ B) = 72 + 43 - 100

⇒ n (A ∩ B) = 115 - 100

⇒ n (A ∩ B) = 15

number of students who speak both the language = 15

number of students who speaks English only = 72 – 15 = 57

number of students who speaks French only = 43 – 15 = 28

Given: n(A) = 70, n(B) = 90 and n (A ∩ B) = x = 25

n(A) = number of trees planted by Parth

n(B) = number of trees planted by Pradnya

n (A ∩ B) = number of trees planted by both of them together

n (A ∪ B) = number of trees planted by Parth or Pradnya

As we know, n (A ∪ B) = n(A) + n(B) - n (A ∩ B)

⇒ n (A ∪ B) = 70 + 90 - 25

⇒ n (A ∪ B) = 160 - 25

⇒ n (A ∪ B) = 35

Given: n(A) = 20, n(B) = 28 and n (A ∪ B) = 36

As we know, n (A ∪ B) = n(A) + n(B) - n (A ∩ B)

⇒ n (A ∩ B) = 20 + 28 - 36

⇒ n (A ∩ B) = 48 - 36

⇒ n (A ∩ B)) = 12

Given: n(A) = 8, n(B) = 6 and n (A ∩ B) = 10

n(A) = number of students having only dog as pet

n(B) = number of students having only cat as pet

n (A ∩ B) = number of trees planted by both of them together

Total students having a dog or cat as their pet animal at home = 8 + 6 + 10

= 24

Hence, number of students do not have a dog or cat as their pet animal at

Home = 28 – 24 = 4

Given:

Given:

Given: is a prime number between 80 and 100}

is an odd number between 90 and 100}

X = {83, 89, 93, 97}

Y = {91, 93, 95, 97, 99}

Given: X = set of all quadrilaterals.

Y = set of all rhombuses.

S = set of all squares.

T = set of all parallelograms.

V = set of all rectangles.

S ⊆ X, because all squares are quadrilaterals.

V ⊆ X, because all rectangles are quadrilaterals.

T ⊆ X, because all parallelograms are quadrilaterals.

S ⊆ Y, because all squares are rhombus.

S ⊆ V, because all squares are rectangles.

S ⊆ T, because all squares are parallelograms.

V ⊆ T, because all rectangles are parallelograms.

Y ⊆ T, because all rhombus are parallelograms.

Given: M is any set

As we know ϕ is null set.

Hence,

(M ∪ ϕ ) = M

(M ∩ ϕ ) = ϕ

(i) U = {1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13}

(ii) A = {1, 2, 3, 5, 7}

(iii) B = {1, 5, 8, 9, 10}

(iv) (A ∪ B ) = {1, 2, 3, 5, 7, 8, 9, 10}

(v) (A ∩ B ) = {1, 5}

Given:

As we know, n (A ∪ B) = n(A) + n(B) - n (A ∩ B)

⇒ n (A ∪ B) = 7 + 13 - 4

⇒ n (A ∪ B) = 20 - 4

⇒ n (A ∪ B) = 16