Explore topic-wise InterviewSolutions in .

Correct option is A) (A ∪ B)’ = A’ ∩ B

Given set A = {0, 2, 4} and φ is a null set.

A ∩ φ = {0, 2, 4} ∩ { } 

= { } ……. (1) 

A ∩ A = {0, 2, 4} ∩ {0, 2, 4} 

= {0, 2,4} …….. (2) 

From (1) and (2), 

We conclude that A ∩ φ = φ and A ∩ A = A

(i) For the set of right triangles, the universal set can be the set of triangles or the set of polygons.
(ii) For the set of isosceles triangles, the universal set can be the set of triangles or the set of polygons or the set of two-dimensional figures.

Correct option is A) parallel

Correct option is D) None

Correct option is B) 90°, 45°, 45°

U = {1, 3, 5, 8, 9, 10, 11, 12, 13, 15} 

A = {1, 11, 13} 

B = {8, 5, 10, 11, 15} 

A’ = {3, 5, 8, 9, 10, 12, 15} 

B’ = {1, 3, 9, 12, 13} 

A ∩ B= {11} 

A’ ∩ B’ = {3, 9, 12} …(i) 

A ∪ B = {1, 5, 8, 10, 11, 13, 15} 

A’ ∪ B’ = { 1, 3, 5, 8, 9, 10, 12, 13, 15} … (ii) 

(A ∩ B)’ = { 1, 3, 5, 8, 9, 10, 12, 13,15} … (iii) 

(A ∪ B)’ = {3, 9, 12} ,..(iv) 

(A ∩ B)’ = A’ ∪ B’ … [From (ii) and (iii)]

(A ∪ B)’ = A’ ∩ B’ … [From (i) and (iv)]

Let A = {L, A, U, G, H}

B = {C, R, Y} 

Now, A ∩ B = φ 

∴ A and B are disjoint sets.

i. A = set of multiples of 5 

∴ A = {5, 10, 15, …} 

B = set of multiples of 7 

∴ B = {7, 14, 21,…} 

C = set of multiples of 12 

∴ C = {12, 24, 36, …} 

Now, set of natural numbers, whole numbers, integers, rational numbers are as follows: 

N = {1, 2, 3, …}, W = {0, 1, 2, 3, …} 

I = {…,-3, -2, -1, 0, 1, 2, 3, …} 

Q = {(p/q) | p,q ∈ I,q ≠ 0} 

Since, set A, B and C are the subsets of sets N, W , I and Q. 

∴ For set A, B and C we can take any one of the set from N, W, I or Q as universal set. 

ii. P = set of integers which are multiples of 4. 

P = {4, 8, 12,…} 

T = set of all even square numbers 

T = {2², 4², 6², …} 

Since, set P and T are the subsets of sets N, W, I and Q. 

∴ For set P and T we can take any one of the set from N, W, I or Q as universal set.

a. All elements of set A are not present in set B, C and D. 

∴ A ⊆ B, 

∴ A ⊆ C,

∴ A ⊆ D 

∴ Statement (a) is false. 

b. All elements of set B are not present in set A, C and D. 

∴ B ⊆ A, 

∴ B ⊆ C, 

∴ B ⊆ D 

∴ Statement (b) is false.

i. A = {x | x = n² , n e N, n < 10} 

ii. B = {x | x = 6n, n e N, n < 9} 

iii. C = {y | y is a letter of the word ‘SMILE’} [Other possible words: ‘SLIME’ , ‘MILES’ , ‘MISSILE’ etc.] 

iv. D = {z | z is a day of the week} 

v. X = {y | y is a letter of the word ‘eat’} [Other possible words: ‘tea’ or ‘ate’]

Let A = {L, A, U, G, H} 

B = {C, R, Y} 

Now, 

A ∩ B = φ 

∴ A and B are disjoint sets.

i. A is a set of even natural numbers less than 10.

Listing method: A = {2, 4, 6, 8}

Rule method: A = {x | x = 2n, n e N, n < 5}

ii. B is a set of letters of the word ‘SCIENCE’. Listing method : B = {S, C, I, E, N}

Rule method: B = {x \ x is a letter of the word ‘SCIENCE’}

i. A = {x | v = n², n e N, n < 10}

ii. B = {x j x = 6n, n e N, n < 9} 

iii. C = {y j y is a letter of the word ‘SMILE’} [Other possible words: ‘SLIME’, ‘MILES’, ‘MISSILE’ etc.] 

iv. D = {z | z is a day of the week} 

v. X = {y | y is a letter of the word ‘eat’} [Other possible words: ‘tea’ or ‘ate’]

i. A = {Chaitra, Vaishakh, Jyestha, Aashadha, Shravana, Bhadrapada, Ashwina, Kartika, Margashirsha, Paush, Magha, Falguna} 

ii. X = {C, O, M, P, L, E, N, T} 

iii. Y = {Nose, Ears, Eyes, Tongue, Skin} 

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

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

We can’t list all elements in the set of rational numbers. We know that rational numbers are infinite.

i.  A = {Chaitra, Vaishakh, Jyestha, Aashadha, Shravana, Bhadrapada,   Ashwina, Kartika, Margashirsha, Paush, Magha, Falguna}

ii. X = {C, O, M, P, L, E, N, T} 

iii. Y = {Nose, Ears, Eyes, Tongue, Skin} 

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

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

Correct option is A) μ

 Y = {x | x is a positive factor of the number 2^{p – 1} (2^p – 1), where 2^p – 1 is a prime number}.

Y- {x | x is a positive factor of the number 2^p-1 (2^p – 1), where 2^p – 1 is a prime number}.

So, the factors of 2p-1 are 1,2,2² ,2³ ,…, 2^p-1 .

Y= {1,2,2² ,2³ , …,2^p-1 ,2^p -1}

Given 

n(A) = 21, n(B) = 26, n(C) = 29 

n(A ∩ B) = 14, n(C ∩ A) = 12 

n(B ∩ C) = 14, n(A ∩ B ∩ C) = 8 

(a)  n(A ∪ B ∪ C) = n(A) + n(B) + N(C) − n(A ∩ B) 

n(B ∩ C) − n(C ∩ A) + n(A ∩ B ∩ C) 

= 21 + 26 + 29 – 14 – 12 – 14 + 8 

(b)  n(C only) = n(C) − n(A ∩ C) − n(B ∩ C) + n(A ∩ B ∩ C) 

= 29 – 12 – 14 + 18 

= 11

(i) {x : x is an odd natural numbers} 

(ii) {x : x is an even natural numbers} 

(iii) {x: x is a natural number and not multiple of 3} 

(iv) {x : x is a positive composite number and x = 1} 

(x) {x : x ∈ N and x is not divisible by 3 and 5} 

(xi) {x: x ∈ N and x is not a perfect square} 

(xii) {x:x ∈ N and x is not a perfect cube} 

(xiii) {x:x ∈ 2V and x ≠ 3} (ix) {x:x ∈ 2V and x ≠ 2} 

(x) {1,2,3,4,5,6} 

(xi) {x:x ∈ N and 2x + 1 ≤ 10} = {1, 2, 3, 4}.

Let 

A = B, then A ∪ B = A and A ∩ B = A 

A ∪ B = A ∩ B 

Thus, A = B …(i) 

Conversely, let 

A ∪ B = A ∩ B 

Now, let 

x ∈ A 

x ∈ (A ∪ B ) [∴ A ∪ B = A ∩ B] 

x ∈ (A ∩ B ) 

(x ∈ A and x ∈ B) 

x ∈ B 

A ⊆ B …(ii) 

 Now, let 

y ∈ A 

y ∈ A ∪ B 

y ∈ A ∩ B[∴ A ∪ B = A ∩ B] 

y ∈ A and y ∈ B 

y ∈ A 

∴ B ⊆ A …(iii) 

From equations (ii) and (iii), we get A = B

Let 

x ∈ {A ∩ (B − C)} 

x ∈ A and x ∈ B and x ∉ C 

(x ∈ A and x ∈ B) and (x ∈ A and  x ∉ C) 

(x ∈ A ∩ B − A ∩ C)

A ∩ (B ∩ C) ⊆ (A ∩ B) − (A ∩ C) …(i) 

Again, let 

y ∈ (A ∩ B) ∩ (A − C) 

y ∈ A and (y ∈ B and y ∉  C) 

y ∈ A and y ∈ B − C 

y ∈ {A ∩ (B − C)} 

(A ∩ B) − (A ∩ C) ⊆ A ∩ (B − C) …(ii) 

From equation (i) and (ii), 

we get 

A ∩ (B − C) = (A ∩ B) - (A ∩ C)

(i) (A ∪ B) ∩ (A ∩ B’) = A 

L.H.S = (A ∪ B) ∩ (A ∩ B’) 

= A ∪ (B ∩ B’)           (By distributive law) 

= A ∪ ∅             (∴ B ∩ B’ = ∅) 

= A 

R.H.S. 

(ii) A - (A ∩ B) = A – B 

L.H.S = A - (A ∩ B) 

= A ∩ (A ∩ B)’              [∴ A – B = (A ∩ B)’] 

= A ∩ (A’ ∪ B)’           (By Demorgan's law) 

= (A ∩ A)’ ∪ (A ∩ B)              (By distributive law) 

= ∅ ∪ A ∩ B’             (∴ A ∩ A’ = ∅) 

= A ∩ B’ 

= A – B 

= R.H.S

We have

A – (A ∩ B) = A ∩ (A ∩ B)′ (since A – B = A ∩ B′)

= A ∩ (A′ ∪ B′) [by De Morgan’s law)

= (A ∩ A′) ∪ (A ∩ B′) [by distributive law]

= φ ∪ (A ∩ B′)

= A ∩ B′ = A – B

Let,

x ∈ {A − (B ∪ C)} 

x ∈ A and x ∉ (B ∪ C). 

(x ∈ A and x ∉ B) and (x ∈ A and x ∉ C)

x ∈ (A − B) x ∈ (A − C) 

x ∈ {(A − B) ∩ (A − C)} 

A− (A − B) ⊆ (A − B) ∩ (A − C) …(i) 

Again, let 

y ∈ (A − B) ∩ (A − C) 

y ∈ (A − B) and y ∈ (A − C) 

(y ∈ A and y ∉ B) and (y ∈ A and y ∉ C 

y ∈ A and y ∉ B ∪ C) 

y ∈ {A − (B − C)} 

(A − B) ∩ (A − C) ⊆ A − (B ∪ C) …(ii) 

From eqs. (i) & (ii) 

A – (B ∪ C) = (A - B) ∩ (A - C)

(i) {2, 3, 5, 7, 11, 13, 17, 19}

(ii) {12, 22, 32, 42} = {1, 4, 9, 16}

(iii) {2, 4, 6, 8}

(iv) {a, b, c, d, e, f, g, h}

(v) {b, a, s, k, e, t}

(vi) {Jaipur, Jodhpur, Jalandhar, Jaunpur}

(vii) {Δ, O, □, O}

(viii) {o, a}

(ix) {0, 1, 4, 9}

B = { x ∶ x ∈ N and x²= x} is not an empty set, because there is only one natural number whose square is equal to the number is self i.e. 1 or B = {1}.

(i) For X = {1, 2, 3, 4, 5, 6}, it is the given that n ∈ X, but 2n ∉ X. 

Let, A = {x | x ∈ X and 2x ∉ X} 

Now, 1 ∉ A as 2.1 = 2 ∈ X 

2 ∉ A as 2.2 = 4 ∈ X 

3 ∉ A as 2.3 = 6 ∈ X 

But 4 ∈ A as 2.4 = 8 ∉ X 

5 ∈ A as 2.5 = 10 ∉ X 

6 ∈ A as 2.6 = 12 ∉ X 

So, A = {4, 5, 6} 

(ii) Let B = {x | x ∈ X and x + 5 = 8} 

Here, B = {3} 

as x = 3 ∈ X and 3 + 5 = 8 and there is no other element belonging to X such that x + 5 = 8. 

(iii) Let C = {x | x ∈ X, x > 4} 

Therefore, C = {5, 6}

We have 

A – (A ∩ B) = A ∩ (A ∩ B)′ (since A – B = A ∩ B′) 

= A ∩ (A′ ∪ B′) [by De Morgan’s law) 

= (A∩A′) ∪ (A∩ B′) [by distributive law] 

= φ ∪ (A ∩ B′) 

= A ∩ B′ = A – B

(i) {x : x is an even natural number less than 12}

(ii) {x : x is a prime number less than 12}

(iii) {x : x is a months of the year whose name starts with letter J}

(iv) {x : x is a vowel in English alphabets}

(v) {x : x is a day of the week whose name starts with letter T}

(vi) {x: x is a perfect square natural number upto 25}

(vii) {x : x is a natural number upto 30 and divisible by 5}.

(i) {1,-2, 3, 4, 6, 8, 12, 24}; {x : x is a natural number which divides 24 completely}

(ii) {21, 23, 25, 27, 29, 31, 33}; {x:x is an odd number between 20 and 35}

(iii) {c, a, l, u, t}; {x: x is a letter used in the word ‘CALCUTTA’}

(iv) {January, February, March,April, May}; {x : x is name of first five months of a year}

(v) {16, 25, 36, 49, 64, 81}; {x : x is a perfect square two digit number}.

No. consider the following sets A, B and C : 

A = {1, 2, 3} 

B = {2, 3, 5} 

C = {4, 5, 6} 

Now (A ∩ B) ∪ C = ({1, 2, 3} ∩ {2, 3, 5}) ∪ {4, 5, 6} 

= {2, 3} ∪ {4, 5, 6} 

= {2, 3, 4, 5, 6} 

And A ∩ (B ∪ C) = {1, 2, 3} ∩ [{2, 3, 5} ∪ {4, 5, 6} 

= {1, 2, 3} ∩ {2, 3, 4, 5, 6} 

= {2, 3} 

Therefore, (A ∩ B) ∪ C ≠ A ∩ (B ∪ C)

Infinite set

Because circle is a collection of infinite points whose distances form the centre is constant called radius.

Factors of 42 = 1, 2, 3, 6, 7, 14, 21, 42 

So roster form = {1, 2, 3, 6, 7, 14, 21, 42} 

The builder form = {x/x ∈ N, x is a factor of 42}

Correct option is B) (A ∩ B) – C

Correct option is B) A’ ∪ B’

Positive Integers = 0, 1, 2, 3, …500 

Positive Integers greater than 500 = 501, 502, 503, … 

There are infinite positive integers which are greater than 500. 

So, the given set is infinite.

A = {2, 3, 5, 7, 11, 13, 17, 19} 

B = {1, 3, 5, 7} 

i) A ∩ B = {2, 3, 5, 7,11, 13,17,19} ∩ {1,3, 5,7} = {3,5,7} . 

ii) B ∩ A = {1, 3, 5, 7} ∩ {2, 3, 5, 7, 11, 13, 17, 19} = {3,5,7} 

iii) A – B = {2, 3,5,7, 11, 13, 17, 19} – {1,3, 5,7} = {2, 11, 13, 17, 19} 

iv) B – A = {1,3,5, 7} – {2, 3, 5, 7, li; 13, 17, 19}= {1} 

∴ We observed A ∩ B = B ∩ A 

A – B ≠ B – A

Here, set of all vowels which precede q are

Since, A, E, I, O these are the vowels they come before q.

∴ B = {A, E, I, O}.