Class 11 Sets Important Questions Solutions

Answer Key

1. 2^x– 1 is always an odd number for all positive integral values of x since 2^x is an even number.

• In particular, 2^x– 1 is an odd number for x = 1, 2, … , 9.
  Therefore, A = {1, 2, 3, 4, 5, 6, 7, 8, 9}

• x²+ 7x – 8 = 0
  (x + 8) (x – 1) = 0
  x = – 8 or x = 1
  Therefore, C = {– 8, 1}

 

2. For “CATARACT”, Distinct letters are
   {C, A, T, R} = {A, C, R, T}
   For “TRACT”, Distinct letters are
   {T, R, A, C} = {A, C, R, T
   The letters needed to spell cataract are equal to the set of letters needed to spell tract.
   Hence, the two sets are equal.

 

3. (i) A = {x | x ∈ N and x is odd}

A = {1, 3, 5, 7, …, 99}
(ii) B = {y | y = x + 2, x ∈ N}

1 ∈ N, y = 1 + 2 = 3

2 ∈ N, y = 2 + 2 = 4, and so on.

Therefore, B = {3, 4, 5, 6, … , 100}

 

4. A = All natural numbers, i.e., {1, 2, 3…..}

B = All even natural numbers, i.e., {2, 4, 6, 8…}

C = All odd natural numbers, i.e., {1, 3, 5, 7……}

D = All prime natural numbers, i.e., {1, 2, 3, 5, 7, 11, …}

(i) A ∩ B

A contains all elements of B.

∴ B ⊂ A = {2, 4, 6, 8…}

         ∴ A ∩ B = B

(ii) A ∩ C

A contains all elements of C.

∴ C ⊂ A = {1, 3, 5…}

         ∴ A ∩ C = C

(iii) A ∩ D

A contains all elements of D.

∴ D ⊂ A = {2, 3, 5, 7..}

         ∴ A ∩ D = D

(iv) B ∩ C

B ∩ C = ϕ

         There is no natural number which is both even and odd at the same time.

(v) B ∩ D

B ∩ D = 2

         {2} is the only natural number which is even and a prime number.

C ∩ D

C ∩ D = {1, 3, 5, 7…}

= D – {2}

         Every prime number is odd except {2}.

 

5. (i) A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

A ∪ B = {x: x ∈ A or x ∈ B}

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

         (ii) A = {1, 2, 3, 4, 5}

C = {7, 8, 9, 10, 11}

A ∪ C = {x: x ∈ A or x ∈ C}

              = {1, 2, 3, 4, 5, 7, 8, 9, 10, 11}

         (iii) B = {4, 5, 6, 7, 8}

C = {7, 8, 9, 10, 11}

B ∪ C = {x: x ∈ B or x ∈ C}

                  = {4, 5, 6, 7, 8, 9, 10, 11}

         (iv)  B = {4, 5, 6, 7, 8}

D = {10, 11, 12, 13, 14}

B ∪ D = {x: x ∈ B or x ∈ D}

                = {4, 5, 6, 7, 8, 10, 11, 12, 13, 14}

(v) A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

C = {7, 8, 9, 10, 11}

A ∪ B = {x: x ∈ A or x ∈ B}

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

A ∪ B ∪ C = {x: x ∈ A ∪ B or x ∈ C}

                        = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

(vi) A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

D = {10, 11, 12, 13, 14}

A ∪ B = {x: x ∈ A or x ∈ B}

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

A ∪ B ∪ D = {x: x ∈ A ∪ B or x ∈ D}

                         = {1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14}

(vii) B = {4, 5, 6, 7, 8}

C = {7, 8, 9, 10, 11}

D = {10, 11, 12, 13, 14}

B ∪ C = {x: x ∈ B or x ∈ C}

= {4, 5, 6, 7, 8, 9, 10, 11}

B ∪ C ∪ D = {x: x ∈ B ∪ C or x ∈ D}

                        = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}

(viii) A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

C = {7, 8, 9, 10, 11}

B ∪ C = {x: x ∈ B or x ∈ C}

= {4, 5, 6, 7, 8, 9, 10, 11}

A ∩ B ∪ C = {x: x ∈ A and x ∈ B ∪ C}

                = {4, 5}

(ix) A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

C = {7, 8, 9, 10, 11}

(A ∩ B) = {x: x ∈ A and x ∈ B}

= {4, 5}

(B ∩ C) = {x: x ∈ B and x ∈ C}

= {7, 8}

(A ∩ B) ∩ (B ∩ C) = {x: x ∈ (A ∩ B) and x ∈ (B ∩ C)}

                                   = ϕ

(x) A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

C = {7, 8, 9, 10, 11}

D = {10, 11, 12, 13, 14}

A ∪ D = {x: x ∈ A or x ∈ D}

= {1, 2, 3, 4, 5, 10, 11, 12, 13, 14}

B ∪ C = {x: x ∈ B or x ∈ C}

= {4, 5, 6, 7, 8, 9, 10, 11}

(A ∪ D) ∩ (B ∪ C) = {x: x ∈ (A ∪ D) and x ∈ (B ∪ C)}

                                 = {4, 5, 10, 11}

 

6. U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}

A′ = {1, 4, 5, 6}

B′ = { 1, 2, 6 }.

A′ ∩ B′ = { 1, 6 }

A ∪ B = { 2, 3, 4, 5 }

(A ∪ B)′ = { 1, 6 }

Therefore, ( A ∪ B )′ = { 1, 6 } = A′ ∩ B′

 

7. (i) A = {x : x is an integer and –3 ≤ x < 7}

Integers are …-5, -4, -3, -2, -2, 0, 1, 2, 3, 4, 5, 6, 7, 8,…..

A = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6}

(ii) B = {x : x is a natural number less than 6}

Natural numbers are 1, 2, 3, 4, 5, 6, 7, ……

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

 

8. U = {1, 2, 3, 4, 5, 6, 7}, A = {2, 4, 6}, B = {3, 5} and C = {1, 2, 4, 7}

(i)  A′ = {1, 3, 5, 7}

C′ = {3, 5, 6}

B ∩ C′ = {3, 5}

A′ ∪ (B ∩ C′) = {1, 3, 5, 7}

(ii) B – A = {3, 5}

A – C = {6}

(B – A) ∪ (A – C) = {3, 5, 6}

 

9. Let U = {x : x ∈ N, x ≤ 9}; A = {x : x is an even number, 0 < x < 10}; B = {2, 3, 5, 7}

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

A = {2, 4, 6, 8}

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

(A U B)’ = {1, 9}

 

10. n (A ∪ B) = 50

n (A) = 28

n (B) = 32

n (A ∪ B) = n (A) + n (B) – n (A ∩ B)

Substituting the values, we get

50 = 28 + 32 – n (A ∩ B)

50 = 60 – n (A ∩ B)

–10 = – n (A ∩ B)

∴ n (A ∩ B) = 10

 

11. n (P) = 40

n (P ∪ Q) = 60

n (P ∩ Q) =10

We know, n (P ∪ Q) = n (P) + n (Q) – n (P ∩ Q)

Substituting the values, we get

60 = 40 + n (Q)–10

60 = 30 + n (Q)

N (Q) = 30

∴ Q has 30 elements.

 

12. Total number of people = 70

Number of people who like Coffee = n (C) = 37

Number of people who like Tea = n (T) = 52

Total number = n (C ∪ T) = 70

Person who likes both would be n (C ∩ T)

n (C ∪ T) = n (C) + n (T) – n (C ∩ T)

Substituting the values, we get

70 = 37 + 52 – n (C ∩ T)

70 = 89 – n (C ∩ T)

n (C ∩ T) =19

∴ There are 19 people who like both coffee and tea.

 

13. Teachers teaching physics or math = 20

Teachers teaching physics and math = 4

Teachers teaching Maths = 12

Let teachers who teach physics be ‘n (P)’ and for Maths be ‘n (M)’

20 teachers who teach physics or math = n (P ∪ M) = 20

4 teachers who teach physics and math = n (P ∩ M) = 4

12 teachers who teach Maths = n (M) = 12

n (P ∪ M) = n (M) + n (P) – n (P ∩ M)

Substituting the values, we get,

20 = 12 + n (P) – 4

20 = 8 + n (P)

n (P) =12

∴ There are 12 physics teachers.

 

14. Let total number of people be n (P) = 950

People who can speak English n (E) = 460

People who can speak Hindi n (H) = 750

(i) How many can speak both Hindi and English.

People who can speak both Hindi and English = n (H ∩ E)

n (P) = n (E) + n (H) – n (H ∩ E)

Substituting the values, we get

950 = 460 + 750 – n (H ∩ E)

950 = 1210 – n (H ∩ E)

n (H ∩ E) = 260

∴ The number of people who can speak both English and Hindi is 260.

(ii) How many can speak Hindi only.

We can see that H is disjoint union of n (H–E) and n (H ∩ E).

(If A and B are disjoint then n (A ∪ B) = n (A) + n (B))

∴ H = n (H–E) ∪ n (H ∩ E)

n (H) = n (H–E) + n (H ∩ E)

750 = n (H – E) + 260

n (H–E) = 490

∴ 490 people can speak only Hindi.

(iii) How many can speak English only.

We can see that E is disjoint union of n (E–H) and n (H ∩ E)

(If A and B are disjoint then n (A ∪ B) = n (A) + n (B))

∴ E = n (E–H) ∪ n (H ∩ E).

n (E) = n (E–H) + n (H ∩ E).

460 = n (H – E) + 260

n (H–E) = 460 – 260 = 200

∴ 200 people can speak only English.

 

15. 15 students do not play any of three games.

n(H ∪ B ∪ C) = 60 – 15 = 45

n(H ∪ B ∪ C) = n(H) + n(B) + n(C) – n(H ∩ B) – n(B ∩ C) – n(C ∩ H) + n(H ∩ B ∩ C)

45  = 23 + 15 + 20 – 7 – 5 – 4 + d

45  = 42 + d

d = 45- 42 = 3

Number of students who play all the three games = 3

Therefore, the number of students who play hockey, basketball and cricket = 3

a + d = 7

a = 7 – 3 = 4

b + d = 4

b = 4 – 3 = 1

a + b + d + e = 23

4 + 1 + 3 + e = 23

e = 15

Similarly, c = 2, g =14, f = 6

Number of students who play hockey but not cricket = a + e

= 4 + 15

= 19

Number of students who play hockey and cricket but not basketball = b = 1

 

 

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