Answer Key
- 2x– 1 is always an odd number for all positive integral values of x since 2x is an even number.
- In particular, 2x– 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}
- 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.
- (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}
- 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}.
- (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}
- 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′
- (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}
- 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}
- 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}
- 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
- 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.
- 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.
- 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.
- 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 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
