Important Questions: Sets

Q1: Write the following sets in roster form:
(i) A = {x : x is an integer and -3 ≤ x < 7}
Ans: A = {x : x is an integer and -3 ≤ x < 7}
Integers satisfying -3 ≤ x < 7 are all whole numbers from -3 up to 6 inclusive.
Therefore, A = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6}

(ii) B = {x : x is a natural number less than 6}
Ans: B = {x : x is a natural number less than 6}
Natural numbers (for this level) start from 1. Natural numbers less than 6 are 1, 2, 3, 4 and 5.
Therefore, B = {1, 2, 3, 4, 5}

Q2: Let X = {1, 2, 3, 4, 5, 6}. If n represent any member of X, express the following as sets:
(i) n ∈ X but 2n ∉ X
Ans: Given X = {1, 2, 3, 4, 5, 6}. We need elements n ∈ X for which 2n ∉ X.
Check each n: 1 → 2·1 = 2 ∈ X
2 → 2·2 = 4 ∈ X
3 → 2·3 = 6 ∈ X
4 → 2·4 = 8 ∉ X
5 → 2·5 = 10 ∉ X
6 → 2·6 = 12 ∉ X
Thus the required set A = {4, 5, 6}

(ii) n + 5 = 8
Ans: Let B = {x | x ∈ X and x + 5 = 8}.
Solve x + 5 = 8 → x = 3. Since 3 ∈ X, B = {3}.

(iii) n is greater than 4
Ans: Let C = {x | x ∈ X, x > 4}.
Elements of X greater than 4 are 5 and 6.
Therefore, C = {5, 6}

Q3: Use the properties of sets to prove that for all the sets A and B, A - (A ∩ B) = A - B
Ans: Start with left-hand side:
A - (A ∩ B) = A ∩ (A ∩ B)′ (using A - X = A ∩ X′)
= A ∩ (A′ ∪ B′) (by De Morgan's law)
= (A ∩ A′) ∪ (A ∩ B′) (by distributive law)
= φ ∪ (A ∩ B′) (since A ∩ A′ = φ)
= A ∩ B′ = A - B.
Hence proved that A - (A ∩ B) = A - B.

Q4: In a class of 60 students, 23 play hockey, 15 play basketball,20 play cricket and 7 play hockey and basketball, 5 play cricket and basketball, 4 play hockey and cricket, 15 do not play any of the three games. Find
(i) How many play hockey, basketball and cricket
(ii) How many play hockey but not cricket
(iii) How many play hockey and cricket but not basketball
Ans:

Q5: In a survey of 600 students in a school, 150 students were found to be drinking Tea and 225 drinking Coffee, 100 were drinking both Tea and Coffee. Find how many students were drinking neither Tea nor Coffee.
Ans: Given total students = 600.
n(T) = 150, n(C) = 225, n(T ∩ C) = 100.
n(T ∪ C) = n(T) + n(C) - n(T ∩ C) = 150 + 225 - 100 = 275.
Number drinking neither = Total - n(T ∪ C) = 600 - 275 = 325.

Q6: Write the following sets in the roster form.
(i) A = {x | x is a positive integer less than 10 and 2x - 1 is an odd number}
Ans: For any integer x, 2x is even, so 2x - 1 is always odd.
All positive integers less than 10 are 1, 2, ..., 9 and each gives 2x - 1 odd.
Therefore, A = {1, 2, 3, 4, 5, 6, 7, 8, 9}

(ii) C = {x : x² + 7x - 8 = 0, x ∈ R}
Ans: Solve x² + 7x - 8 = 0.
Factorise: (x + 8)(x - 1) = 0.
Thus x = -8 or x = 1.
Therefore, C = {-8, 1}

Q7: Given that N = {1, 2, 3, ..., 100}, then
(i) Write the subset A of N, whose elements are odd numbers.
Ans: A = {x | x ∈ N and x is odd}.
Odd numbers in N up to 100 are 1, 3, 5, ..., 99.
Therefore, A = {1, 3, 5, 7, ..., 99}

(ii) Write the subset B of N, whose elements are represented by x + 2, where x ∈ N.
Ans: B = {y | y = x + 2, x ∈ N and y ∈ N, y ≤ 100}.
Since x ∈ {1, 2, ..., 100} and y = x + 2 must be ≤ 100, x can be at most 98.
Thus y runs from 3 to 100.
Therefore, B = {3, 4, 5, 6, ..., 100}

Q8: Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}.
Find A′, B′, A′ ∩ B′, A ∪ B and hence show that ( A ∪ B )′ = A′∩ B′.
Ans: Given U = {1, 2, 3, 4, 5, 6}.
A = {2, 3} → A′ = U \ A = {1, 4, 5, 6}.
B = {3, 4, 5} → B′ = U \ B = {1, 2, 6}.
A′ ∩ B′ = {1, 6}.
A ∪ B = {2, 3, 4, 5} → (A ∪ B)′ = U \ (A ∪ B) = {1, 6}.
Therefore (A ∪ B)′ = A′ ∩ B′ = {1, 6} (De Morgan's law illustrated).

Q9: Let U = {1, 2, 3, 4, 5, 6, 7}, A = {2, 4, 6}, B = {3, 5} and C = {1, 2, 4, 7}, find
(i) A′ ∪ (B ∩ C′)
(ii) (B - A) ∪ (A - C)
Ans: Given U = {1, 2, 3, 4, 5, 6, 7}.
A = {2, 4, 6} → A′ = {1, 3, 5, 7}.
C = {1, 2, 4, 7} → C′ = {3, 5, 6}.
B = {3, 5} → B ∩ C′ = {3, 5}.
(i) A′ ∪ (B ∩ C′) = {1, 3, 5, 7}.
(ii) B - A = B \ A = {3, 5} (no overlap with A).
A - C = A \ C = {6} (since 2 and 4 are in C).
(B - A) ∪ (A - C) = {3, 5, 6}.

Q10: Let U = {x : x ∈ N, x ≤ 9}; A = {x : x is an even number, 0 < x < 10}; B = {2, 3, 5, 7}. Write the set (A U B)'.
Ans: U = {1, 2, 3, 4, 5, 6, 7, 8, 9}.
A = {2, 4, 6, 8}.
B = {2, 3, 5, 7}.
A ∪ B = {2, 3, 4, 5, 6, 7, 8}.
(A ∪ B)′ = U \ (A ∪ B) = {1, 9}.