Active Maths 2 (Strands 1–5): Ch 1 Solutions Chapter 1 Exercise 1.1 Q. 1. (i) Not well defined, as the meaning of the term ‘inexpensive’ is subjective. (ii) Well defined. (iii) Not well defined, as the meaning of the term ‘large‘ is subjective. (iv) Well defined. Q. 2. (i) A ∪ B = {0, 2, 3, 4, 5, 6, 7, 8} (ii) A ∩ B = {2} (iii) A’ = {0, 1, 4, 6, 8} (iv) B’ = {1, 3, 5, 7} (v) A \ B = {3, 5, 7} (vi) B \ A = {0, 4, 6, 8} (vii) (A ∪ B)’ = {1} (viii) A’ ∪ B = {0, 1, 2, 4, 6, 8} Q. 3. (a) U A B • 21 • 25 • 20 • 22 • 24 • 23 (b) (i) A ∪ B = {21, 23, 25} (ii) A ∩ B = {23} (iii) A’ = {20, 22, 24} (iv) B’ = {20, 21, 22, 24, 25} (v) (A ∩ B)’ = {20, 21, 22, 24, 25} (vi) A’ ∪ B’ = {20, 21, 22, 24, 25} (vii) B \ A = {} or ∅ (viii) (A \ B) ∪ (B \ A) = {21, 25} (c) A \ B’ = {23} B \ A’ = {23} ∴ A \ B’ = B \ A’ Answer is yes (d) As B = A ∩ B ⇒ B ⊂ A (e) Case 1 # (A ∪ B) = 0 ⇒ # (A ∩ B) = 0 In this case # (A ∪ B) = # (A ∩ B) Case 2 # (A ∪ B) > 0 and # (A ∩ B) = 0 In this case # (A ∪ B) > # (A ∩ B) Case 3 # (A ∪ B) > 0 and # (A ∩ B) > 0 As A ∩ B is a subset of A ∪ B, # (A ∪ B) > # (A ∩ B) Q. 4. (i) 2 3 = 8 (ii) {}, {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, {x, y, z} (iii) {x, y, z} Q. 5. (i) A ∪ B (iii) (A ∩ B)’ (ii) A ∩ B (iv) B \ A Q. 6. (i) U X Y (ii) U X Y (iii) X Y U (iv) U X Y 1
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Active Maths 2 (Strands 1–5): Ch 1 Solutions
Chapter 1 Exercise 1.1
Q. 1. (i) Not well defined, as the meaning of the term ‘inexpensive’ is subjective.
(ii) Well defined.
(iii) Not well defined, as the meaning of the term ‘large‘ is subjective.
(iv) Well defined.
Q. 2. (i) A ∪ B = {0, 2, 3, 4, 5, 6, 7, 8}
(ii) A ∩ B = {2}
(iii) A’ = {0, 1, 4, 6, 8}
(iv) B’ = {1, 3, 5, 7}
(v) A \ B = {3, 5, 7}
(vi) B \ A = {0, 4, 6, 8}
(vii) (A ∪ B)’ = {1}
(viii) A’ ∪ B = {0, 1, 2, 4, 6, 8}
Q. 3. (a) U
A B
• 21
• 25
• 20 • 22 • 24
• 23
(b) (i) A ∪ B = {21, 23, 25}
(ii) A ∩ B = {23}
(iii) A’ = {20, 22, 24}
(iv) B’ = {20, 21, 22, 24, 25}
(v) (A ∩ B)’ = {20, 21, 22, 24, 25}
(vi) A’ ∪ B’ = {20, 21, 22, 24, 25}
(vii) B \ A = {} or ∅
(viii) (A \ B) ∪ (B \ A) = {21, 25}
(c) A \ B’ = {23}
B \ A’ = {23}
∴ A \ B’ = B \ A’
Answer is yes
(d) As B = A ∩ B
⇒ B ⊂ A
(e) Case 1 # (A ∪ B) = 0 ⇒ # (A ∩ B) = 0 In this case # (A ∪ B) = # (A ∩ B)
Case 2 # (A ∪ B) > 0 and # (A ∩ B) = 0 In this case # (A ∪ B) > # (A ∩ B)
Case 3 # (A ∪ B) > 0 and # (A ∩ B) > 0 As A ∩ B is a subset of A ∪ B,
# (A ∪ B) > # (A ∩ B)
Q. 4. (i) 23 = 8
(ii) {}, {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, {x, y, z}
(iii) {x, y, z}
Q. 5. (i) A ∪ B (iii) (A ∩ B)’
(ii) A ∩ B (iv) B \ A
Q. 6. (i) U
X Y
(ii) U
X Y
(iii)
X Y
U
(iv) U
X Y
1
2 Active Maths 2 (Strands 1–5): Ch 1 Solutions
Q. 7. (a) U = {1, 2, 3, 4, 5, 6, 7, 8}
O = {1, 3, 5, 7}
P = {3, 5, 7}
(b)
• 4 • 6 • 8• 2
U
O P
• 3
• 5
• 7
• 1
(c) (i) 4
(ii) 3
(iii) 3
(iv) 1
(v) 4
(vi) 4
(d) = ; =
Q. 8. (i)
271
U(500)
I(103) S(146)
2083 126
83 + 20 + 126 = 229
500 − 229 = 271
(ii) 83 + 126 = 209
209 ____ 500 × 100 ____ 1 % = 41.8%
Q. 9. (i)
170
U(400)
G(150) V(100)
20130 80
150 + 100 + 170 = 420
420 − 400 = 20
(ii) 20
(iii) 80 ____ 400 × 100 ____ 1 % = 20%
(iv) 130 + 80 = 210
210 ____ 400 × 100 ____ 1 % = 52.5%
Q. 10. (i) {0} is the set containing the element 0 whereas ∅ is the set containing zero elements.
(ii) 0 is the element ‘zero’ whereas {0} is the set containing the element 0.
(iii) X and Y are disjoint (mutually exclusive) sets.