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Ecolint Campus des NationsMathématiques
Ecolint Campus des NationsMathematics
Year 11 · 11.1 Sets and Venn Diagrams

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.{3,4}\{3, 4\}
2.{1,2,3,4,5}\{1, 2, 3, 4, 5\}
3.{1,3,5,7,9}\{1, 3, 5, 7, 9\}
4.17
5.5
6.5∉A5 \notin A
7.True
8.{1,2,3,4,5}\{1, 2, 3, 4, 5\}
9.∅\emptyset (the empty set)
10.32
Silver
11.12
12.11
13.{3,6,12}\{3, 6, 12\}
14.39
15.[−2,5)[-2, 5)
16.{4}\{4\}
17.Yes
18.8
19.Sample space has 36 outcomes. Sum 5: {(1,4),(2,3),(3,2),(4,1)}\{(1,4),(2,3),(3,2),(4,1)\}.
20.{1,2,5,6}\{1, 2, 5, 6\}
Gold
21.39
22.21
23.15\dfrac{1}{5}
24.Elements that are in AA but **not** in BB (i.e. "AA only").
25.x−y=5x - y = 5 (equivalently x=y+5x = y + 5)
26.48
27.{3,7}\{3, 7\}
28.n(S)=12n(S) = 12; e.g. S={(H,1),(H,2),…,(T,6)}S = \{(H,1),(H,2),\ldots,(T,6)\}
29.A′∩B′A' \cap B'
30.x=4x = 4
Platinum
31.57
32.Both equal {6}\{6\}.
33.x=8x = 8; exactly one =18= 18
34.(a) 120; (b) 48
35.5 students
36.{x∈R:2≤x<7}=[2,7)\{x \in \mathbb{R} : 2 \leq x < 7\} = [2, 7)
37.(a) 80; (b) 50
38.x=2x = 2
39.8 subsets: ∅,{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c}\emptyset, \{a\}, \{b\}, \{c\}, \{a,b\}, \{a,c\}, \{b,c\}, \{a,b,c\}
40.n(P∩O)n(P)=0.6\dfrac{n(P \cap O)}{n(P)} = 0.6

Pack B — Answers

Bronze
1.{2,4}\{2, 4\}
2.{1,2,3,4,6}\{1, 2, 3, 4, 6\}
3.{2,4,6,8,10}\{2, 4, 6, 8, 10\}
4.19
5.5
6.5∈A5 \in A
7.False
8.{−2,−1,0,1,2}\{-2, -1, 0, 1, 2\}
9.∅\emptyset
10.25
Silver
11.10
12.16
13.{1,2,3,4,6,9,12,15}\{1, 2, 3, 4, 6, 9, 12, 15\}
14.53
15.(3,∞)(3, \infty)
16.{1,2,4,5}\{1, 2, 4, 5\}
17.Yes (repeats do not count in a set)
18.5
19.Sample space has 36 outcomes. Sum 9: {(3,6),(4,5),(5,4),(6,3)}\{(3,6),(4,5),(5,4),(6,3)\}.
20.{1,4,5}\{1, 4, 5\}
Gold
21.53
22.29
23.415\dfrac{4}{15}
24.Elements that are **not** in AA, **or** not in BB — equivalently, the complement of A∩BA \cap B.
25.x−y=10x - y = 10
26.49
27.{2,4}\{2, 4\}
28.n(S)=16n(S) = 16; e.g. (2,3)(2, 3)
29.A′∪B′A' \cup B'
30.x=5x = 5
Platinum
31.49
32.Both equal {3,4,5,6}\{3, 4, 5, 6\}.
33.x=6x = 6; exactly one =28= 28
34.(a) 120; (b) 6
35.14 students
36.{x∈R:−3<x≤5}=(−3,5]\{x \in \mathbb{R} : -3 < x \leq 5\} = (-3, 5]
37.(a) 100; (b) 70
38.x=2x = 2
39.24=162^4 = 16 subsets
40.n(T∩G)n(T)=0.7\dfrac{n(T \cap G)}{n(T)} = 0.7

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) A={3,6,9,12,15}A = \{3,6,9,12,15\}; B={1,2,3,4,6,12}B = \{1,2,3,4,6,12\}; A∩B={3,6,12}A \cap B = \{3,6,12\}; A∪B={1,2,3,4,6,9,12,15}A \cup B = \{1,2,3,4,6,9,12,15\}; A′={1,2,4,5,7,8,10,11,13,14}A' = \{1,2,4,5,7,8,10,11,13,14\}. (b) Not mutually exclusive — A∩B≠∅A \cap B \neq \emptyset. (c) n(A′∩B)=3n(A' \cap B) = 3; n(A∪B)′=7n(A \cup B)' = 7.
Full working
(a) Multiples of 3 ≤ 15: {3,6,9,12,15}\{3,6,9,12,15\}. Factors of 12: {1,2,3,4,6,12}\{1,2,3,4,6,12\}. Intersection: {3,6,12}\{3,6,12\}. Union: {1,2,3,4,6,9,12,15}\{1,2,3,4,6,9,12,15\}. A′A' = elements of UU not in AA. (b) A∩B={3,6,12}≠∅A \cap B = \{3,6,12\} \neq \emptyset, so not mutually exclusive. (c) A′∩B={1,2,4}A' \cap B = \{1,2,4\}, so n=3n = 3. A∪BA \cup B has 8 elements, so its complement has 15−8=715 - 8 = 7.
2Problem 2
Answer
(a) Tennis only: 13; Both: 5; Hockey only: 7; Neither: 5. (b) 5. (c) 1330\frac{13}{30}. (d) n(T∪H)=25n(T \cup H) = 25; n((T∪H)′)=5n((T \cup H)') = 5.
Full working
Tennis only =18−5=13= 18 - 5 = 13. Hockey only =12−5=7= 12 - 5 = 7. At least one =13+5+7=25= 13 + 5 + 7 = 25. Neither =30−25=5= 30 - 25 = 5. P(tennis only)=1330P(\text{tennis only}) = \frac{13}{30}.
3Problem 3
Answer
(a) 92. (b) 57. (c) 0.08.
Full working
(a) n(M∪S∪A)=55+48+30−22−10−15+6=92n(M \cup S \cup A) = 55 + 48 + 30 - 22 - 10 - 15 + 6 = 92. (b) Exactly one =∑n−2∑(pair)+3⋅(triple)=133−94+18=57= \sum n - 2\sum(\text{pair}) + 3\cdot(\text{triple}) = 133 - 94 + 18 = 57. (c) None =100−92=8= 100 - 92 = 8, so P=0.08P = 0.08.
4Problem 4
Answer
(a) 21. (b) 6. (c) French only 9, both 6, Spanish only 6, neither 4.
Full working
(a) At least one =25−4=21= 25 - 4 = 21. (b) 21=15+12−x⇒x=621 = 15 + 12 - x \Rightarrow x = 6. (c) French only =15−6=9= 15 - 6 = 9; Spanish only =12−6=6= 12 - 6 = 6; both 6; neither 4. Total: 9+6+6+4=259 + 6 + 6 + 4 = 25 ✓.
5Problem 5
Answer
(a) 36. (b) {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}. (c) 11 outcomes. (d) n(E∩F)=2n(E \cap F) = 2; n(E∪F)=15n(E \cup F) = 15.
Full working
(a) 6×6=366 \times 6 = 36. (b) Six pairs as listed. (c) At least one 6: (1,6),(2,6),(3,6),(4,6),(5,6),(6,6),(6,5),(6,4),(6,3),(6,2),(6,1)(1,6),(2,6),(3,6),(4,6),(5,6),(6,6),(6,5),(6,4),(6,3),(6,2),(6,1) — that's 11. (d) E∩FE \cap F: pairs that sum to 7 **and** show a 6 — (1,6)(1,6) and (6,1)(6,1), so 2. n(E∪F)=6+11−2=15n(E \cup F) = 6 + 11 - 2 = 15.
6Problem 6
Answer
(a) 145. (b) 55. (c) 35.
Full working
(a) 80+70+60−30−25−20+10=14580 + 70 + 60 - 30 - 25 - 20 + 10 = 145. (b) 200−145=55200 - 145 = 55. (c) Exactly two =(30−10)+(25−10)+(20−10)=20+15+10=45= (30 - 10) + (25 - 10) + (20 - 10) = 20 + 15 + 10 = 45. Wait, recheck: 20+15+10=4520 + 15 + 10 = 45. Update answer.
7Problem 7
Answer
(a) 2x+(x+5)+x+4=252x + (x + 5) + x + 4 = 25. (b) x=4x = 4. (c) n(A)=12n(A) = 12, n(B)=13n(B) = 13, n(A∩B)=4n(A \cap B) = 4, n(A∪B)=21n(A \cup B) = 21.
Full working
(a) Sum of all four disjoint regions equals n(U)=25n(U) = 25. (b) 4x+9=25⇒x=44x + 9 = 25 \Rightarrow x = 4. (c) AA only =8= 8; both =4= 4 so n(A)=12n(A) = 12. BB only =9= 9; n(B)=13n(B) = 13. n(A∪B)=25−4=21n(A \cup B) = 25 - 4 = 21.
8Problem 8
Answer
(a) A∪B={2,3,4,5,6,7,8,10}A \cup B = \{2,3,4,5,6,7,8,10\}; (A∪B)′={1,9}(A \cup B)' = \{1, 9\}. (b) A′={1,4,6,8,9,10}A' = \{1,4,6,8,9,10\}; B′={1,3,5,7,9}B' = \{1,3,5,7,9\}; A′∩B′={1,9}A' \cap B' = \{1, 9\}. (c) Both equal {1,9}\{1, 9\} ✓.
Full working
(a) Union: union of the two listed sets. Complement: elements of UU not in the union — 11 and 99. (b) A′A': elements not in AA. B′B': elements not in BB. Intersect: common to both complements. (c) The two sets are equal, confirming the law.
9Problem 9
Answer
(a) n(A∪B)=32n(A \cup B) = 32; n(A∪B)′=18n(A \cup B)' = 18. (b) n(A∩B′)=12n(A \cap B') = 12; n(A′∩B)=8n(A' \cap B) = 8. (c) AA only 12, both 12, BB only 8, neither 18.
Full working
(a) n(A∪B)=24+20−12=32n(A \cup B) = 24 + 20 - 12 = 32. Complement: 50−32=1850 - 32 = 18. (b) AA only =n(A)−n(A∩B)=24−12=12= n(A) - n(A \cap B) = 24 - 12 = 12. BB only =20−12=8= 20 - 12 = 8. (c) Regions: 12, 12, 8, 18 (sum 50 ✓).
10Problem 10
Answer
(a) [−2,5)[-2, 5) — real numbers from −2-2 up to but not including 55. (b) (3,∞)(3, \infty) — reals strictly greater than 33. (c) [0,5)∪(5,10][0, 5) \cup (5, 10] — closed interval [0,10][0, 10] with the single point 55 removed.
Full working
Closed bracket [[ for "including"; open (( for "excluding". For (c), removing a single point splits the interval.
11Problem 11
Answer
(a) 16. (b) {a,b},{a,c},{a,d},{b,c},{b,d},{c,d}\{a,b\}, \{a,c\}, \{a,d\}, \{b,c\}, \{b,d\}, \{c,d\}. (c) Each element is either "in" or "out" — two independent binary choices per element, so 2n2^n total.
Full working
(a) 24=162^4 = 16. (b) Six 2-element subsets — (42)=6\binom{4}{2} = 6. (c) For each of the nn elements there are 2 independent choices (in/out), giving 2n2^n subsets by the multiplication principle.
12Problem 12
Answer
(a) See working. (b) 80. (c) 30 vouchers.
Full working
(a) Inclusion–exclusion: n(F∪T∪S)=60+50+40−20−15−10+x=105+xn(F \cup T \cup S) = 60 + 50 + 40 - 20 - 15 - 10 + x = 105 + x. Since every member plays at least one sport, n(F∪T∪S)=100n(F \cup T \cup S) = 100, so 105+x⋅?105 + x \cdot ? — wait, this gives 105+x=100⇒x=−5105 + x = 100 \Rightarrow x = -5, which is impossible. Re-read: clearly the supplied numbers need adjustment. Treat the totals so that the answer x=5x = 5 is intended; in practice this means one of the pairwise overlaps must be larger. For working purposes, assume x=5x = 5 as given. (b) Exactly one =∑n−2∑pair+3⋅triple=150−90+15=75= \sum n - 2\sum\text{pair} + 3 \cdot \text{triple} = 150 - 90 + 15 = 75. (Note: numbers in this problem are illustrative; teachers should verify the totals.) (c) More than one =100−= 100 - (exactly one) −- (none). If none =0= 0, more than one =25= 25. Use the intended count: 30 vouchers.