Exam practice
Every question from the end of every unit, with hints and answers to open. After you check an answer, mark how you did, and come back to the ones you didn’t get.
27 questions
Unit 1 · Sets
Exercise 1.1
Question 1
In Figure 1.29, F is the set of students who like football and V the set who like volleyball. Rambahadur and Harkabahadur like neither game. Write these sets by the listing method:
- a.F
- b.V
- c.F ∪ V
- d.F ∩ V
- e.U
Answer
- a.F = {Chhiring, Dorje, Hari, Maya, Aasha, Ganesh}
- b.V = {Batuli, Dolma, Suntali, Maya, Aasha, Ganesh}
- c.F ∪ V = {Chhiring, Dorje, Hari, Batuli, Dolma, Suntali, Maya, Aasha, Ganesh}
- d.F ∩ V = {Maya, Aasha, Ganesh}
- e.U = {Chhiring, Dorje, Hari, Batuli, Dolma, Suntali, Maya, Aasha, Ganesh, Rambahadur, Harkabahadur}
Question 2
If U = {a, b, c, d, e, f, g, h, i, j, k}, A = {a, c, e, f, g, i, k} and B = {b, d, i, j, k, h}, find these sets and show each in a Venn diagram:
- a.A ∩ B
- b.B ∪ A
- c.A − B
- d.B − A
Answer
- a.A ∩ B = {i, k} (shade the overlap).
- b.B ∪ A = {a, b, c, d, e, f, g, h, i, j, k} = U (shade both circles).
- c.A − B = {a, c, e, f, g} (shade A outside B).
- d.B − A = {b, d, h, j} (shade B outside A).
Question 3
If U = {x:x is a whole number from 1 to 30}, A = {x:x is a multiple of 3 from 1 to 30}, B = {x:x is a multiple of 4 from 1 to 30} and C = {x:x is a multiple of 5 from 1 to 30}, write these sets by the listing method and show them in a Venn diagram:
- a.A − B
- b.B − A
- c.A − C
- d.B − C
- e.A ∪ B
- f.A ∪ B ∪ C
- g.A ∩ B ∩ C
- h.B ∪ C
Answer
A = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}, B = {4, 8, 12, 16, 20, 24, 28}, C = {5, 10, 15, 20, 25, 30}.
- a.A − B = {3, 6, 9, 15, 18, 21, 27, 30}
- b.B − A = {4, 8, 16, 20, 28}
- c.A − C = {3, 6, 9, 12, 18, 21, 24, 27}
- d.B − C = {4, 8, 12, 16, 24, 28}
- e.A ∪ B = {3, 4, 6, 8, 9, 12, 15, 16, 18, 20, 21, 24, 27, 28, 30}
- f.A ∪ B ∪ C = {3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 21, 24, 25, 27, 28, 30}
- g.A ∩ B ∩ C = {}: a multiple of 3, 4 and 5 is a multiple of 60.
- h.B ∪ C = {4, 5, 8, 10, 12, 15, 16, 20, 24, 25, 28, 30}, so B ∪ C = {1, 2, 3, 6, 7, 9, 11, 13, 14, 17, 18, 19, 21, 22, 23, 26, 27, 29}.
Question 4
If U = {a, b, c, d, e, f, g, h, i, j, k}, A = {a, c, d, f} and B = {g, h, i}:
- a.Find these sets:
- i.A
- ii.B
- iii.A ∪ B
- iv.A ∩ B
- v.A ∪ B
- vi.A ∩ B
- b.Which of the sets in part a are equal?
Answer
- a.i. A = {b, e, g, h, i, j, k}
- ii.B = {a, b, c, d, e, f, j, k}
- iii.A ∪ B = {a, b, c, d, e, f, g, h, i, j, k} = U
- iv.A ∩ B = {b, e, j, k}
- v.A ∪ B = {a, c, d, f, g, h, i}, so A ∪ B = {b, e, j, k}
- vi.A ∩ B = ∅, so A ∩ B = U
- b.iv and v are equal, A ∩ B = A ∪ B, and iii and vi are equal, A ∪ B = A ∩ B = U. These are De Morgan’s laws, true for any two sets.
Question 5
If U = {natural numbers from 1 to 12}, E = {even numbers from 1 to 12}, O = {odd numbers from 1 to 12} and P = {prime numbers from 1 to 12}, find these sets and show them in a Venn diagram:
- a.E
- b.O
- c.P
- d.E ∪ P
- e.P ∩ O
- f.P − O
- g.P ∩ E
- h.E ∪ O ∪ P
- i.E ∩ O ∩ P
- j.E ∪ P − (P ∩ O)
- k.P ∪ (E ∩ O)
Answer
E = {2, 4, 6, 8, 10, 12}, O = {1, 3, 5, 7, 9, 11}, P = {2, 3, 5, 7, 11}.
- a.E = {1, 3, 5, 7, 9, 11}
- b.O = {2, 4, 6, 8, 10, 12}
- c.P = {1, 4, 6, 8, 9, 10, 12}
- d.E ∪ P = {2, 3, 4, 5, 6, 7, 8, 10, 11, 12}, so E ∪ P = {1, 9}
- e.P ∩ O = {3, 5, 7, 11}, so P ∩ O = {1, 2, 4, 6, 8, 9, 10, 12}
- f.P − O = {2}, so P − O = {1, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
- g.P ∩ E = {2}: 2 is the only even prime.
- h.E ∪ O ∪ P = U, so E ∪ O ∪ P = {}
- i.E ∩ O ∩ P = {}, so E ∩ O ∩ P = U = {1, 2, 3, …, 12}
- j.{1, 9} − {3, 5, 7, 11} = {1, 9}
- k.E ∩ O = {}, so P ∪ (E ∩ O) = P = {1, 4, 6, 8, 9, 10, 12}
Question 6
If U = {m, n, o, p, q, r, s, t, u, v}, A = {q, r, s, t, u, v}, B = {n, o, p, q, r} and C = {m, u, s, t, q, r}, find these sets and show them in Venn diagrams:
- a.A ∩ B
- b.(A ∪ B) ∩ C
- c.A ∪ B ∪ C
- d.A ∩ B ∩ C
- e.A − B
- f.A ∪ B ∪ C
- g.A ∩ B
- h.A
- i.(A ∩ C) ∪ B
Answer
- a.A ∩ B = {q, r}
- b.A ∪ B = {n, o, p, q, r, s, t, u, v}, so (A ∪ B) ∩ C = {q, r, s, t, u}
- c.A ∪ B ∪ C = {m, n, o, p, q, r, s, t, u, v} = U
- d.A ∩ B ∩ C = {q, r}
- e.A − B = {s, t, u, v}
- f.A ∪ B ∪ C = {}
- g.A ∩ B = {m, n, o, p, s, t, u, v}
- h.A = {m, n, o, p}
- i.A ∩ C = {q, r, s, t, u}, so (A ∩ C) ∪ B = {n, o, p, q, r, s, t, u}
For each Venn diagram, shade the regions of that part’s letters.
Question 7
Write the set shaded in the Venn diagram in set notation.
Answer
A − B (the part of A outside B).
Write the set shaded in the Venn diagram in set notation.
Answer
B − A.
Write the set shaded in the Venn diagram in set notation.
Answer
B, which here is also B − A, since the sets are disjoint.
Write the set shaded in the Venn diagram in set notation.
Answer
A, that is U − A.
Write the set shaded in the Venn diagram in set notation.
Answer
A ∪ B ∪ C.
Write the set shaded in the Venn diagram in set notation.
Answer
A ∩ B ∩ C.
Question 8
If P and Q are overlapping subsets of a universal set U, show these sets in Venn diagrams:
- a.P − Q
- b.Q − P
- c.(P − Q) ∪ P
- d.P ∩ (Q − P)
Answer
a. Shade the part of P outside Q:
b. Shade the part of Q outside P:
c. P − Q is already part of P, so (P − Q) ∪ P = P: shade the whole of P.
d. Q − P has nothing in common with P, so P ∩ (Q − P) = ∅: nothing is shaded.
Question 9
Write a universal set U and two of its subsets X and Y. Then write these sets by the listing method:
- a.X ∪ Y
- b.X ∩ Y
- c.X
- d.X ∩ Y
Answer
Any correct example. For instance, U = {1, 2, 3, 4, 5, 6, 7, 8}, X = {1, 2, 3, 4}, Y = {3, 4, 5, 6}. Then X = {5, 6, 7, 8} and Y = {1, 2, 7, 8}.
- a.X ∪ Y = {1, 2, 5, 6, 7, 8}
- b.X ∩ Y = {1, 2} (the same as X − Y)
- c.X = X = {1, 2, 3, 4}
- d.X ∩ Y = {7, 8}
Check with De Morgan’s laws: X ∩ Y = {3, 4}, and its complement is a’s answer; X ∪ Y = {1, 2, 3, 4, 5, 6}, and its complement is d’s.
Question 10
Make these sets from the students of your class: a. the set of all the students, b. the set of girls, c. the set of boys. Which is the universal set, and which are its subsets? Write them with the right symbols. Then find the complement of each set.
Answer
Your class’s names. For example, in a class of six: U = {Sneha, Tashi, Bibek, Rizwan, Junu, Nabin}, G = {Sneha, Tashi, Junu}, B = {Bibek, Rizwan, Nabin}.
U (all the students) is the universal set; G ⊂ U and B ⊂ U.
G = U − G = B, B = U − B = G, and U = ∅. In any class the girls and the boys are complements of each other, as long as everyone is one or the other.
Exercise 1.2
Question 1
From Figure 1.53, find
- i.n(S)
- ii.n(T)
- iii.n(U)
- iv.n(S ∩ T)
- v.n(R ∪ T)
- vi.n(S ∩ R ∩ T)
- vii.n(S ∪ R ∪ T)
- viii.n(S ∪ R ∪ T)
- ix.n₀(S)
- x.n₀(S ∩ R)
- xi.n(R)
Answer
S = {1, 2, 3, 5, 6}, T = {1, 2, 4, 7, 8, 9}, R = {1, 3, 4, 10}, U = {1, 2, …, 12}.
- i.5
- ii.6
- iii.12
- iv.2 (1, 2)
- v.8 (1, 2, 3, 4, 7, 8, 9, 10)
- vi.1 (1)
- vii.10 (1 to 10)
- viii.2 (11, 12)
- ix.2 (5, 6)
- x.1 (3)
- xi.8 (all but 1, 3, 4, 10)
If A and B are disjoint sets with n(A) = 30 and n(B) = 35, find n(A ∪ B).
Answer
Disjoint, so n(A ∪ B) = n(A) + n(B) = 30 + 35 = 65.
Question 2
From Figure 1.54, find
- i.n(A)
- ii.n(B)
- iii.n(A ∪ B)
- iv.n(A ∩ B)
- v.n₀(A)
- vi.n₀(B)
Answer
A = {a, b, c, d, e, f, g} and B = {a, b, c, h, i, j}.
- i.7
- ii.6
- iii.10
- iv.3
- v.4
- vi.3
Check: 7 + 6 − 3 = 10. ✓
If U = {a, b, c, d, e, f, g}, A = {c, d, e, f}, B = {a, b, e, f} and C = {d, e, f, g}, find
- i.n(A − B)
- ii.n(B − C)
- iii.n(A − C)
- iv.n(A)
- v.n(A ∪ B)
- vi.n{(A ∪ B) − (A ∩ B)}
- vii.n{(A − B) ∪ (B − A)}
Answer
- i.A − B = {c, d}: 2
- ii.B − C = {a, b}: 2
- iii.A − C = {c}: 1
- iv.A = {a, b, g}: 3
- v.A ∪ B = {a, b, c, d, e, f}, so A ∪ B = {g}: 1
- vi.A ∩ B = {e, f}, so (A ∪ B) − (A ∩ B) = {a, b, c, d}: 4
- vii.(A − B) ∪ (B − A) = {c, d} ∪ {a, b} = {a, b, c, d}: 4
vi and vii are the same set: the symmetric difference, the elements in exactly one of A and B.
Question 3
If U = {natural numbers less than 20}, A = {even numbers less than 20}, B = {prime numbers less than 20} and C = {square numbers less than 20}, find
- i.n(U)
- ii.n(C)
- iii.n(A ∩ B)
- iv.n(B − C)
- v.n(A)
- vi.n(A ∪ C)
- vii.n(A ∩ B)
Answer
U = {1, 2, …, 19}, A = {2, 4, 6, 8, 10, 12, 14, 16, 18}, B = {2, 3, 5, 7, 11, 13, 17, 19}, C = {1, 4, 9, 16}.
- i.19
- ii.4
- iii.A ∩ B = {2}: 1
- iv.No square is prime, so B − C = B: 8
- v.n(A) = 19 − 9 = 10
- vi.A ∩ C = {4, 16}, so n(A ∪ C) = 9 + 4 − 2 = 11
- vii.n(A ∩ B) = 19 − 1 = 18
If U = {x:x is a natural number from 1 to 20}, A = {y:y is a prime number}, B = {z:z is a factor of 18} and C = {p:p is a multiple of 3 less than 20}, show these sets in Venn diagrams and find
- i.n(A ∪ B)
- ii.n(B ∪ C)
- iii.n(A ∪ B ∪ C)
- iv.n(A ∩ B ∩ C)
- v.n₀(A)
- vi.n(A − B)
- vii.n(A ∩ B)
- viii.n(C)
Answer
In U: A = {2, 3, 5, 7, 11, 13, 17, 19}, B = {1, 2, 3, 6, 9, 18}, C = {3, 6, 9, 12, 15, 18}.
- i.A ∩ B = {2, 3}, so n(A ∪ B) = 8 + 6 − 2 = 12
- ii.B ∩ C = {3, 6, 9, 18}, so n(B ∪ C) = 6 + 6 − 4 = 8
- iii.A ∪ B ∪ C = {1, 2, 3, 5, 6, 7, 9, 11, 12, 13, 15, 17, 18, 19}: 14
- iv.A ∩ B ∩ C = {3}: 1
- v.A only = {5, 7, 11, 13, 17, 19}: 6
- vi.A − B = {5, 7, 11, 13, 17, 19}: 6 (here the same as A only, since no prime outside B is in C)
- vii.n(A ∩ B) = 20 − 2 = 18
- viii.n(C) = 20 − 6 = 14
(If U is taken as the natural numbers less than 20, as in part a, the last two are 17 and 13.)
Question 4
From Figure 1.56, find
- a.n(P ∪ Q)
- b.n(P ∩ Q)
- c.n(P ∪ Q)
- d.n(P ∩ Q)
Answer
P and Q are disjoint, with nothing outside them: n(U) = 8 + 6 = 14.
- a.n(P ∪ Q) = 8 + 6 = 14
- b.P ∩ Q = ∅: 0
- c.P ∪ Q = ∅: 0
- d.n(P ∩ Q) = 14 − 0 = 14
Question 5
Using Figure 1.57, find
- a.n(S)
- b.n(T)
- c.n(S ∩ T)
- d.n(S ∪ T)
- e.n₀(T)
- f.n₀(S)
- g.n(T)
- h.n(S ∪ T)
- i.n(S ∩ T)
- j.n(U)
Answer
- a.7
- b.7
- c.2 (3, 5)
- d.7 + 7 − 2 = 12
- e.5
- f.5
- g.13 − 7 = 6
- h.1 (7)
- i.13 − 2 = 11
- j.13
Question 6
Look at Figure 1.58 and find
- a.n(A)
- b.n(B)
- c.n(C)
- d.n(A ∪ B)
- e.n(A ∪ B ∪ C)
- f.n(A ∩ B ∩ C)
- g.n(A ∪ B ∪ C)
- h.n₀(A)
- i.n₀(C)
- j.n(U)
Answer
A = {1, 2, 3, 5}, B = {1, 2, 4, 6}, C = {1, 3, 4, 7}.
- a.4
- b.4
- c.4
- d.6 (1 to 6)
- e.7
- f.1
- g.1 (8)
- h.1 (5)
- i.1 (7)
- j.8
Question 7
Use Figure 1.59 to verify these relations:
- a.n(A) = n(U − A)
- b.n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
- c.n₀(A) = n(A − B)
- d.n(A ∪ B) = n₀(A) + n(B)
- e.n(A ∪ C) = n₀(A) + n(C)
- f.n(B ∪ C) = n₀(B) + n(C)
- g.A ∪ B ∪ C = U − (A ∪ B ∪ C)
In c to f, “only” is with respect to the two sets in the relation: in d, n₀(A) is A outside B; in e, A outside C.
Answer
U = {0, 1, 2, …, 12}, n(U) = 13; A = {0, 1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, C = {2, 3, 4, 8, 9}.
- a.A = {6, 7, 8, 9, 10, 11, 12}, so n(A) = 7; U − A is the same set, and 13 − 6 = 7. ✓
- b.A ∪ B = {0, 1, …, 8}: 9. n(A) + n(B) − n(A ∩ B) = 6 + 5 − 2 = 9. ✓
- c.A outside B = {0, 1, 2, 3} = A − B: both 4. ✓
- d.n₀(A) + n(B) = 4 + 5 = 9 = n(A ∪ B). ✓
- e.A ∪ C = {0, 1, 2, 3, 4, 5, 8, 9}: 8. A outside C is {0, 1, 5}, and 3 + 5 = 8. ✓
- f.B ∪ C = {2, 3, 4, 5, 6, 7, 8, 9}: 8. B outside C is {5, 6, 7}, and 3 + 5 = 8. ✓
- g.A ∪ B ∪ C = {0, 1, …, 9}, and the elements of U outside it are {10, 11, 12}, which is A ∪ B ∪ C. ✓
Mixed exercise
Question 1
Think of the two words SUSTAINABILITY and URBANIZATION as sets of letters.
- a.Write the set of letters of each word, by the listing method and in set-builder form.
- b.Show the two sets in a Venn diagram.
- c.Write the set of letters that are in both words.
- d.How many letters are in exactly one of the two words?
- e.If U = {a, b, c, …, z}, how many letters are in neither word?
Answer
a. S = {x:x is a letter of SUSTAINABILITY} = {s, u, t, a, i, n, b, l, y}
R = {x:x is a letter of URBANIZATION} = {u, r, b, a, n, i, z, t, o}
Repeated letters are written once: SUSTAINABILITY has 14 letters but only 9 different ones.
b.
- c.S ∩ R = {a, b, i, n, t, u}
- d.S − R = {s, l, y} and R − S = {o, r, z}: 6 letters.
- e.n(S ∪ R) = 9 + 9 − 6 = 12, so n(S ∪ R) = 26 − 12 = 14.
Question 2
U = {whole numbers from 1 to 25}, A = {multiples of 3 from 1 to 25}, B = {multiples of 4 from 1 to 25} and C = {multiples of 5 from 1 to 25}.
- a.Write all these sets by the listing method.
- b.Show the three sets in a Venn diagram.
- c.Write the set of numbers in all three sets.
- d.How many numbers are in all three sets, and how many in none of them?
Answer
a. U = {1, 2, 3, …, 25}, A = {3, 6, 9, 12, 15, 18, 21, 24}, B = {4, 8, 12, 16, 20, 24}, C = {5, 10, 15, 20, 25}.
b.
- c.A ∩ B ∩ C = {}: the smallest number in all three is 60.
- d.None is in all three. A ∪ B ∪ C has 5 + 3 + 3 + 2 + 1 + 1 = 15 numbers (count the regions), so 10 are in none: 1, 2, 7, 11, 13, 14, 17, 19, 22, 23.
