These eight original questions cover the whole set operations section of SPM Mathematics. Work each one on paper first, then compare with the answer.
Questions 1 and 2 are notation, 3 to 5 are two-set diagrams, 6 is a three-set diagram, and 7 and 8 translate words and symbols.
Practice questions
Question 1. ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {even numbers} and B = {multiples of 4}. List A ∩ B, A ∪ B and B′.
Answer
A = {2, 4, 6, 8, 10} and B = {4, 8}. Every multiple of 4 is also even, so B is inside A.
A ∩ B = {4, 8} and A ∪ B = {2, 4, 6, 8, 10}. B′ = {1, 2, 3, 5, 6, 7, 9, 10}.
Question 2. Using the sets in question 1, list A′ ∩ B and (A ∩ B)′. Explain why the answers differ.
Answer
A′ = {1, 3, 5, 7, 9}. None of these is in B, so A′ ∩ B = { }, the empty set.
(A ∩ B)′ = ξ with 4 and 8 removed = {1, 2, 3, 5, 6, 7, 9, 10}.
The prime acts on A alone in the first, and on the whole intersection in the second.
Question 3. In a class of 45, 28 read comics (C), 20 read novels (N) and 9 read both. How many read neither?
Answer
n(C ∪ N) = 28 + 20 − 9 = 39. Neither = 45 − 39 = 6.
Check with regions: comics only 19, both 9, novels only 11, neither 6, and 19 + 9 + 11 + 6 = 45.
Question 4. In a group of 30, 18 have a library card (L) and 15 have a bookshop card (B). Everyone has at least one card. How many have both?
Answer
Everyone is in the union, so n(L ∪ B) = 30. Then 30 = 18 + 15 − n(L ∩ B), which gives n(L ∩ B) = 33 − 30 = 3.
Check: library only 15, both 3, bookshop only 12, and 15 + 3 + 12 = 30.
Question 5. For a class of 40, n(F) = 22, n(B) = 17 and n(F ∩ B) = 8. A student writes 22 in the F-only region, 17 in the B-only region and 8 in the overlap, then finds “neither” by subtraction. What goes wrong, and what is the correct value of neither?
Answer
The student’s circles hold 22 + 17 + 8 = 47, which is more than 40, so “neither” comes out as −7. A negative count signals double counting.
The set totals already include the overlap, so F only = 22 − 8 = 14 and B only = 17 − 8 = 9. Inside the circles: 14 + 8 + 9 = 31, so neither = 40 − 31 = 9.
Question 6. In a class of 50 students, n(P) = 20 for Physics, n(C) = 22 for Chemistry and n(B) = 18 for Biology. Also n(P ∩ C) = 8, n(P ∩ B) = 6, n(C ∩ B) = 7, n(P ∩ C ∩ B) = 3, and 8 students take none. Find how many take exactly two of the three subjects, and confirm the total.
Answer
Pairs only: P and C = 8 − 3 = 5, P and B = 6 − 3 = 3, C and B = 7 − 3 = 4. Exactly two subjects: 5 + 3 + 4 = 12.
Singles only: P = 20 − (5 + 3 + 3) = 9, C = 22 − (5 + 4 + 3) = 10, B = 18 − (3 + 4 + 3) = 8.
Total inside: 9 + 10 + 8 + 5 + 3 + 4 + 3 = 42, and 42 + 8 outside = 50, which matches the class size.
Question 7. Write in set notation: (a) students who play chess (H) but not draughts (D), (b) students who play neither.
Answer
(a) H ∩ D′. (b) (H ∪ D)′.
Writing H′ ∪ D′ for (b) is a common slip, because that means not in H or not in D, which includes students who play one of the games.
Question 8. With ξ = {1, 2, …, 12}, X = {factors of 12} and Y = {odd numbers}, find n(X ∩ Y′) and n(X ∪ Y).
Answer
X = {1, 2, 3, 4, 6, 12} and Y = {1, 3, 5, 7, 9, 11}.
X ∩ Y′ are the elements of X that are even: {2, 4, 6, 12}, so n = 4.
X ∩ Y = {1, 3}, so n(X ∪ Y) = 6 + 6 − 2 = 10.
If you got some wrong
Errors in questions 1 and 2 point back to reading union, intersection and complement notation. Questions 3 to 5 link to two-set Venn diagram questions.
Question 6 uses the method in three-set Venn diagram questions. Questions 7 and 8 are covered in translating word statements into set notation.
The mistake log tool helps you note which slips repeat. If they do, online one-to-one Mathematics tuition lets a teacher work on them with you.