Grade 7 Maths Questions on Set Theory
With Solutions and Explanations
Detailed solutions and full explanations to grade 7 Maths questions on set theory are presented.

Problem 1
Give the cardinality of set A and B defined by
A = {a, b, c, d} and B = {1, 4, 7, 9, 10, 12, 23}
Solution
The cardinality of a set is equal to the number of all distinct elements in the set. Set A has 4 distinct elements and set B has 7 distinct , hence
 A  = 4 , the notation  A  means cardinality of A
and  B  = 7

Problem 2
Give the cardinality of
a) the set of all months beginning with M.
b) the set of all vowels in the English alphabet.
Solution
let us find the elements of the set S of all months beginning with M
S = {March , May}
Hence cardinality of S is
 S  = 2
let us find the elements of the set V of all vowels in the English alphabet
V = {a,e,i,o,u}
Hence cardinality of S is
 S  = 5

Problem 3
Sets A and B are defined by
A = {3, 5, 7, 8} and B = {x, y, z}
Answer by true or false
a) 3 ∈ A
b) 3 ∈ B
c) x ∉ A
d) z ∈ B
e) 8 ∈ B
Solution
The symbol ∈ means "is an element of".
a) 3 ∈ A , true because 3 is in an element of set A.
b) 3 ∈ B, false because 3 is not an element of B.
c) x ∉ A , true because x is not an element of A.
d) z ∈ B , true because z is an element of B.
e) 8 ∈ B , false because 8 is not an element of B.

Problem 4
List all terms in each set
a) The set of all positive even numbers less than or equal to 10
b) The set of all letters in the word "AUSTRALIA".
c) The set of all whole numbers greater than 3 and smaller than 16, and divisible by 3.
d) The set of all whole numbers greater than 5 and smaller than 35, and divisible by 5.
e) The set of all prime numbers divisible by 3.
f) The set of all numbers whose absolute value is equal to 7.
Solution
a) {2,4,6,8,10}
b) {A,U,S,T,R,L,I}
c) {6,9,12,15}
d) {10,15,20,25,30}
e) {3}
f) {7,7}

Problem 5
Set A, B, C and D are defined by:
A = {2,3,4,5,6,7}
B = {3,5,7}
C = {3,5,7,20,25,30}
D = {20,25,30}
Answer by true or false
a) A ⊂ B
b) B ⊂ A
c) B ⊄ C
d) C ⊂ D
e) D ⊄ A
Solution
a) A ⊂ B means A is a subset of B and is true if all elements of A are also elements of B. Elements 2,4 and 6 of A are not elements of B and therefore A ⊂ B is false.
b) B ⊂ A is true since all elements of B are also elements of A.
c) B ⊄ C is false. Since all elements of B are also elements of C, then B is a subset of C.
d) C ⊂ D is false since elements 3,5 and 7 are elements of C but not of D.
d) D ⊄ A is true since elements 20,25 and 30 are elements of D but not of A.

Problem 6
Use set A, B, C and D defined in question 5 to find
a) A ⋃ B
b) A ⋂ B
c) B ⋂ C
d) C ⋃ B
e) D ⋂ C
f) (A ⋂ B) ⋂ C
g) (A ⋃ B) ⋂ (C ⋃ D)
h) (A ⋃ B) ⋃ (C ⋃ D)
Solution
a) A ⋃ B is the set of all elements of A and B and each element is listed once only. Hence
A ⋃ B = {2,3,4,5,6,7} = A
b) A ⋂ B is the set of all elements that are common to set A and set B. Hence
A ⋂ B = {3,5,7} = B
c) B ⋂ C is the set of all elements that are common to set B and set C. Hence
B ⋂ C = {3,5,7} = B
d) C ⋃ B is the set of all elements in C and B, each element listed once. Hence
C ⋃ B = {3,5,7,20,25,30} = C
e) D ⋂ C is the set of all elements that are common to set D and set C. Hence
D ⋂ C = {20,25,30} = D
f) We first determine A ⋂ B
A ⋂ B = B
Next, we determine
(A ⋂ B) ⋂ C = B ⋂ C = B
g) We first determine (A ⋃ B)
(A ⋃ B) = A
Next, we determine (C ⋃ D)
(C ⋃ D) = C
We now have
(A ⋃ B) ⋂ (C ⋃ D)
= A ⋂ C = {3,5,7} = B
h) We first determine (A ⋃ B)
(A ⋃ B) = A
Next we determine (C ⋃ D)
(C ⋃ D) = C
We now have
(A ⋃ B) ⋃ (C ⋃ D)
= A ⋃ C = {2,3,4,5,6,7,20,25,30}
More References and links
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