Given the universal set U = \{x \in \mathbb{Z} : 1 \le x \le 15\}, let A = \{x \in U : x \… — JAMB UTME

Given the universal set $U = \{x \in \mathbb{Z} : 1 \le x \le 15\}$ , let $A = \{x \in U : x \text{ is a multiple of } 3\}$ and $B = \{x \in U : x \text{ is an even number}\}$ . What is the cardinality of $(A \cup B)'$ ?

$5$ Answer
B
$10$
C
$3$
D
$8$

Answer

The cardinality of

(A \cup B)'

5

The universal set contains

15

elements. The set of multiples of

3

within

U

contains

5

elements, and the set of even numbers contains

7

elements. Two numbers (

6

and

12

) belong to both sets. Subtracting the overlapping count gives

10

unique elements in the union. Subtracting

10

from the universal set total of

15

yields

5

elements in the complement.

Step-by-Step Solution

List elements of sets

U

A

, and

B

, and determine their cardinalities

U = \{1, 2, 3, \dots, 15\}

, so

n(U) = 15

A = \{3, 6, 9, 12, 15\}

(

n(A) = 5

B = \{2, 4, 6, 8, 10, 12, 14\}

(

n(B) = 7

Establishing explicit set memberships allows accurate counting of set operations.

Find the intersection

A \cap B

and compute the union cardinality

n(A \cup B)

A \cap B = \{6, 12\}

, so

n(A \cap B) = 2

. Using inclusion-exclusion:

n(A \cup B) = n(A) + n(B) - n(A \cap B) = 5 + 7 - 2 = 10

The union includes all elements that are either multiples of 3, even, or both, avoiding double counting.

Calculate the complement cardinality

n((A \cup B)')

relative to

U

n((A \cup B)') = n(U) - n(A \cup B) = 15 - 10 = 5

. Explicitly,

(A \cup B)' = \{1, 5, 7, 11, 13\}

The complement set

(A \cup B)'

consists of all elements in the universal set

U

that are not in

A \cup B

Key Concept

Complement of Set Union and Inclusion-Exclusion Principle

Estimated Time:1m 0s

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