Sets and Set Operations Practice Questions — JAMB UTME

Question 1Question →

Given the universal set $\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 15\}$ , with subsets $P = \{x \in \mathcal{U} : x \text{ is a prime number}\}$ and $Q = \{x \in \mathcal{U} : x \text{ is an odd number}\}$ , what is the number of elements in the set $(P \cup Q)'$ ?

Show answer & explanation

Answer: 6; six; 6 elements

Answer

The number of elements in

(P \cup Q)'

is 6.

The set

P \cup Q

consists of all numbers from 1 to 15 that are either prime or odd:

\{1, 2, 3, 5, 7, 9, 11, 13, 15\}

. The complement

(P \cup Q)'

relative to

\mathcal{U}

contains all elements of

\mathcal{U}

that are neither prime nor odd, which are the even composite numbers:

\{4, 6, 8, 10, 12, 14\}

. Counting these elements gives 6.

Step-by-Step Solution

1

List all elements of the universal set

\mathcal{U}

, subset

P

, and subset

Q

.

\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}

,

P = \{2, 3, 5, 7, 11, 13\}

, and

Q = \{1, 3, 5, 7, 9, 11, 13, 15\}

.

Listing the explicit elements allows precise execution of set union and complement operations.

2

Find the union

P \cup Q

.

P \cup Q = \{1, 2, 3, 5, 7, 9, 11, 13, 15\}

.

The union combines all unique elements that belong to either set

P

, set

Q

, or both.

3

Determine the complement set

(P \cup Q)'

relative to

\mathcal{U}

and count its elements.

(P \cup Q)' = \{4, 6, 8, 10, 12, 14\}

, which contains 6 elements.

The complement set consists of all elements in

\mathcal{U}

that are not present in

P \cup Q

.

Key Concept

Complement of Set Union

Question 2Question →

Given the universal set $\mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ , set $A = \{1, 2, 4, 5, 8\}$ , and set $B = \{2, 3, 5, 7, 9\}$ , what is $(A \cup B)'$ ?

\{6, 10\}

\{1, 4, 8\}

\{3, 6, 7, 9, 10\}

\{2, 5\}

Show answer & explanation

Answer: $\{6, 10\}$

Answer

The set

\{6, 10\}

Combining set

A

and set

B

gives

A \cup B = \{1, 2, 3, 4, 5, 7, 8, 9\}

. The complement

(A \cup B)'

contains the elements of the universal set

\mathcal{E}

that are not in this union, which are

6

and

10

, giving

\{6, 10\}

.

Step-by-Step Solution

1

Find the union of set

A

and set

B

(

A \cup B

)

A \cup B = \{1, 2, 3, 4, 5, 7, 8, 9\}

The union combines all distinct elements present in set

A

, set

B

, or both.

2

Determine the complement of

(A \cup B)

with respect to the universal set

\mathcal{E}

(A \cup B)' = \mathcal{E} \setminus (A \cup B) = \{6, 10\}

The complement consists of all elements in the universal set

\mathcal{E}

that are not present in

A \cup B

.

Key Concept

Set Union and Set Complement

Question 3Question →

If set $A = \{a, b, c, d, e\}$ and set $B = \{c, d, e, f, g\}$ , what is the number of elements in the set $(A \setminus B) \cup (B \setminus A)$ ?

Show answer & explanation

Answer: 4; four

Answer

The number of elements in

(A \setminus B) \cup (B \setminus A)

is 4.

The set difference

A \setminus B

consists of elements in

A

that are not in

B

, which gives

\{a, b\}

. Similarly,

B \setminus A

consists of elements in

B

that are not in

A

, giving

\{f, g\}

. The union

(A \setminus B) \cup (B \setminus A)

is

\{a, b, f, g\}

, which has a cardinality of 4.

Step-by-Step Solution

1

Find the relative difference

A \setminus B

A \setminus B = \{a, b\}

Remove elements of

B

present in

A

.

2

Find the relative difference

B \setminus A

B \setminus A = \{f, g\}

Remove elements of

A

present in

B

.

3

Take the union of the two set differences

(A \setminus B) \cup (B \setminus A) = \{a, b, f, g\}

Combine elements from both set differences.

4

Count the number of elements in the resulting set

4 elements

The set

\{a, b, f, g\}

contains 4 distinct elements.

Key Concept

Symmetric Difference of Two Sets

Estimated Time:45s

Question 4Question →

Given the universal set

\mathcal{E} = \{x \in \mathbb{Z} : 1 \le x \le 20\}

, and subsets:

A = \{x \in \mathcal{E} : x \text{ is a multiple of } 3\}

B = \{x \in \mathcal{E} : x \text{ is a multiple of } 4\}

C = \{x \in \mathcal{E} : x \text{ is a prime number}\}

What is the number of elements in the set $(A \cup B)' \cap C$ ?

7

8

5

10

Show answer & explanation

Answer: 7

Answer

7

The universal set contains integers from 1 to 20. Subset

A

has multiples of 3, subset

B

has multiples of 4, and subset

C

contains the prime numbers

\{2, 3, 5, 7, 11, 13, 17, 19\}

. The set

A \cup B

contains all multiples of 3 or 4 within the range, namely

\{3, 4, 6, 8, 9, 12, 15, 16, 18, 20\}

. The complement

(A \cup B)'

consists of elements in the universal set not in

A \cup B

:

\{1, 2, 5, 7, 10, 11, 13, 14, 17, 19\}

. Taking the intersection of

(A \cup B)'

with

C

filters this list to only the prime numbers:

\{2, 5, 7, 11, 13, 17, 19\}

, giving a cardinality of 7.

Step-by-Step Solution

1

List the elements of the universal set and the given subsets.

\mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}

A = \{3, 6, 9, 12, 15, 18\}

B = \{4, 8, 12, 16, 20\}

C = \{2, 3, 5, 7, 11, 13, 17, 19\}

Explicit listing allows accurate evaluation of set operations.

2

Find the union

A \cup B

.

A \cup B = \{3, 4, 6, 8, 9, 12, 15, 16, 18, 20\}

The union combines all elements present in either

A

or

B

.

3

Determine the complement

(A \cup B)'

relative to

\mathcal{E}

.

(A \cup B)' = \{1, 2, 5, 7, 10, 11, 13, 14, 17, 19\}

The complement contains all elements of the universal set

\mathcal{E}

that are not in

A \cup B

.

4

Find the intersection

(A \cup B)' \cap C

and count its cardinality.

(A \cup B)' \cap C = \{2, 5, 7, 11, 13, 17, 19\}

, which contains 7 elements.

The intersection yields elements common to both

(A \cup B)'

and prime set

C

.

Key Concept

Set Complement and Intersections

Question 5Question →

In a group of 100 candidates preparing for an entrance examination, 48 registered for Mathematics, 45 for Physics, and 40 for Chemistry. If 18 candidates registered for both Mathematics and Physics, 15 for both Physics and Chemistry, 20 for both Mathematics and Chemistry, and 8 registered for none of these three subjects, how many candidates registered for all three subjects?

Show answer & explanation

Answer: 12; 12 candidates; 12 students

Answer

12 candidates registered for all three subjects.

By the Principle of Inclusion-Exclusion, the total number of candidates taking at least one subject is

100 - 8 = 92

. Summing the individual totals gives

48 + 45 + 40 = 133

. Subtracting the pairwise intersections gives

133 - (18 + 15 + 20) = 80

. Adding the intersection of all three sets must equal 92, yielding

92 - 80 = 12

.

Step-by-Step Solution

1

Determine the total number of candidates who registered for at least one of the three subjects.

n(M \cup P \cup C) = 100 - 8 = 92

Subtracting the number of candidates who registered for none of the subjects from the universal set gives the cardinality of the union.

2

Apply the Principle of Inclusion-Exclusion for three sets.

n(M \cup P \cup C) = n(M) + n(P) + n(C) - n(M \cap P) - n(P \cap C) - n(M \cap C) + n(M \cap P \cap C)

This formula relates the individual set sizes, pair intersections, and triple intersection to the union.

3

Substitute the known values into the inclusion-exclusion equation.

92 = 48 + 45 + 40 - 18 - 15 - 20 + n(M \cap P \cap C)

Insert the given cardinalities into the formula.

4

Simplify and solve for

n(M \cap P \cap C)

.

92 = 80 + n(M \cap P \cap C) \implies n(M \cap P \cap C) = 92 - 80 = 12

Isolate the unknown triple intersection term.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

Question 6Question →

Given the universal set $\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 30\}$ , let $A = \{x \in \mathcal{U} : x \text{ is a multiple of } 4\}$ and $B = \{x \in \mathcal{U} : x \text{ is a perfect square}\}$ . What is the number of elements in $(A \cup B)'$ ?

20

18

10

22

Show answer & explanation

Answer: 20

Answer

The cardinality of

(A \cup B)'

is 20.

The universal set has 30 elements. Set

A

has 7 elements (multiples of 4 up to 30) and set

B

has 5 elements (perfect squares up to 30). The numbers 4 and 16 belong to both sets, so the intersection has 2 elements. By the principle of inclusion-exclusion, the union

A \cup B

contains

7 + 5 - 2 = 10

elements. The complement

(A \cup B)'

contains all elements in the universal set that are not in the union, giving

30 - 10 = 20

elements.

Step-by-Step Solution

1

Determine the elements and cardinality of the universal set

\mathcal{U}

.

\mathcal{U} = \{1, 2, 3, \dots, 30\}

, so

n(\mathcal{U}) = 30

.

The universal set bounds all possible elements under consideration.

2

List the elements of set

A

and set

B

, and find their individual cardinalities.

A = \{4, 8, 12, 16, 20, 24, 28\} \implies n(A) = 7

.

B = \{1, 4, 9, 16, 25\} \implies n(B) = 5

.

Identify multiples of 4 and perfect squares within the range 1 to 30.

3

Find the intersection

A \cap B

and compute the cardinality of the union

A \cup B

.

A \cap B = \{4, 16\} \implies n(A \cap B) = 2

.

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

.

Apply the principle of inclusion-exclusion to avoid double-counting elements belonging to both sets.

4

Calculate the cardinality of the complement

(A \cup B)'

.

n((A \cup B)') = n(\mathcal{U}) - n(A \cup B) = 30 - 10 = 20

.

The complement set consists of all elements in the universal set that are not in

A \cup B

.

Key Concept

Complement of Set Union and Inclusion-Exclusion Principle

Question 7Question →

In a agricultural survey of 120 farmers in a community, 65 grow maize, 50 grow yam, and 42 grow cassava. Furthermore, 24 grow both maize and yam, 18 grow both maize and cassava, and 15 grow both yam and cassava. If 8 farmers grow none of these three crops, find the number of farmers who grow all three crops.

Show answer & explanation

Answer: 12

Answer

12 farmers grow all three crops.

Using the 3-set inclusion-exclusion principle, the total number of farmers growing at least one crop is

120 - 8 = 112

. Expanding

|M \cup Y \cup C| = |M| + |Y| + |C| - (|M \cap Y| + |M \cap C| + |Y \cap C|) + |M \cap Y \cap C|

gives

112 = 65 + 50 + 42 - 24 - 18 - 15 + x

. Simplifying yields

112 = 100 + x

, which gives

x = 12

.

Step-by-Step Solution

1

Determine the cardinality of the union of all three sets.

|M \cup Y \cup C| = 112

Subtract the farmers who grow none of the crops from the universal set size (

120 - 8 = 112

).

2

Set up the Inclusion-Exclusion equation for three sets.

112 = 65 + 50 + 42 - (24 + 18 + 15) + x

Inclusion-exclusion states that

|A \cup B \cup C| = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)

.

3

Solve for the unknown value

x

representing the intersection of all three sets.

x = 12

Simplifying gives

112 = 100 + x

, which leads directly to

x = 12

.

Key Concept

Principle of Inclusion-Exclusion for 3 Sets

Question 8Question →

Let the universal set $\mathcal{E} = \{x \in \mathbb{Z} : 1 \le x \le 25\}$ . Two subsets $A$ and $B$ of $\mathcal{E}$ are defined as $A = \{x \in \mathcal{E} : x \text{ is a perfect square}\}$ and $B = \{x \in \mathcal{E} : x \text{ is an odd number}\}$ . What is the cardinality of $(A \cup B)'$ ?

10

12

15

7

Show answer & explanation

Answer: 10

Answer

10

The correct answer is 10 because the universal set has 25 elements. The union

A \cup B

contains all 13 odd numbers in the range along with the 2 even perfect squares (

4

and

16

), giving

n(A \cup B) = 15

. Subtracting this from the universal set size gives

25 - 15 = 10

.

Step-by-Step Solution

1

Identify the elements of the universal set and subsets A and B

\mathcal{E} = \{1, 2, 3, \dots, 25\}

with

n(\mathcal{E}) = 25

.

A = \{1, 4, 9, 16, 25\}

B = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25\}

Explicitly listing elements clarifies overlapping regions.

2

Find the union set

A \cup B

and its cardinality

A \cup B = \{1, 3, 4, 5, 7, 9, 11, 13, 15, 16, 17, 19, 21, 23, 25\}

, so

n(A \cup B) = 15

.

The union includes all odd numbers from 1 to 25 plus the even perfect squares (4 and 16).

3

Calculate the cardinality of the complement

(A \cup B)'

n((A \cup B)') = n(\mathcal{E}) - n(A \cup B) = 25 - 15 = 10

.

The complement set contains all elements in

\mathcal{E}

that are neither odd nor perfect squares, which are the 10 non-square even numbers:

\{2, 6, 8, 10, 12, 14, 18, 20, 22, 24\}

.

Key Concept

Set Complements and De Morgan's Laws / Set Operations

Estimated Time:1m 30s

Question 9Question →

Let the universal set $\mathcal{U} = \{x \in \mathbb{Z}^+ : x \le 60\}$ . Subsets $A$ , $B$ , and $C$ of $\mathcal{U}$ are defined as:

$A = \{x \in \mathcal{U} : x \text{ is a multiple of } 3\}$
$B = \{x \in \mathcal{U} : x \text{ is a multiple of } 4\}$
$C = \{x \in \mathcal{U} : x \text{ is a factor of } 60\}$

What is the cardinality of the set $(A \Delta B) \setminus C'$ , where $A \Delta B$ represents the symmetric difference of $A$ and $B$ , and $C'$ is the complement of $C$ in $\mathcal{U}$ ?

6

8

12

19

Show answer & explanation

Answer: 6

Answer

6

Using the set identity

X \setminus Y' = X \cap Y

, the expression

(A \Delta B) \setminus C'

simplifies directly to

(A \Delta B) \cap C

. The set

C

consists of all factors of 60:

\{1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60\}

. The symmetric difference

A \Delta B

consists of numbers that are divisible by 3 or 4, but not both (excluding multiples of 12). Checking the 12 factors of 60 against this rule, the numbers 3, 6, 15, and 30 are multiples of 3 but not 4, while 4 and 20 are multiples of 4 but not 3. The numbers 12 and 60 are multiples of both 3 and 4, so they are excluded. Thus, the resulting set is

\{3, 4, 6, 15, 20, 30\}

, which contains exactly 6 elements.

Step-by-Step Solution

1

Simplify the set expression using complement and relative difference identities.

(A \Delta B) \setminus C' = (A \Delta B) \cap (C')' = (A \Delta B) \cap C

By definition of set difference,

X \setminus Y = X \cap Y'

. Replacing

Y

with

C'

gives

X \cap (C')' = X \cap C

.

2

List all elements in set

C

(factors of 60).

C = \{1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60\}

60 prime factorizes as

2^2 \times 3^1 \times 5^1

, giving

(2+1)(1+1)(1+1) = 12

positive factors.

3

Recall the definition of symmetric difference

A \Delta B

.

A \Delta B = (A \cup B) \setminus (A \cap B)

An element belongs to

A \Delta B

if it is a multiple of 3 or a multiple of 4, but NOT a multiple of both 3 and 4 (i.e. not a multiple of 12).

4

Evaluate which elements of

C

belong to

A \Delta B

.

The elements in

C

that are multiples of 3 only (not 12) are

\{3, 6, 15, 30\}

. The elements in

C

that are multiples of 4 only (not 12) are

\{4, 20\}

. Elements

\{12, 60\}

are multiples of 12, so they are excluded.

Testing each factor of 60 against the criteria for symmetric difference.

5

Form the set

(A \Delta B) \cap C

and find its cardinality.

(A \Delta B) \cap C = \{3, 4, 6, 15, 20, 30\}

, so

n((A \Delta B) \cap C) = 6

.

Counting the elements in the intersection.

Key Concept

Symmetric Difference and Set Complement Identities

Estimated Time:2m 30s

Question 10Question →

Let the universal set $\mathcal{U} = \{x \in \mathbb{Z} : 10 \le x \le 99\}$ . Consider two subsets $P$ and $Q$ of $\mathcal{U}$ , where $P = \{10a + b \in \mathcal{U} : a > b\}$ and $Q = \{10a + b \in \mathcal{U} : a \cdot b \text{ is odd}\}$ , with $a$ and $b$ representing the tens and units digits of $x$ respectively. What is the cardinality of the set $P \setminus Q$ ?

Show answer & explanation

Answer: 35; 35 elements

Answer

35

To evaluate

n(P \setminus Q)

, we calculate

n(P) - n(P \cap Q)

. Set

P

consists of all two-digit numbers where the tens digit exceeds the units digit, giving

1 + 2 + \dots + 9 = 45

numbers. Set

Q

consists of numbers with two odd digits. The intersection

P \cap Q

contains pairs of distinct odd digits

(a,b)

where

a > b

, which corresponds to

\binom{5}{2} = 10

pairs. Thus,

n(P \setminus Q) = 45 - 10 = 35

.

Step-by-Step Solution

1

Calculate the cardinality of set

P

, denoted as

n(P)

.

n(P) = 45

For a two-digit integer

10a + b

,

a \in \{1, 2, \dots, 9\}

and

b \in \{0, 1, \dots, 9\}

. The condition

a > b

yields

0

valid values of

b

for

a=1

(1 choice: 10), 2 choices for

a=2

(20, 21), up to 9 choices for

a=9

(90 through 98). The total is

\sum_{a=1}^9 a = \frac{9 \times 10}{2} = 45

.

2

Calculate the cardinality of the intersection

P \cap Q

, denoted as

n(P \cap Q)

.

n(P \cap Q) = 10

Set

Q

contains numbers where the product

a \cdot b

is odd, which requires both

a

and

b

to be odd digits chosen from

\{1, 3, 5, 7, 9\}

. For

P \cap Q

, we need

a > b

with distinct odd digits. The number of such pairs

(a, b)

is given by choosing 2 distinct odd digits out of 5:

\binom{5}{2} = 10

.

3

Calculate the cardinality of the set difference

P \setminus Q

.

n(P \setminus Q) = 35

Using the identity

n(P \setminus Q) = n(P) - n(P \cap Q)

, we get

45 - 10 = 35

.

Key Concept

Cardinality of set difference and set-builder operations

Estimated Time:2m 30s

Question 11Question →

Let the universal set be $\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 36\}$ . Subsets $A$ , $B$ , and $C$ of $\mathcal{U}$ are defined as follows:
$A = \{x \in \mathcal{U} : x \text{ is a perfect square}\}$ ,
$B = \{x \in \mathcal{U} : x \text{ is a multiple of } 3\}$ , and
$C = \{x \in \mathcal{U} : x \text{ is an even number}\}$ .
What is the cardinality of the set $(A \cup B)' \cap C$ ?

8

9

10

12

Show answer & explanation

Answer: 10

Answer

The cardinality of the set

(A \cup B)' \cap C

is 10.

The set

(A \cup B)' \cap C

represents elements in

C

that are neither in

A

nor in

B

. Set

C

contains 18 even integers up to 36. Among these 18 numbers, three are perfect squares (4, 16, 36) and six are multiples of 3 (6, 12, 18, 24, 30, 36). Since 36 satisfies both conditions, the total number of even numbers in

A \cup B

is

3 + 6 - 1 = 8

. Subtracting these 8 elements from 18 leaves 10 elements.

Step-by-Step Solution

1

Identify the elements of set C (even numbers in the universal set)

C = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36\}

, so

n(C) = 18

.

The complement set operation is restricted to elements within set C.

2

Find the elements of set C that belong to set A (even perfect squares)

C \cap A = \{4, 16, 36\}

, so

n(C \cap A) = 3

.

Identify which elements in C must be excluded due to being perfect squares.

3

Find the elements of set C that belong to set B (even multiples of 3, i.e., multiples of 6)

C \cap B = \{6, 12, 18, 24, 30, 36\}

, so

n(C \cap B) = 6

.

Identify which elements in C must be excluded due to being multiples of 3.

4

Find the overlap between the excluded sets within C

C \cap A \cap B = \{36\}

, so

n(C \cap A \cap B) = 1

.

Avoid double-counting 36 when subtracting excluded elements.

5

Calculate

n(C \cap (A \cup B))

using inclusion-exclusion and find the remaining complement elements in C

n(C \cap (A \cup B)) = 3 + 6 - 1 = 8

. Therefore,

n((A \cup B)' \cap C) = 18 - 8 = 10

.

Subtracting the 8 excluded even numbers from the total 18 even numbers yields the final cardinality.

Key Concept

Set Complement and Relative Difference with Inclusion-Exclusion

Estimated Time:2m 0s

Question 12Question →

Let the universal set be $\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 40\}$ . Subsets $P$ , $Q$ , and $R$ of $\mathcal{U}$ are defined as follows:
$P = \{x \in \mathcal{U} : x \text{ is a multiple of } 3\}$ ,
$Q = \{x \in \mathcal{U} : x \text{ is a multiple of } 4\}$ , and
$R = \{x \in \mathcal{U} : x \text{ is a perfect square\}}$ .
What is the cardinality of the set $(P \cup Q)' \cap R'$ ?

18

22

20

17

Show answer & explanation

Answer: 18

Answer

18

By De Morgan's Law, the expression

(P \cup Q)' \cap R'

is equivalent to

(P \cup Q \cup R)'

. The universal set contains 40 integers. The set

P \cup Q \cup R

consists of all integers from 1 to 40 that are multiples of 3, multiples of 4, or perfect squares, giving a total of 22 distinct elements. Subtracting this from 40 leaves 18 elements that belong to neither

P

,

Q

, nor

R

.

Step-by-Step Solution

1

Apply De Morgan's Laws to simplify the required set expression

(P \cup Q)' \cap R' = (P \cup Q \cup R)'

The intersection of the complements of two sets is equal to the complement of their union.

2

List and count the elements belonging to sets

P

,

Q

, and

R

within

\mathcal{U} = \{1, 2, \dots, 40\}

P

has 13 elements (multiples of 3);

Q

adds 7 elements not in

P

(multiples of 4 excluding 12, 24, 36);

R

adds 2 elements not in

P \cup Q

(

1

and

25

)

Counting unique elements avoids double-counting shared elements across sets.

3

Calculate the cardinality of

P \cup Q \cup R

n(P \cup Q \cup R) = 13 + 7 + 2 = 22

Summing the non-overlapping subsets gives the total size of the union.

4

Subtract the union cardinality from the universal set cardinality

n(\mathcal{U}) = 40

n((P \cup Q \cup R)') = 40 - 22 = 18

The cardinality of the complement is the size of the universal set minus the size of the set itself.

Key Concept

De Morgan's Laws and Set Complements

Question 13Question →

Let the universal set be $\mathcal{U} = \{x \in \mathbb{Z}^+ : 1 \le x \le 200\}$ . Subsets $A$ , $B$ , and $C$ of $\mathcal{U}$ are defined as follows:
- $A = \{x \in \mathcal{U} : x \text{ is a multiple of } 5\}$
- $B = \{x \in \mathcal{U} : x \text{ is a multiple of } 7\}$
- $C = \{x \in \mathcal{U} : x \text{ is a multiple of } 10\}$

Find the number of elements in the set $(A \cap B') \setminus C$ .

Show answer & explanation

Answer: 17; seventeen

Answer

The number of elements in the set

(A \cap B') \setminus C

is 17.

The set

(A \cap B') \setminus C

consists of elements in

A

that are neither in

B

nor in

C

. Taking

A \setminus C

isolates the 20 odd multiples of 5 between 1 and 200. From these 20 numbers, removing those divisible by 7 leaves only the odd multiples of 35 (35, 105, and 175) to be subtracted, giving

20 - 3 = 17

.

Step-by-Step Solution

1

Simplify the set expression

(A \cap B') \setminus C

.

(A \cap B') \setminus C = (A \setminus C) \setminus B

By set algebra, taking elements in

A

that are not in

B

and then excluding elements in

C

is equivalent to first removing elements of

C

from

A

, and then removing any remaining elements that belong to

B

.

2

Determine the cardinality of

A \setminus C

.

n(A \setminus C) = 20

Set

A

contains all multiples of 5 up to 200, of which there are

\lfloor 200/5 \rfloor = 40

. Set

C

contains all multiples of 10 up to 200, which are the even multiples of 5, amounting to

\lfloor 200/10 \rfloor = 20

. Therefore,

A \setminus C

consists of the odd multiples of 5 up to 200, which gives

40 - 20 = 20

elements.

3

Find which elements of

A \setminus C

are also in set

B

.

The common elements are 35, 105, and 175 (3 elements).

Elements in both

A

and

B

are multiples of

\text{lcm}(5, 7) = 35

. The multiples of 35 up to 200 are 35, 70, 105, 140, and 175. However, 70 and 140 are multiples of 10 (belonging to set

C

) and have already been removed. Thus, only the odd multiples of 35 (35, 105, and 175) remain in

A \setminus C

and also belong to

B

.

4

Subtract these common elements to calculate

n((A \setminus C) \setminus B)

.

20 - 3 = 17

Removing the 3 common elements from the 20 elements of

A \setminus C

leaves 17 elements.

Key Concept

Set difference, intersection with complement, and evaluation of cardinalities using divisibility properties.

Question 14Question →

Let the universal set be $\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ and set $P = \{2, 3, 5, 7\}$ . Which of the following is the complement of set $P$ , denoted by $P'$ ?

\{1, 4, 6, 8, 9, 10\}

\{1, 3, 5, 7, 9\}

\{4, 6, 8, 10\}

\{2, 3, 5, 7\}

Show answer & explanation

Answer: $\{1, 4, 6, 8, 9, 10\}$

Answer

\{1, 4, 6, 8, 9, 10\}

The complement of set

P

, denoted as

P'

, consists of every element in the universal set

\mathcal{U}

that is not present in set

P

. Removing

2, 3, 5, 7

from

\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}

leaves

\{1, 4, 6, 8, 9, 10\}

.

Step-by-Step Solution

1

Identify elements of the universal set

\mathcal{U}

and set

P

\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}

and

P = \{2, 3, 5, 7\}

The complement set requires comparing set

P

against the universal set.

2

Remove elements of set

P

from

\mathcal{U}

to find

P'

P' = \mathcal{U} \setminus P = \{1, 4, 6, 8, 9, 10\}

By definition, the complement

P'

consists of all elements in

\mathcal{U}

that do not belong to

P

.

Key Concept

Set Complement

Question 15Question →

In a survey of 120 final year secondary school students in Lagos, 47 offer Physics ( $P$ ), 57 offer Chemistry ( $C$ ), and 48 offer Biology ( $B$ ). Records show that 17 students offer both Physics and Chemistry, 20 offer both Chemistry and Biology, and 15 offer both Physics and Biology. If the number of students who offer none of these three subjects is three times the number of students who offer all three subjects, find the total number of students who offer exactly one of the three subjects.

Show answer & explanation

Answer: 63

Answer

63 students offer exactly one of the three subjects.

By applying the principle of inclusion-exclusion for three sets, the total union is

|P \cup C \cup B| = 47 + 57 + 48 - (17 + 20 + 15) + x = 100 + x

, where

x

is the number of students taking all three subjects. Setting the universal set total to 120 gives

(100 + x) + 3x = 120

, solving to

x = 5

. Subtracting the relevant intersections yields 20 students for Physics only, 25 for Chemistry only, and 18 for Biology only. Summing these gives 63.

Step-by-Step Solution

1

Define unknown variables and set up the inclusion-exclusion formula for the union of three sets.

|P \cup C \cup B| = 100 + x

, where

x = |P \cap C \cap B|

.

The Principle of Inclusion-Exclusion states that

|P \cup C \cup B| = |P| + |C| + |B| - (|P \cap C| + |C \cap B| + |P \cap B|) + |P \cap C \cap B|

.

2

Formulate and solve an equation for the total number of students in the universal set.

120 = (100 + x) + 3x \implies 4x = 20 \implies x = 5

.

The total number of students equals those taking at least one subject plus those taking none (

3x

).

3

Determine the number of students taking exactly two subjects.

Physics and Chemistry only =

17 - 5 = 12

; Chemistry and Biology only =

20 - 5 = 15

; Physics and Biology only =

15 - 5 = 10

.

Subtract the triple intersection count (

x = 5

) from each pairwise intersection count.

4

Calculate the number of students offering only Physics, only Chemistry, and only Biology.

Physics only =

47 - (12 + 10 + 5) = 20

; Chemistry only =

57 - (12 + 15 + 5) = 25

; Biology only =

48 - (10 + 15 + 5) = 18

.

Subtract the sum of the two-subject-only regions and the three-subject region from each set's total cardinality.

5

Sum the counts of students offering exactly one subject.

20 + 25 + 18 = 63

.

The total offering exactly one subject is the sum of the three disjoint single-subject regions.

Key Concept

3-Set Principle of Inclusion-Exclusion and Venn Diagram Region Decomposition

Question 16Question →

Let the universal set be $\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 12\}$ . Consider the subsets $P = \{x : x \text{ is a factor of } 12\}$ and $Q = \{x : x \text{ is an even number}, 1 \le x \le 12\}$ . Which of the following sets represents $(P \cup Q)'$ ?

\{5, 7, 9, 11\}

\{2, 4, 6, 12\}

\{5, 7, 11\}

\{1, 3, 5, 7, 9, 11\}

Show answer & explanation

Answer: $\{5, 7, 9, 11\}$

Answer

\{5, 7, 9, 11\}

The set of factors of 12 within the given range is {1, 2, 3, 4, 6, 12} and the set of even numbers up to 12 is {2, 4, 6, 8, 10, 12}. Their union contains all numbers that are either even or factors of 12, which is {1, 2, 3, 4, 6, 8, 10, 12}. Subtracting this union from the universal set {1, 2, ..., 12} yields {5, 7, 9, 11}.

Step-by-Step Solution

1

List all elements of the universal set and the subsets

P

and

Q

.

\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}

,

P = \{1, 2, 3, 4, 6, 12\}

, and

Q = \{2, 4, 6, 8, 10, 12\}

.

Identify the element listings explicitly from the given definitions.

2

Find the union of sets

P

and

Q

, denoted by

P \cup Q

.

P \cup Q = \{1, 2, 3, 4, 6, 8, 10, 12\}

.

Combine all unique elements belonging to either

P

,

Q

, or both.

3

Find the complement

(P \cup Q)'

relative to the universal set

\mathcal{U}

.

(P \cup Q)' = \mathcal{U} \setminus (P \cup Q) = \{5, 7, 9, 11\}

.

Select all elements in

\mathcal{U}

that do not appear in

P \cup Q

.

Key Concept

Complement of a Set Union

Estimated Time:50s

Question 17Question →

In a survey of 160 music enthusiasts, 80 listen to Afrobeat, 70 listen to Highlife, and 65 listen to Reggae. It was found that 30 listen to both Afrobeat and Highlife, 25 listen to both Highlife and Reggae, and 28 listen to both Afrobeat and Reggae. If 15 enthusiasts listen to none of these three genres, how many enthusiasts listen to exactly two of these genres?

Show answer & explanation

Answer: 44

Answer

The number of enthusiasts who listen to exactly two of the three genres is 44.

The correct answer is 44. By subtracting the 15 non-listeners from 160, the union of the three sets contains 145 members. Solving the 3-set inclusion-exclusion equation yields 13 enthusiasts who listen to all three genres. Subtracting 13 from each given pairwise intersection gives 17, 12, and 15 for the regions representing listeners of exactly two genres. Summing these three counts yields 44.

Step-by-Step Solution

1

Calculate the cardinality of the union of all three sets.

n(A \cup H \cup R) = 160 - 15 = 145

Subtracting the number of enthusiasts listening to none of the genres from the universal set.

2

Apply the Principle of Inclusion-Exclusion for 3 sets to find the triple intersection.

n(A \cap H \cap R) = 13

Substituting known cardinalities into

n(A \cup H \cup R) = n(A)+n(H)+n(R) - [n(A \cap H)+n(H \cap R)+n(A \cap R)] + n(A \cap H \cap R)

.

3

Calculate the count of enthusiasts in each 'exactly two genres' region.

Afrobeat and Highlife only = 17; Highlife and Reggae only = 12; Afrobeat and Reggae only = 15.

Subtracting the triple intersection count (

13

) from each pairwise intersection.

4

Sum the counts of the three distinct two-genre regions.

17 + 12 + 15 = 44

Combining all mutually exclusive regions representing enthusiasts of exactly two genres.

Key Concept

Principle of Inclusion-Exclusion for three sets and Venn diagram region partitioning.

Question 18Question →

Let the universal set be $\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 25\}$ . If $A = \{x \in \mathcal{U} : x \text{ is a multiple of } 3\}$ and $B = \{x \in \mathcal{U} : x \text{ is a perfect square}\}$ , what is the cardinality of $(A \cup B)'$ ?

13

12

14

17

Show answer & explanation

Answer: 13

Answer

13

The total number of elements in the universal set is 25. Set A has 8 elements (multiples of 3) and set B has 5 elements (perfect squares). The element 9 belongs to both sets. Therefore, the union of A and B contains 8 + 5 - 1 = 12 elements. Subtracting this from the universal set size gives 25 - 12 = 13 elements in the complement.

Step-by-Step Solution

1

Determine the elements of the universal set and its cardinality.

\mathcal{U} = \{1, 2, 3, \dots, 25\}

, so

n(\mathcal{U}) = 25$.

The universal set contains all integers from 1 to 25 inclusive.

2

List the elements of set A and set B.

A = \{3, 6, 9, 12, 15, 18, 21, 24\}

(so

n(A) = 8

) and

B = \{1, 4, 9, 16, 25\}

(so

n(B) = 5

).

Set A contains multiples of 3 within the domain, and set B contains perfect squares within the domain.

3

Find the intersection

A \cap B

and calculate the cardinality of the union

A \cup B

.

A \cap B = \{9\}

, so

n(A \cap B) = 1

. Thus,

n(A \cup B) = n(A) + n(B) - n(A \cap B) = 8 + 5 - 1 = 12

.

The principle of inclusion-exclusion avoids double-counting the common element 9.

4

Calculate the cardinality of the complement

(A \cup B)'

.

n((A \cup B)') = n(\mathcal{U}) - n(A \cup B) = 25 - 12 = 13

.

The complement consists of all elements in the universal set that do not belong to the union of A and B.

Key Concept

Complement of Set Union and Inclusion-Exclusion Principle

Question 19Question →

Let the universal set be $\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 20\}$ . Consider two subsets of $\mathcal{U}$ given by $A = \{x \in \mathcal{U} : x \text{ is a multiple of } 4\}$ and $B = \{x \in \mathcal{U} : x \text{ is a factor of } 20\}$ . What is the number of elements in the complement of $(A \cup B)$ , denoted as $n((A \cup B)')$ ?

9

11

13

15

Show answer & explanation

Answer: 11

Answer

11

The universal set contains 20 elements. The set of multiples of 4 within this range has 5 elements, and the set of factors of 20 has 6 elements. Their shared elements are 4 and 20 (2 elements). Thus, the union contains

5 + 6 - 2 = 9

elements. Subtracting this from 20 gives 11 elements in the complement.

Step-by-Step Solution

1

Identify the elements of the universal set and the subsets

A

and

B

.

\mathcal{U} = \{1, 2, 3, \dots, 20\}

with

n(\mathcal{U}) = 20

.

A = \{4, 8, 12, 16, 20\}

with

n(A) = 5

.

B = \{1, 2, 4, 5, 10, 20\}

with

n(B) = 6

.

Listing the elements clearly determines the cardinality of each individual set.

2

Find the intersection and union of sets

A

and

B

.

A \cap B = \{4, 20\}

, so

n(A \cap B) = 2

.

A \cup B = \{1, 2, 4, 5, 8, 10, 12, 16, 20\}

, so

n(A \cup B) = 5 + 6 - 2 = 9

.

Applying the principle of inclusion-exclusion avoids double-counting common elements.

3

Calculate the cardinality of the complement

(A \cup B)'

.

n((A \cup B)') = n(\mathcal{U}) - n(A \cup B) = 20 - 9 = 11

.

The complement contains all elements in the universal set that are not in the union.

Key Concept

Set Complements and Inclusion-Exclusion Principle

Question 20Question →

In a secondary school class of 45 students, 28 study Chemistry, 25 study Physics, and 6 study neither of the two subjects. How many students study both Chemistry and Physics?

Show answer & explanation

Answer: 14; 14 students

Answer

14 students study both Chemistry and Physics.

First, find the number of students taking at least one subject by subtracting those taking neither from the total: 45 - 6 = 39. Then, apply the formula n(C ∪ P) = n(C) + n(P) - n(C ∩ P). Substituting the values gives 39 = 28 + 25 - n(C ∩ P), which simplifies to n(C ∩ P) = 53 - 39 = 14.

Step-by-Step Solution

1

Calculate the number of students taking at least one of the two subjects.

n(Chemistry ∪ Physics) = 45 - 6 = 39

Students who study neither subject must be subtracted from the total class population to find the cardinality of the union.

2

Set up the two-set inclusion-exclusion formula.

n(Chemistry ∪ Physics) = n(Chemistry) + n(Physics) - n(Chemistry ∩ Physics)

The sum of individual set cardinalities overcounts elements present in both sets.

3

Substitute the known values into the equation and solve for the intersection.

39 = 28 + 25 - n(Chemistry ∩ Physics) ⇒ n(Chemistry ∩ Physics) = 53 - 39 = 14

Subtracting 39 from 53 gives the exact number of students taking both subjects.

Key Concept

Two-set inclusion-exclusion principle and complement of a set

Estimated Time:1m 30s

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