Sets and Set Operations

22 questions

All types22 Multiple choice12 Numeric4 Short answer6

Let the universal set be $\mathcal{U} = \{x \in \mathbb{Z}^+ : 1 \le x \le 15\}$ . Consider two subsets of $\mathcal{U}$ defined as $A = \{x \in \mathcal{U} : x \text{ is prime}\}$ and $B = \{x \in \mathcal{U} : x \text{ is an odd integer greater than } 1\}$ . What is the number of elements in $(A \cap B)' \setminus (A \cup B)'$ ?

3

5

7

10

Show answer & explanation

Answer: $3$

Answer

Evaluating

(A \cap B)'

gives

\{1, 2, 4, 6, 8, 9, 10, 12, 14, 15\}

and evaluating

(A \cup B)'

gives

\{1, 4, 6, 8, 10, 12, 14\}

. Subtracting

(A \cup B)'

from

(A \cap B)'

leaves

\{2, 9, 15\}

, which has 3 elements.

Step-by-Step Solution

List the elements of the universal set

\mathcal{U}

and subsets

A

and

B

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

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

, and

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

Explicitly writing the elements helps ensure set operations are performed accurately.

Find the intersection

A \cap B

and union

A \cup B

A \cap B = \{3, 5, 7, 11, 13\}

and

A \cup B = \{2, 3, 5, 7, 9, 11, 13, 15\}

These intermediate set operations are required to evaluate their complements.

Determine the complements

(A \cap B)'

and

(A \cup B)'

relative to

\mathcal{U}

(A \cap B)' = \{1, 2, 4, 6, 8, 9, 10, 12, 14, 15\}

and

(A \cup B)' = \{1, 4, 6, 8, 10, 12, 14\}

The complement of a set consists of all elements in

\mathcal{U}

not present in that set.

Calculate the set difference

(A \cap B)' \setminus (A \cup B)'

and count its elements.

(A \cap B)' \setminus (A \cup B)' = \{2, 9, 15\}

, which contains 3 elements.

Set difference removes all elements of

(A \cup B)'

from

(A \cap B)'

, leaving elements present in

A \cup B

but not in

A \cap B

Key Concept

Set Complements and Relative Difference

Estimated Time:1m 30s

Question 22Question →

In a sports academy of 120 athletes, 70 play football, 60 play basketball, and 50 play tennis. If 10 athletes play none of these three sports and 15 athletes play all three sports, how many athletes play exactly two of these sports?

Show answer & explanation

Answer: 40

Answer

40 athletes play exactly two of the sports.

The correct answer is 40. Subtracting the 10 athletes who play no sports from the total of 120 leaves 110 athletes playing at least one sport. Using inclusion-exclusion, the sum of pairwise intersections is

S_2 = 70 + 60 + 50 + 15 - 110 = 85

. Since

S_2

contains the region of all three sports counted three times, subtracting

3 \times 15 = 45

gives 40 athletes who play exactly two sports.

Step-by-Step Solution

Determine the cardinality of the union of all three sets

n(F \cup B \cup T) = 120 - 10 = 110

Athletes who play none of the three sports are excluded from the total universal set.

Apply the Principle of Inclusion-Exclusion for three sets to find the sum of 2-set intersections

S_2 = n(F) + n(B) + n(T) + n(F \cap B \cap T) - n(F \cup B \cup T) = 70 + 60 + 50 + 15 - 110 = 85

The formula relates the total union, individual set cardinalities, pairwise intersections, and triple intersection.

Subtract three times the triple intersection from

S_2

to isolate regions corresponding to exactly two sports

Exactly two sports =

S_2 - 3 \times n(F \cap B \cap T) = 85 - 3(15) = 40

Each pairwise intersection sum

S_2

includes the triple intersection region three times.

Key Concept

Three-set inclusion-exclusion principle and region cardinality decomposition

Estimated Time:1m 30s

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Sets and Set Operations

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