Mock Test 2 — Sets

Let $A = \{x \in \mathbb{R} : \sin x = 0\}$ and $B = \{x \in \mathbb{R} : \cos x = 0\}$. The intersection set $A \cap B$ is equal to:

If $P$ is the power set of a null set $\emptyset$, then the total number of elements inside the second-tier nested power set $\mathcal{P}(\mathcal{P}(\emptyset))$ is:

The simplified form of the set expression $[A \cap (A \cup B')] \cup (A \cap B)$ is exactly:

Let three finite sets have cardinalities $|A| = 10$, $|B| = 12$, and $|C| = 15$. If all pairs are mutually disjoint, the cardinality of the combined union $|A \cup B \cup C|$ is:

If $A = \{x \in \mathbb{R} : x^2 - 5x + 6 \le 0\}$ and $B = \{x \in \mathbb{R} : x^2 - 7x + 12 \le 0\}$, then the intersection interval $A \cap B$ is:

The total number of subsets of a finite set $A$ is $56$ more than the total number of subsets of a finite set $B$. Find the number of elements inside set $A$.

Given the algebraic operation equation $(A - B) \cup (B - A) = A \cup B$, we can logically deduce that:

The complement expansion expression $[(A \cap B) \cup (A \cap B')]'$ simplifies directly to:

If a universal set has cardinality $|U| = 50$, and two subsets satisfy $|A| = 25$, $|B| = 20$, and $|(A \cup B)'| = 10$, then the intersection cardinality $|A \cap B|$ is:

The set expression $A - (B \cup C)$ can be distributed and rewritten as:

A three-set survey tracks reader metrics across three local newspapers ($A, B, C$): $40\%$ read $A$, $50\%$ read $B$, and $30\%$ read $C$. The pairs overlap by $15\%$ for $A \cap B$, $10\%$ for $B \cap C$, and $12\%$ for $A \cap C$. If $5\%$ of the readers read all three newspapers, find the integer percentage of the population that reads at least one newspaper.

Find the number of distinct elements inside the power set collection $\mathcal{P}(A)$ if set $A = \{1, 1, 2, 2, 3\}$.

Two sets are defined as $A = \{2, 4, 6\}$ and $B = \{x \in \mathbb{Z} : |x - 4| \le 2\}$. Find the cardinality of their intersection set $|A \cap B|$.

Assertion (A): The Distributive Law states that the intersection operation distributes over a union: $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.
Reason (R): Set algebra is fundamentally governed by the same axiomatic laws as boolean propositional logic, meaning dual distribution identities are preserved.

Assertion (A): If $A \triangle B = \emptyset$, then the two sets $A$ and $B$ must be completely identical ($A = B$).
Reason (R): The symmetric difference captures elements that belong to exactly one of the sets, so a result of $\emptyset$ means no elements belong to one set without also belonging to the other.