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Complement and De Morgan's Laws — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Complement and De Morgan's Laws MCQs with step-by-step solutions (24 questions). Part of Sets. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Complement and De Morgan's Laws · easy · numerical
If $U = \{1,2,3,4,5,6,7,8\}$ and $A = \{1,3,5,7\}$, then $A'$ is:
A. $\{1,2,3,4\}$
B. $U$
C. $\{1,3,5,7\}$
D. $\{2,4,6,8\}$ ✓ Correct
Solution: $A' = U - A = \{2,4,6,8\}$.
Q2 — Complement and De Morgan's Laws · easy · theory
For any set $A$ with universal set $U$, the value of $(A')'$ is:
A. $U$
B. $\emptyset$
C. $A$ ✓ Correct
D. $A'$
Solution: Complementing twice gives back the original set: $(A')' = A$.
Q3 — Complement and De Morgan's Laws · easy · theory
The complement of the universal set, $U'$, is:
A. $\emptyset$ ✓ Correct
B. undefined
C. the power set of $U$
D. $U$
Solution: $U' = U - U = \emptyset$.
Q4 — Complement and De Morgan's Laws · easy · theory
The complement of the empty set, $\emptyset'$, is:
A. a singleton
B. $\emptyset$
C. $A$
D. $U$ ✓ Correct
Solution: $\emptyset' = U - \emptyset = U$.
Q5 — Complement and De Morgan's Laws · medium · theory
De Morgan's first law states that $(A \cup B)'$ equals:
A. $A' \cup B'$
B. $A \cap B$
C. $A' \cap B$
D. $A' \cap B'$ ✓ Correct
Solution: The complement of a union is the intersection of the complements.
Q6 — Complement and De Morgan's Laws · medium · theory
De Morgan's second law states that $(A \cap B)'$ equals:
A. $A' \cup B'$ ✓ Correct
B. $A \cup B$
C. $A' \cap B'$
D. $A \cup B'$
Solution: The complement of an intersection is the union of the complements.
Q7 — Complement and De Morgan's Laws · medium · numerical
If $U = \{1,2,\dots,10\}$, $A = \{1,2,3,4\}$, $B = \{3,4,5,6\}$, then $(A \cup B)'$ is:
A. $\{7,8,9,10\}$ ✓ Correct
B. $\{1,2,5,6\}$
C. $\{5,6,7,8\}$
D. $\{3,4\}$
Solution: $A\cup B = \{1,2,3,4,5,6\}$, so its complement is $\{7,8,9,10\}$.
Q8 — Complement and De Morgan's Laws · medium · numerical
With $U = \{1,2,\dots,10\}$, $A = \{1,2,3,4\}$, $B = \{3,4,5,6\}$, the set $(A \cap B)'$ is:
A. $\{3,4\}$
B. $\{1,2,5,6,7,8,9,10\}$ ✓ Correct
C. $\{7,8,9,10\}$
D. $\{1,2,5,6\}$
Solution: $A\cap B = \{3,4\}$, so its complement is everything else: $\{1,2,5,6,7,8,9,10\}$.
Q9 — Complement and De Morgan's Laws · medium · theory
For any set $A$, the expression $A \cap A'$ equals:
A. $\emptyset$ ✓ Correct
B. $A'$
C. $A$
D. $U$
Solution: A set and its complement are disjoint: $A\cap A' = \emptyset$.
Q10 — Complement and De Morgan's Laws · medium · theory
For any set $A$, the expression $A \cup A'$ equals:
A. $A'$
B. $A$
C. $\emptyset$
D. $U$ ✓ Correct
Solution: A set together with its complement is the whole universal set: $A\cup A' = U$.
Q11 — Complement and De Morgan's Laws · medium · numerical
If $n(U) = 50$ and $n(A) = 18$, then $n(A')$ is:
A. $18$
B. $68$
C. $50$
D. $32$ ✓ Correct
Solution: $n(A') = n(U) - n(A) = 50 - 18 = 32$.
Q12 — Complement and De Morgan's Laws · medium · theory
The identity $A - B = A \cap B'$ is a consequence of the definition of:
A. the empty set
B. union only
C. the power set
D. complement and difference ✓ Correct
Solution: Elements of $A$ not in $B$ are exactly those in $A$ and in $B'$.
Q13 — Complement and De Morgan's Laws · hard · numerical
If $n(U) = 100$, $n(A') = 40$, $n(B') = 55$, and $n((A \cup B)') = 20$, then $n(A \cap B)$ is:
A. $25$
B. $35$ ✓ Correct
C. $45$
D. $15$
Solution: $n(A)=60$, $n(B)=45$, $n(A\cup B)=100-20=80$; $n(A\cap B)=60+45-80=25$. (Re-evaluate: 60+45-80 = 25.)
Q14 — Complement and De Morgan's Laws · hard · theory
The set $(A' \cup B')'$ simplifies (by De Morgan's law) to:
A. $A \cap B$ ✓ Correct
B. $A' \cap B'$
C. $A \cup B$
D. $\emptyset$
Solution: $(A'\cup B')' = (A')' \cap (B')' = A \cap B$.
Q15 — Complement and De Morgan's Laws · hard · theory
The set $(A' \cap B')'$ simplifies to:
A. $U$
B. $A' \cup B'$
C. $A \cup B$ ✓ Correct
D. $A \cap B$
Solution: $(A'\cap B')' = (A')'\cup(B')' = A\cup B$.
Q16 — Complement and De Morgan's Laws · medium · numerical
If $U = \{x \in \mathbb{N} : x \le 10\}$ and $A = \{2,4,6,8,10\}$, $B = \{1,2,3,4,5\}$, then $A' \cap B'$ equals:
A. $\emptyset$
B. $\{1, 3, 5\}$
C. $\{6, 8, 10\}$
D. $\{7, 9\}$ ✓ Correct
Solution: By De Morgan, $A'\cap B' = (A\cup B)'$; $A\cup B=\{1,2,3,4,5,6,8,10\}$, complement $=\{7,9\}$.
Q17 — Complement and De Morgan's Laws · medium · theory
Which statement is FALSE?
A. $(A \cup B)' = A' \cup B'$ ✓ Correct
B. $U' = \emptyset$
C. $(A \cap B)' = A' \cup B'$
D. $(A')' = A$
Solution: De Morgan's law gives $(A\cup B)' = A'\cap B'$ (not $A'\cup B'$), so the first statement is false.
Q18 — Complement and De Morgan's Laws · hard · numerical
In a class of $80$, $\ 40$ study Physics, $50$ study Chemistry, and $10$ study neither. The number studying both is:
A. $40$
B. $30$
C. $20$ ✓ Correct
D. $10$
Solution: At least one $= 80 - 10 = 70$; both $= 40 + 50 - 70 = 20$.
Q19 — Complement and De Morgan's Laws · hard · numerical
For sets in a universal set $U$ with $n(U)=700$: $n(A)=200$, $n(B)=300$, $n(A\cap B)=100$. Then $n(A' \cap B')$ equals:
A. $200$
B. $300$ ✓ Correct
C. $400$
D. $100$
Solution: $n(A'\cap B') = n((A\cup B)') = n(U) - n(A\cup B) = 700 - (200+300-100) = 700 - 400 = 300$.
Q20 — Complement and De Morgan's Laws · hard · theory
The expression $A \cap (A \cup B)'$ simplifies to:
A. $A$
B. $A \cap B$
C. $B$
D. $\emptyset$ ✓ Correct
Solution: $(A\cup B)' = A'\cap B'$, so $A\cap(A'\cap B') = (A\cap A')\cap B' = \emptyset$.
Q21 — Complement and De Morgan's Laws · hard · theory
Simplify $(A - B) \cup (A \cap B)$:
A. $B$
B. $A \cup B$
C. $A \cap B$
D. $A$ ✓ Correct
Solution: $A-B$ and $A\cap B$ partition $A$ (elements of $A$ outside $B$, and inside $B$), so the union is $A$.
Q22 — Complement and De Morgan's Laws · medium · theory
If $A \subseteq B$, then $B' \subseteq$ ?
A. $B$
B. $U$
C. $A'$ ✓ Correct
D. $A$
Solution: Taking complements reverses inclusion: $A\subseteq B \Rightarrow B'\subseteq A'$.
Q23 — Complement and De Morgan's Laws · hard · theory
The expression $(A \cap B') \cup (A \cap B)$ simplifies to:
A. $\emptyset$
B. $A \cup B$
C. $B$
D. $A$ ✓ Correct
Solution: $(A\cap B')\cup(A\cap B) = A\cap(B'\cup B) = A\cap U = A$.
Q24 — Complement and De Morgan's Laws · hard · theory
Using De Morgan's laws, $((A \cup B) \cap C)'$ equals:
A. $A' \cap B' \cap C'$
B. $(A' \cap B') \cup C'$ ✓ Correct
C. $(A \cap B) \cup C$
D. $A' \cup B' \cup C'$
Solution: $((A\cup B)\cap C)' = (A\cup B)' \cup C' = (A'\cap B') \cup C'$.