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Operations on Sets — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Operations on Sets MCQs with step-by-step solutions (26 questions). Part of Sets. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Operations on Sets · easy · numerical
If $A = \{1, 2, 3, 4\}$ and $B = \{3, 4, 5, 6\}$, then $A \cup B$ is:
A. $\{1, 2, 5, 6\}$
B. $\{3, 4\}$
C. $\{1, 2, 3, 4, 5, 6\}$ ✓ Correct
D. $\{1, 2, 3, 4\}$
Solution: The union collects all elements from either set: $\{1,2,3,4,5,6\}$.
Q2 — Operations on Sets · easy · numerical
If $A = \{1, 2, 3, 4\}$ and $B = \{3, 4, 5, 6\}$, then $A \cap B$ is:
A. $\{3, 4\}$ ✓ Correct
B. $\emptyset$
C. $\{1, 2, 3, 4, 5, 6\}$
D. $\{1, 2, 5, 6\}$
Solution: The intersection is the common elements: $\{3,4\}$.
Q3 — Operations on Sets · easy · numerical
If $A = \{1, 2, 3, 4\}$ and $B = \{3, 4, 5, 6\}$, then $A - B$ is:
A. $\{5, 6\}$
B. $\{3, 4\}$
C. $\{1, 2\}$ ✓ Correct
D. $\{1, 2, 5, 6\}$
Solution: $A - B$ keeps elements of $A$ not in $B$: $\{1,2\}$.
Q4 — Operations on Sets · medium · numerical
For the sets above, $B - A$ equals:
A. $\emptyset$
B. $\{5, 6\}$ ✓ Correct
C. $\{1, 2\}$
D. $\{3, 4\}$
Solution: $B - A$ keeps elements of $B$ not in $A$: $\{5,6\}$.
Q5 — Operations on Sets · medium · theory
Two sets $A$ and $B$ are called disjoint if:
A. $A \subseteq B$
B. $A \cap B = \emptyset$ ✓ Correct
C. $A \cup B = \emptyset$
D. $A = B$
Solution: Disjoint sets share no elements, i.e. $A\cap B = \emptyset$.
Q6 — Operations on Sets · medium · theory
The identity $A - B = A \cap B'$ holds because:
A. $B' = B$
B. $A - B$ is always empty
C. difference equals union
D. removing $B$ from $A$ keeps exactly the elements of $A$ outside $B$ ✓ Correct
Solution: Elements of $A$ not in $B$ are those in $A$ and in $B'$, so $A - B = A\cap B'$.
Q7 — Operations on Sets · medium · theory
For any set $A$, $A \cup \emptyset$ equals:
A. $A$ ✓ Correct
B. $\emptyset$
C. $U$
D. $A'$
Solution: Union with the empty set adds nothing: $A \cup \emptyset = A$.
Q8 — Operations on Sets · medium · theory
For any set $A$, $A \cap \emptyset$ equals:
A. $U$
B. $A$
C. $\emptyset$ ✓ Correct
D. $A'$
Solution: Nothing is common with the empty set: $A \cap \emptyset = \emptyset$.
Q9 — Operations on Sets · easy · theory
If $A \subseteq B$, then $A \cup B$ equals:
A. $A \cap B$
B. $\emptyset$
C. $A$
D. $B$ ✓ Correct
Solution: When $A$ is contained in $B$, the union is the larger set $B$.
Q10 — Operations on Sets · easy · theory
If $A \subseteq B$, then $A \cap B$ equals:
A. $B$
B. $\emptyset$
C. $A \cup B$
D. $A$ ✓ Correct
Solution: When $A\subseteq B$, the common part is all of $A$: $A\cap B = A$.
Q11 — Operations on Sets · medium · numerical
The symmetric difference $A \triangle B = (A - B) \cup (B - A)$ for $A = \{1,2,3\}$, $B = \{2,3,4\}$ is:
A. $\{1, 4\}$ ✓ Correct
B. $\emptyset$
C. $\{2, 3\}$
D. $\{1, 2, 3, 4\}$
Solution: $A-B=\{1\}$, $B-A=\{4\}$, so $A\triangle B = \{1,4\}$.
Q12 — Operations on Sets · medium · theory
The distributive law of union over intersection states:
A. $A \cup (B \cap C) = A \cap (B \cup C)$
B. $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ ✓ Correct
C. $A \cup (B \cap C) = (A \cap B) \cup (A \cap C)$
D. $A \cap (B \cup C) = (A \cup B) \cap C$
Solution: Union distributes over intersection: $A\cup(B\cap C) = (A\cup B)\cap(A\cup C)$.
Q13 — Operations on Sets · medium · theory
The distributive law of intersection over union states:
A. $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ ✓ Correct
B. $A \cap (B \cup C) = (A \cup B) \cap (A \cup C)$
C. $A \cap (B \cup C) = (A \cap B) \cap (A \cap C)$
D. $A \cap (B \cup C) = A \cup (B \cap C)$
Solution: Intersection distributes over union: $A\cap(B\cup C) = (A\cap B)\cup(A\cap C)$.
Q14 — Operations on Sets · medium · theory
For any set $A$, $A \cup A$ and $A \cap A$ are respectively (idempotent laws):
A. $\emptyset$ and $A$
B. $U$ and $A$
C. $A$ and $A$ ✓ Correct
D. $A$ and $\emptyset$
Solution: By idempotency, $A\cup A = A$ and $A\cap A = A$.
Q15 — Operations on Sets · medium · theory
Which law is expressed by $A \cup B = B \cup A$?
A. Associative law
B. De Morgan's law
C. Distributive law
D. Commutative law ✓ Correct
Solution: Order does not matter for union: this is commutativity.
Q16 — Operations on Sets · medium · theory
The associative law for intersection is:
A. $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
B. $(A \cap B) \cap C = A \cap (B \cap C)$ ✓ Correct
C. $A \cap B = B \cap A$
D. $(A \cap B) \cup C = A \cap (B \cup C)$
Solution: Grouping does not affect the result of repeated intersection.
Q17 — Operations on Sets · medium · numerical
If $A = \{x : x \text{ is a multiple of } 2,\ x \le 12\}$ and $B = \{x : x \text{ is a multiple of } 3,\ x \le 12\}$, then $A \cap B$ is:
A. $\{2, 3\}$
B. $\{6, 12\}$ ✓ Correct
C. $\{6\}$
D. $\{4, 6, 12\}$
Solution: Common multiples of 2 and 3 up to 12 are multiples of 6: $\{6,12\}$.
Q18 — Operations on Sets · hard · numerical
If $A = \{x \in \mathbb{R} : 1 \le x \le 5\}$ and $B = \{x \in \mathbb{R} : 3 \le x \le 8\}$, then $A \cap B$ is:
A. $[3, 5]$ ✓ Correct
B. $[1, 3]$
C. $[1, 8]$
D. $(3, 5)$
Solution: Overlap of $[1,5]$ and $[3,8]$ is $[3,5]$.
Q19 — Operations on Sets · hard · numerical
For the intervals $A = [1, 5]$ and $B = [3, 8]$, the union $A \cup B$ is:
A. $(1, 8)$
B. $[1, 3] \cup [5, 8]$
C. $[1, 8]$ ✓ Correct
D. $[3, 5]$
Solution: The intervals overlap, so their union is the single interval $[1,8]$.
Q20 — Operations on Sets · hard · theory
For any sets, $(A - B) \cap (B - A)$ equals:
A. $A \cap B$
B. $A \cup B$
C. $\emptyset$ ✓ Correct
D. $A \triangle B$
Solution: $A-B$ lies outside $B$ and $B-A$ lies inside $B$; they can share no element, so the intersection is empty.
Q21 — Operations on Sets · medium · numerical
If $A = \{a, b, c, d\}$ and $B = \{b, d, e\}$, then $n(A \cup B) + n(A \cap B)$ equals:
A. $7$ ✓ Correct
B. $5$
C. $9$
D. $6$
Solution: $A\cup B = \{a,b,c,d,e\}$ ($n=5$), $A\cap B=\{b,d\}$ ($n=2$); sum $=7$. (Also equals $n(A)+n(B)=4+3$.)
Q22 — Operations on Sets · hard · theory
For any three sets, $A \cap (B - C)$ equals:
A. $(A \cap B) - (A \cap C)$ ✓ Correct
B. $(A - B) \cap C$
C. $A \cap B \cap C$
D. $(A \cap B) - C$ only when $C \subseteq A$
Solution: $A\cap(B-C) = A\cap B\cap C' = (A\cap B) - (A\cap C)$.
Q23 — Operations on Sets · medium · numerical
In a class of $60$ students, $35$ like cricket, $30$ like football and $10$ like neither game. The number of students who like both cricket and football is:
A. $15$ ✓ Correct
B. $5$
C. $25$
D. $20$
Solution: Students who like at least one game $= 60 - 10 = 50$. Using $n(C \cup F) = n(C) + n(F) - n(C \cap F)$: $50 = 35 + 30 - n(C \cap F)$, so $n(C \cap F) = 15$.
Q24 — Operations on Sets · hard · theory
If $X = \{4^n - 3n - 1 : n \in \mathbb{N}\}$ and $Y = \{9(n - 1) : n \in \mathbb{N}\}$, then $X \cup Y$ is equal to:
A. $Y$ ✓ Correct
B. $X$
C. $X \cap Y$
D. $\mathbb{N}$
Solution: By the binomial theorem, $4^n - 3n - 1 = (1+3)^n - 3n - 1 = \binom{n}{2}3^2 + \binom{n}{3}3^3 + \dots$, which is always a multiple of $9$ (it equals $0, 9, 54, \dots$). So every element of $X$ is a multiple of $9$, while $Y$ is the set of ALL non-negative multiples of $9$. Thus $X \subseteq Y$, giving $X \cup Y = Y$.
Q25 — Operations on Sets · medium · numerical
In a school of $200$ students, $120$ study Physics, $90$ study Chemistry and $40$ study neither subject. The number of students who study both Physics and Chemistry is ____.
Solution: Students studying at least one subject $= 200 - 40 = 160$. Using $n(P \cup C) = n(P) + n(C) - n(P \cap C)$: $160 = 120 + 90 - n(P \cap C)$, so $n(P \cap C) = 50$.
Q26 — Operations on Sets · medium · numerical
In a group of $70$ people, $37$ like coffee, $52$ like tea, and each person likes at least one of the two drinks. The number of people who like both coffee and tea is ____.
Solution: Since everyone likes at least one drink, $n(C \cup T) = 70$. Then $n(C \cap T) = n(C) + n(T) - n(C \cup T) = 37 + 52 - 70 = 19$.