XOR and XNOR questions can look unrelated, but most reduce to three checks: are the inputs different or the same, is the number of 1s odd or even, and is one expression the complement of another? Those checks also drive Boolean-identity simplification, parity tests, comparators, XOR swapping, and adder-subtractor circuits.
Attempt each question before reading its explanation. Use the XOR and XNOR gate relation learn hub to revisit the core rule before retrying a missed question. Four questions have exact question-bank deep links; practise the remaining eight from the linked learn hub.
XOR and XNOR gate relations: the 60-second map
For two inputs, XOR is 1 when the inputs differ. XNOR is 1 when they are the same.
A | B | A XOR B | A XNOR B |
|---|---|---|---|
0 | 0 | 0 | 1 |
0 | 1 | 1 | 0 |
1 | 0 | 1 | 0 |
1 | 1 | 0 | 1 |
The Boolean forms and identities to keep ready are:
A XOR B = A'B + AB'A XNOR B = AB + A'B' = (A XOR B)'A XOR A = 0A XOR 0 = AA XOR 1 = A'XOR over many inputs is 1 when the number of 1s is odd.
For A=1, B=0, XOR is 1 because the inputs differ, while XNOR is 0. For A=1, B=1, XOR is 0 and XNOR is 1. Equality makes XNOR 1 and XOR 0; inequality reverses both outputs. Reapply that rule when expressions use complemented inputs or several XOR terms.
XOR and XNOR truth-table MCQs
Q1. XOR output for different and same inputs (DSSSB 2018)
The XOR gate gives ______ as output when both inputs are different and ______ when both inputs are same.
(a) 0, 0
(b) 0, 1
(c) 1, 0
(d) 1, 1
Answer: (c) 1, 0. The 01 and 10 rows produce 1, while the equal-input rows 00 and 11 produce 0. XOR therefore acts as an inequality detector for two bits.
Q2. Gate that is high for the same input (BPSC 2023)
The logic gate that provides high output for same input is
(a) NOT
(b) X-NOR
(c) XOR
(d) More than one of the above
(e) None of the above
Answer: (b) X-NOR. XNOR gives 1 for both equal-input rows, 00 and 11, while XOR gives 0. NOT is a unary gate, so it does not compare two inputs.
XOR Boolean identities and cancellation MCQs
Q3. Convert a sum-product expression to XOR (UGC NET 2011)
(A + B)(AB)’ is equivalent to
(a) A ⊕ B
(b) A ∪ B
(c) (A ⊕ B) ∪ A
(d) (A ∪ B) ⊕ A
Answer: (a) A ⊕ B. By De Morgan's law, (AB)' = A' + B', so (A+B)(A'+B') = AA' + AB' + BA' + BB'. Since AA'=0 and BB'=0, the result is AB' + A'B, exactly the XOR form.
Next, use the Boolean Algebra and K-Map MCQs to practise canonical forms and simplification.
Q4. Use XOR cancellation (EMRS 2023)
If ⊕ represents XOR Gate in Boolean algebra and A ⊕ B = C, then B ⊕ C equals:
(a) 1
(b) A̅
(c) A
(d) 0
Answer: (c) A. Substitute C=A⊕B: B⊕C = B⊕A⊕B = A⊕(B⊕B) = A⊕0 = A. This uses XOR's associativity and commutativity, followed by self-cancellation.
Q5. Odd-parity evaluation (UGC NET 2016)
Which of the following logic expressions is incorrect ?
(a) 1 ⊕ 0 = 1
(b) 1 ⊕ 1 ⊕ 1 = 1
(c) 1 ⊕ 1 ⊕ 0 = 1
(d) 1 ⊕ 1 = 0
Answer: (c) 1 ⊕ 1 ⊕ 0 = 1. First, 1⊕1=0; then 0⊕0=0, so the printed equality is false. An even number of 1s gives XOR 0, while an odd number gives XOR 1.
XOR operator deduction MCQs
Q6. Identify an operator from its four defining cases (GATE 2016)
Consider the Boolean operator # with the following properties:
x # 0 = x, x # 1 = x̄, x # x = 0 and x # x̄ = 1. Then x # y is equivalent to
(a) xȳ + x̄y
(b) xȳ + x̄ȳ
(c) x̄y + xy
(d) xy + x̄ȳ
Answer: (a) xȳ + x̄y. For (x,y)=00,01,10,11, the stated properties give 0,1,1,0. That is XOR, whose sum-of-products form is xȳ+x̄y; the worked GATE 2016 operator question carries this question.
Q7. XOR/XNOR identities and a chained-XNOR trap (GATE 2018)
Let ⊕ and ⊙ denote the Exclusive OR and Exclusive NOR operations, respectively. Which one of the following is NOT CORRECT
(a)
¬(P ⊕ Q) = P ⊙ Q(b)
P' ⊕ Q = P ⊙ Q(c)
P' ⊕ Q' = P ⊕ Q(d)
P ⊕ P' ⊕ Q = P ⊙ P' ⊙ Q'
Answer: (d) P ⊕ P' ⊕ Q = P ⊙ P' ⊙ Q'. Take P=0, Q=0: the left side is 0⊕1⊕0=1, but the binary XNOR chain on the right is (0⊙1)⊙1 = 0⊙1 = 0. Options (a) to (c) follow directly from the complement rule, as also shown on the worked GATE 2018 XOR-XNOR identity question.
XNOR equivalence and parity MCQs
Q8. Recognise expressions for XNOR (GATE 2013; BARC 2013)
Which one of the following expressions does NOT represent exclusive NOR of x and y?
(a)
xy + x' y'(b)
x ⊕ y'(c)
x' ⊕ y(d)
x' ⊕ y'
Answer: (d) x' ⊕ y'. Complementing both inputs leaves their equality or difference relation unchanged, so x'⊕y' = x⊕y, which is XOR rather than XNOR. Complementing exactly one XOR input, as in (b) or (c), produces XNOR; see the worked GATE 2013 XNOR-equivalence question.
Q9. Four-variable XOR parity relation (GATE 2016)
Let x1 ⊕ x2 ⊕ x3 ⊕ x4 = 0, where x1, x2, x3 and x4 are Boolean variables and ⊕ is XOR. Which option must always be true?
(a)
x1x2x3x4 = 0(b)
x1x3 + x2 = 0(c)
x1' ⊕ x3' = x2' ⊕ x4'(d)
x1 + x2 + x3 + x4 = 0
Answer: (c) x1' ⊕ x3' = x2' ⊕ x4'. The condition gives x1⊕x3 = x2⊕x4, and a'⊕b' = a⊕b on both sides. For (x1,x2,x3,x4)=(1,0,1,0), the four-way XOR is 0, while option (c) gives 0⊕0=0 and 1⊕1=0; the worked GATE 2016 parity question shows the same result.
XOR circuit application MCQs
Q10. Swap two values with XOR (UGC NET 2005)
Consider the following sequence of instructions:
a = a ⊕ b,
b = a ⊕ b,
a = b ⊕ a.
This sequence
(a) retains the value of a and b
(b) complements the value of a and b
(c) swap a and b
(d) negates values of a and b
Answer: (c) swap a and b. Start with a=5=0101 and b=3=0011: line 1 makes a=0101⊕0011=0110=6; line 2 makes b=0110⊕0011=0101=5; line 3 makes a=0101⊕0110=0011=3. The final pair is (a,b)=(3,5), so the original values have exchanged places.
Q11. Turn an adder into an adder-subtractor (DSSSB 2021)
By including which logical gate, binary adder circuit is used for both binary addition and subtraction?
(a) Ex-AND gate
(b) AND gate
(c) Ex-OR gate
(d) OR gate
Answer: (c) Ex-OR gate. With control input M, each bit passes through Bi⊕M: at M=0, Bi⊕0=Bi, so the circuit adds; at M=1, Bi⊕1=Bi', and carry-in 1 makes A+B'+1=A-B in two's complement. For B=0101, the controlled XOR bank emits 0101 at M=0 and 1010 at M=1 before carry-in is applied.
The Combinational Circuits guide is the useful next read for adders, subtractors, multiplexers, and decoders.
Q12. Build a two-bit equality comparator with XNOR
Let x1x0 represent a 2-bit binary number with value 00, 01, 10 or 11. Similarly, y1y0 represents another 2-bit number. A circuit takes x1x0 and y1y0 as inputs and produces 1 only when the two numbers are equal. Which expression represents this logic?
(a) (x0 ⊙ x1) (y0 ⊙ y1)
(b) (x0 ⊕ x1) (y0 ⊕ y1)
(c) (x0 ⊙ y0) (x1 ⊙ y1)
(d) (x0 ⊕ y0) (x1 ⊕ y1)
Answer: (c) (x0 ⊙ y0)(x1 ⊙ y1). For x=10, y=10, the low-bit XNOR is 0⊙0=1, the high-bit XNOR is 1⊙1=1, and their AND is 1. For x=10, y=11, the low-bit comparison is 0⊙1=0, so the final AND output is 0.
XOR and XNOR MCQs: score, review, and next step
The questions address same-versus-different recognition, identities, parity, operator deduction, complement relations, and software and circuit applications. After 48 hours, retry only the group you missed instead of memorising answers immediately.
Use this score as a study diagnostic, not an exam-score prediction:
10 to 12: Move to mixed Digital Electronics MCQs.
7 to 9: Revise XOR identities and parity once, then retry the missed questions.
0 to 6: Rebuild from the four-row truth table before attempting another set.
For a full subject sequence, continue with GATE Guidance by Sanchit Sir. Then use the GATE CS Exam Preparation category to choose broader preparation coverage.
