GATE 2023 IN Q.17: solution and answer

GATE 2023 · INQ.17 · 1 mark · MCQ

Match the Exclusive-OR (XOR) operations (i) to (iv) with the results (p) to (s), where $X$ is a Boolean input.

| Operation | Result |
|---|---|
| (i) $X \oplus X$ | (p) $1$ |
| (ii) $X \oplus \overline{X}$ | (q) $0$ |
| (iii) $X \oplus 0$ | (r) $\overline{X}$ |
| (iv) $X \oplus 1$ | (s) $X$ |

(A)
(i) - (q), (ii) - (r), (iii) - (s), (iv) - (p)
(B)
(i) - (q), (ii) - (r), (iii) - (p), (iv) - (s)
(C)
(i) - (p), (ii) - (s), (iii) - (q), (iv) - (r)
(D)
(i) - (q), (ii) - (p), (iii) - (s), (iv) - (r)

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Answer and solution

ANSWER

(D): (i)-(q), (ii)-(p), (iii)-(s), (iv)-(r). So $X \oplus X = 0$ , $X \oplus \overline{X} = 1$ , $X \oplus 0 = X$ and $X \oplus 1 = \overline{X}$ .

Official answer: DSource: Official answer key, GATE 2023 IN (IIT Kanpur)

Definition of XOR (complement written as ): $A \oplus B = A \overline{B} + \overline{A} B$ , which is $1$ exactly when the two inputs differ.
(i) $X \oplus X = X \overline{X} + \overline{X} X = 0 + 0 = 0$ , since $X \overline{X} = 0$ . The two inputs are always equal, so the result is (q).
(ii) $X \oplus \overline{X} = X \cdot \overline{\overline{X}} + \overline{X} \cdot \overline{X} = X \cdot X + \overline{X} \cdot \overline{X} = X + \overline{X} = 1$ . The two inputs always differ, so the result is (p).
(iii) $X \oplus 0 = X \cdot \overline{0} + \overline{X} \cdot 0 = X \cdot 1 + 0 = X$ . XOR with $0$ passes the input through unchanged, so the result is (s).
(iv) $X \oplus 1 = X \cdot \overline{1} + \overline{X} \cdot 1 = X \cdot 0 + \overline{X} = \overline{X}$ . XOR with $1$ inverts the input (a controlled inverter), so the result is (r).
The matching is (i)-(q), (ii)-(p), (iii)-(s), (iv)-(r), which is option (D).

KEY IDEA

XOR outputs $1$ exactly when its inputs differ. So a signal XORed with itself gives $0$ , and XORed with its complement gives $1$ . XOR with a constant makes a controlled inverter: a control of $0$ passes $X$ through unchanged, and a control of $1$ inverts it to $\overline{X}$ .

WATCH OUT FOR

Mixing up the two constant cases: taking $X \oplus 1 = 1$ or $X \oplus 0 = 0$ by treating XOR like OR or AND. Another mistake is swapping (i) and (ii), writing $X \oplus X = 1$ because "a signal matches itself". Options (A), (B) and (C) each contain one of these slips.

From the GATE 2023 IN paper. Every question of this paper · Download the full paper · All 22 Boolean logic questions, with notes

GATE 2023 IN Q.17: solution and answer

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