GATE 2025 EC Q.24: solution and answer

GATE 2025 · ECQ.24 · 1 mark · MCQ

A full adder and an XOR gate are used to design a digital circuit with inputs $X$ , $Y$ , and $Z$ , and output $F$ , as shown below. The input $Z$ is connected to the carry-in input of the full adder.

If the input $Z$ is set to logic '1', then the circuit functions as ________ with $X$ and $Y$ as inputs.

GATE 2025 EC question 24: Three inputs X, Y, Z enter from the top. — Figure from the GATE 2025 EC paper. It is part of the question.

(A)
an adder
(B)
a subtractor
(C)
a multiplier
(D)
a binary to Gray code converter

Answer and solution

ANSWER

With $Z = 1$ the XOR gate feeds $\overline{Y}$ into the full adder and the carry-in is 1. The adder therefore computes $X + \overline{Y} + 1 = X - Y$ , so the circuit works as a subtractor. The answer is (B).

Official answer: BSource: Official answer key, GATE 2025 EC (IIT Roorkee)

Read the circuit: the full adder's operands are $A = X$ and $B = Y \oplus Z$ , and its carry-in is $C_{in} = Z$ .
XOR acts as a controlled inverter: $Y \oplus 1 = \overline{Y}$ (an overline means complement). So with $Z = 1$ we get $B = \overline{Y}$ and $C_{in} = 1$ .
The full adder now computes $X + \overline{Y} + 1$ . Since $\overline{Y} + 1$ is the 2's complement of $Y$ , this is $X - Y$ , which is subtraction by 2's complement.
Check the sum bit: $S = X \oplus \overline{Y} \oplus 1 = X \oplus Y$ . This is exactly the difference bit $D$ of a subtractor.
Check the carry bit using $C_{o u t} = A B + C_{in} (A \oplus B)$ : $C_{o u t} = X \overline{Y} + (X \oplus \overline{Y}) \cdot 1 = X \overline{Y} + X Y + \overline{X} \overline{Y} = X + \overline{Y}$ . Its complement is $\overline{C_{o u t}} = \overline{X} Y$ , which is the borrow of $X - Y$ .
Truth table for $(X, Y) \to (C_{o u t}, S)$ : $00 \to 10$ , $01 \to 01$ , $10 \to 11$ , $11 \to 10$ . So $S = X \oplus Y$ , and the borrow $= \overline{C_{o u t}}$ is 1 only for $X = 0, Y = 1$ . That is the subtractor's truth table.
(When $Z = 0$ , $B = Y$ and $C_{in} = 0$ , so the same circuit is an adder. $Z$ is the add/subtract control.) With $Z = 1$ the circuit is a subtractor, so the answer is (B).

KEY IDEA

An XOR gate with one input tied to a control line $M$ passes $Y$ through when $M = 0$ and inverts it when $M = 1$ . If the same $M$ also drives the carry-in, the full adder computes $X + Y$ for $M = 0$ and $X + \overline{Y} + 1 = X - Y$ (2's complement) for $M = 1$ . This is the standard adder-subtractor, and the difference bit is $X \oplus Y$ while the borrow is $\overline{C_{o u t}}$ .

WATCH OUT FOR

Answering "an adder" because the block is labelled full adder, without noticing that the XOR gate complements $Y$ when $Z = 1$ and that $Z$ also supplies the carry-in of 1. A second slip is reading $C_{o u t}$ itself as the borrow. In 2's-complement subtraction the borrow is the complement of the carry-out: here $C_{o u t} = X + \overline{Y}$ , and borrow $= \overline{X} Y$ .

From the GATE 2025 EC paper. Every question of this paper · Download the full paper · All 12 Combinational questions, with notes

GATE 2025 EC Q.24: solution and answer

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