GATE 2025 EC Q.23: solution and answer

GATE 2025 · ECQ.23 · 1 mark · MCQ

A 3-input majority logic gate has inputs $X, Y,$ and $Z$ . The output $F$ of the gate is logic '1' if two or more of the inputs are logic '1'. The output $F$ is logic '0' if two or more of the inputs are logic '0'.

Which one of the following options is a Boolean expression of the output $F$ ?

(A)
$X Y + Y Z + Z X$
(B)
$X \oplus Y \oplus Z$
(C)
$X + Y + Z$
(D)
$X Y Z$

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Answer and solution

ANSWER

(A) $X Y + Y Z + Z X$ . The majority function is 1 exactly for $X Y Z = 011, 101, 110, 111$ , i.e. $\sum m (3, 5, 6, 7)$ , and that minimises to $X Y + Y Z + Z X$ .

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

Truth table from the definition. $F = 1$ exactly when at least two of the three inputs are 1. With the minterm index read as $X Y Z$ in binary, that happens in rows $X Y Z = 011, 101, 110, 111$ . So $F = \sum m (3, 5, 6, 7)$ . In every other row at least two inputs are 0, so $F = 0$ there.
K-map grouping of the 1s. Minterm 7 ( $X Y Z = 111$ ) is adjacent to each of the other three 1s. Each pair differs in exactly one variable, and that variable drops out: $m_{3} + m_{7} = Y Z$ ( $X$ drops), $m_{5} + m_{7} = X Z$ ( $Y$ drops), $m_{6} + m_{7} = X Y$ ( $Z$ drops). The three pairs cover every 1 and nothing else, so $F = X Y + Y Z + Z X .$ This is option (A). No complemented literals appear anywhere in this question.
Check (B) at a single-1 input. At $X Y Z = 100$ : $X \oplus Y \oplus Z = 1 \oplus 0 \oplus 0 = 1$ . But only one input is 1, so $F$ must be 0. (B) is the odd-parity function (the sum output of a full adder), not the majority function.
Check (C) at the same input. At $X Y Z = 100$ : $X + Y + Z = 1$ , but $F$ must be 0. OR needs only one input high.
Check (D) at a two-1 input. At $X Y Z = 110$ : $X Y Z = 1 \cdot 1 \cdot 0 = 0$ , but two inputs are 1, so $F$ must be 1. AND needs all three inputs high.
Each of (B), (C) and (D) gives the wrong output in at least one row. (A) was derived from the full truth table, so it is correct in all 8 rows. Answer: (A).

KEY IDEA

A 3-input majority function is 1 when at least two inputs are 1, so it is $\sum m (3, 5, 6, 7)$ . Each 1 in that set pairs with $m_{7} = X Y Z$ , so the minimal sum of products is one AND term per pair of inputs: $F = X Y + Y Z + Z X$ . This is also the carry output of a full adder.

WATCH OUT FOR

Picking $X \oplus Y \oplus Z$ because it also comes from the full adder. That expression is the SUM output (odd parity), which is 1 when exactly one input is 1. The majority function is the CARRY output, $X Y + Y Z + Z X$ . Testing any single-1 input such as $X Y Z = 100$ rules out both XOR and OR at once.

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GATE 2025 EC Q.23: solution and answer

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