Logic Gates Questions and Answers

Answer: D

Correct answer: NAND

By De Morgan's law \( \overline{A}+\overline{B} = \overline{A\cdot B} \), which is the NAND function. It is the expression for a NAND written with the inversion pushed inwards onto the variables.

Answer: A

Correct answer: Y = A + B + C + D

OR is written with plus signs. The dot form would give AND, and the other two symbols are not Boolean operators.

Answer: D

Correct answer: 16

A truth table needs one row per input combination, and four inputs give \( 2^4 = 16 \).

Answer: C

Correct answer: 16

The number of rows depends only on the number of inputs, not on the function: \( 2^4 = 16 \) for any four-input gate.

Answer: B

Correct answer: X = A + B + C

OR is written as a sum, so a three-input OR sums all three variables.

Answer: D

Correct answer: A = 0, B = 0, C = 0

NOR is the complement of OR, and OR is 0 only when every input is 0. So the NOR output is HIGH in that single case out of eight.

Answer: C

Correct answer: NOR gate

NOR gives 0 whenever any input is 1, since OR would give 1 there. An OR gate would give a HIGH output in that case, and NAND goes LOW only when all inputs are HIGH.

Answer: B

Correct answer: all inputs are HIGH

NAND is the complement of AND, and AND is 1 only when every input is 1. So NAND is 0 in exactly that case and 1 in all the others.

Answer: B

Correct answer: False

The statement has the two the wrong way round. A truth table shows how the output level responds to all possible input combinations, not the reverse.

Answer: B

Correct answer: False

NOR gives 1 when all inputs are LOW, so a LOW input on its own does not force the output LOW - a NOR output is LOW when any input is HIGH.