NIMCET 2010 — Computer PYQ

Q: How do I solve NIMCET 2010 Computer questions?

Practice NIMCET 2010 Computer previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of NIMCET Computer questions?

NIMCET Computer questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are NIMCET previous year papers available for free?

Yes. Mathem Solvex provides free access to NIMCET previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

NIMCET 2010 — Computer PYQ

NIMCET | Computer | 2010

Which gate is equivalent to (NOR) OR (XOR)?

Choose the correct answer:

Explanation

Solution

The correct answer is: (a) NAND gate

1. Write the individual expressions:

$A \text{ NOR } B = \overline{A + B}$
$A \text{ XOR } B = A\overline{B} + \overline{A}B$

2. Combine them using the OR operation:

Y = (\overline{A + B}) + (A\overline{B} + \overline{A}B)

3. Simplify using De Morgan's Law and Boolean identities:

Using De Morgan's Law: $\overline{A + B} = \overline{A} \cdot \overline{B}$
Substitute back: $Y = (\overline{A}\overline{B}) + (A\overline{B} + \overline{A}B)$
Rearrange the terms: $Y = \overline{A}\overline{B} + \overline{A}B + A\overline{B}$
Factor out $\overline{A}$ from the first two terms: $Y = \overline{A}(\overline{B} + B) + A\overline{B}$
Since $(\overline{B} + B) = 1$ : $Y = \overline{A}(1) + A\overline{B}$
$Y = \overline{A} + A\overline{B}$
Using the Distributive Law ( $X + \overline{X}Z = X + Z$ ):

$Y = \overline{A} + \overline{B}$

4. Final Result:

According to De Morgan's Law, $\overline{A} + \overline{B} = \overline{A \cdot B}$ .

The expression $\overline{A \cdot B}$ represents the NAND gate.