Logo
Jeeify
MCQMediumJEE 2026Logic Gates

JEE Physics 2026 Question with Solution

The correct truth table for the given input data of the following logic gate is:

Logic circuit with inputs A and B entering an AND gate followed by a NOT gate, inputs C and D entering an OR gate, both outputs feeding an AND gate followed by a NOT gate to produce Y output.
  • A

    ABCDY11011001101010011111\begin{array}{|c|c|c|c|c|}\hline A & B & C & D & Y \\ \hline 1 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}

  • B

    ABCDY11011001101010111110\begin{array}{|c|c|c|c|c|}\hline A & B & C & D & Y \\ \hline 1 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \\ \hline \end{array}

  • C

    ABCDY11010001101010111111\begin{array}{|c|c|c|c|c|}\hline A & B & C & D & Y \\ \hline 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}

  • D

    ABCDY11010001111010111111\begin{array}{|c|c|c|c|c|}\hline A & B & C & D & Y \\ \hline 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The circuit has inputs A,B,C,DA, B, C, D. Inputs AA and BB pass through an AND gate followed by a NOT gate, while inputs CC and DD pass through an OR gate. These two outputs then enter an AND gate followed by a NOT gate.

Find: The correct truth table for output YY.

Step 1: Identify individual logic operations.

From the given circuit, the upper branch gives NAND of AA and BB, so its output is A⋅B‾\overline{A \cdot B}. The lower branch gives OR of CC and DD, so its output is C+DC + D.

Step 2: Write the Boolean expression.

The final gate is an AND followed by NOT, therefore

Y=(A⋅B‾⋅(C+D))‾Y = \overline{\left(\overline{A \cdot B} \cdot (C + D)\right)}

Using De Morgan's theorem,

Y=(A⋅B)+(C+D)‾Y = (A \cdot B) + \overline{(C + D)}

Step 3: Evaluate the given rows.

Checking the listed input combinations row-wise, the resulting output values match the truth table shown in option D.

Conclude: The correct option is D.

Boolean Simplification Trick

Given: The circuit is a combination of NAND, OR, and final NAND stages.

Find: Which option matches the truth table.

A quick approach is to simplify the circuit before checking rows. First write

Y=(A⋅B‾⋅(C+D))‾Y = \overline{\left(\overline{A \cdot B} \cdot (C + D)\right)}

Now apply De Morgan's theorem directly:

Y=(A⋅B)+(C+D)‾Y = (A \cdot B) + \overline{(C + D)}

This works because the complement of a product becomes the sum of complements.

Now compare this simplified form with the rows in the options instead of tracing the entire circuit gate by gate each time. The matching table is option D.

Therefore, the correct option is D.

Common mistakes

  • Treating the first branch as A⋅BA \cdot B instead of A⋅B‾\overline{A \cdot B} is incorrect because the AND gate is followed by a NOT gate. Always include the inversion after identifying the gate output.

  • Ignoring the final NOT gate leads to using A⋅B‾⋅(C+D)\overline{A \cdot B} \cdot (C + D) directly as output. This is wrong because the last stage is a NAND operation. First form the AND output, then complement it.

  • Applying De Morgan's theorem incorrectly to (A⋅B‾⋅(C+D))‾\overline{\left(\overline{A \cdot B} \cdot (C + D)\right)} can produce a wrong expression. The correct simplification is Y=(A⋅B)+(C+D)‾Y = (A \cdot B) + \overline{(C + D)}, not a product form.

Practice more Logic Gates questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Logic Gates

Related questions