Combinational Logic Circuits and Data Selectors Practice Questions for GATE CS

Solve 143+ Combinational Logic Circuits and Data Selectors practice questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Combinational Logic Circuits and Data Selectors

Chapter Roadmap

Combinational Logic Circuits and Data Selectors

Your structured learning path to mastering digital building blocks.

1. Logic Gate Circuit Analysis and Hazards

Master propagation delays, static and dynamic hazards, and timing analysis.

2. Decoders, Multiplexers and Address Selection

Understand internal architecture, enable lines, and addressing logic.

3. Boolean Function Realization Using Multiplexers

Implement arbitrary logic functions using minimal multiplexer hardware.

4. Cascaded Multiplexer and Decoder Circuits

Analyze complex, multi-stage systems built from smaller modules.

Weightage Hint: Expect ~8 questions. Mastery of MUX realization and hazard elimination is critical.

Logic Gate Circuit Analysis and Hazards

Topic Introduction

Logic Gate Circuit Analysis and Hazards

Bridging the gap between ideal Boolean algebra and real-world timing.

What you will learn here

The concept of propagation delay and how it affects circuit timing.
The definition and causes of static and dynamic hazards in combinational logic.
How to identify potential hazards using Karnaugh maps.
The standard method for eliminating static hazards by adding redundant prime implicants.
Step-by-step techniques for tracing signal transitions through delayed gates.

Combinational Logic Circuits and Data Selectors: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Digital Logic MCQ

Assertion (A): When implementing the function $f (x, y, z) = \sum m (0, 1, 3, 4, 5, 6)$ using a 4-to-1 multiplexer with $y$ and $z$ as select lines, the data input $I_{2}$ is connected to the variable $x$ .

Reason (R): Minterm 2 (binary 010) is absent from the function, and minterm 6 (binary 110) is present, so the transition from 0 to 1 in the implementation table requires the MSB variable $x$ .

Which of the following is correct?

A.
Both A and R are true and R is the correct explanation of A
B.
Both A and R are true but R is NOT the correct explanation of A
C.
A is true but R is false
D.
A is false but R is true

Correct Answer:

Step-by-Step Solution

Key idea: This is a construction problem testing the implementation table method with assertion-reason format.

Step 1: Verify Assertion (A):

Select lines: $S_{1} = y$ , $S_{0} = z$
For $I_{2}$ : $S_{1} S_{0} = 10$ (binary 2)
This corresponds to minterms where $y z = 10$ : minterm 2 (010) and minterm 6 (110)
Minterm 2 is absent (0), minterm 6 is present (1)
Implementation table: top=0, bottom=1 $\to$ input = MSB = $x$
Assertion A is True ✓

Step 2: Verify Reason (R):

Minterm 2 (010) is absent ✓
Minterm 6 (110) is present ✓
Transition 0→1 requires MSB variable $x$ ✓
Reason R is True ✓

Step 3: Check if R explains A:

Yes, the absence of minterm 2 and presence of minterm 6 directly leads to $I_{2} = x$

Answer: Both A and R are true, and R is the correct explanation of A (Option A)

Question 2 · Digital Logic MCQ

When identifying static-1 hazards in a Sum-of-Products circuit using a Karnaugh map, a hazard exists during a single-variable transition between two adjacent cells if:

A.
both cells contain 0 and are covered by the same prime implicant loop
B.
both cells contain 1 and are covered by the same prime implicant loop
C.
both cells contain 1 but are not covered by the same prime implicant loop
D.
both cells contain 0 but are not covered by the same prime implicant loop

Correct Answer:

Step-by-Step Solution

Key idea: Static-1 hazards occur when adjacent 1-cells are not covered by a common prime implicant.

Step 1: Recall the condition for static-1 hazards in SOP circuits:

Output should remain at 1 during transition
Both cells must contain 1

Step 2: Recall the hazard condition:

The two adjacent 1-cells must NOT be covered by the same prime implicant loop

Step 3: Combine conditions:

Both cells contain 1 AND are not in the same loop

Answer: Option C

Question 3 · Digital Logic MCQ

Which of the following statements correctly identifies the condition for a Static-1 hazard on a Karnaugh map?

A.
Two adjacent 0s not covered by the same prime implicant loop
B.
Any two 1s in the K-map that are not adjacent
C.
Two adjacent 1s not covered by the same prime implicant loop
D.
Two diagonally adjacent 1s not covered by the same loop

Correct Answer:

Step-by-Step Solution

Key idea: This is a bounding question where you must identify the exact constraints for a Static-1 hazard.

Step 1: Recall that Static-1 hazards occur in SOP circuits when the output should stay at 1 but briefly drops to 0.

Step 2: On a K-map, this happens when two adjacent 1s are not covered by the same prime implicant loop.

Step 3: The constraints are:

Must be 1s (not 0s)
Must be adjacent (not diagonal or non-adjacent)
Must not be in the same loop

Step 4: Match with the options.

Answer: Option C

Question 4 · Digital Logic MCQ

Consider the following statement: "When implementing a 3-variable Boolean function using a 4-to-1 multiplexer with the implementation table method, the data inputs can only be connected to logic 0 or logic 1."

Is this statement true or false?

A.
True, because only constants are needed
B.
False, because the MSB variable or its complement may be required
C.
True, because external gates are not allowed
D.
False, because all variables must appear on data inputs

Correct Answer:

Step-by-Step Solution

Key idea: This is a bounding question testing the implementation table method constraints.

Step 1: Recall the implementation table method for $2^{n - 1}$ -to-1 MUX.

Step 2: For a 3-variable function with a 4-to-1 MUX, 2 variables are select lines, and 1 variable (MSB) determines data inputs.

Step 3: Data inputs can be: 0, 1, MSB variable, or complement of MSB variable.

Step 4: The statement says "only 0 or 1", which is false because variables can appear.

Answer: False, because the MSB variable or its complement may be required (Option B)

Question 5 · Digital Logic MCQ

Which of the following statements correctly describes the possible values for data inputs when implementing an $n$ -variable function using a $2^{n - 1}$ -to-1 multiplexer via the Implementation Table method?

A.
Any arbitrary boolean variable
B.
Only logic 0 or logic 1
C.
Logic 0, logic 1, the MSB variable, or its complement
D.
The LSB variable or its complement, but never a constant

Correct Answer:

Step-by-Step Solution

Key idea: This is a bounding question testing the constraints of the Implementation Table method.

Step 1: Recall the Implementation Table method for a $2^{n - 1}$ -to-1 MUX.

Step 2: The table compares the function values for MSB = 0 and MSB = 1.

Step 3: The possible outcomes for each column are:

(0, 0) $\to$ Logic 0
(1, 1) $\to$ Logic 1
(0, 1) $\to$ MSB variable
(1, 0) $\to$ Complement of MSB variable

Step 4: Match this with the options.

Answer: Logic 0, logic 1, the MSB variable, or its complement (Option C)

Question 6 · Digital Logic MCQ

A student claims that dynamic hazards can occur in 2-level logic circuits. According to the topic introduction, what is the minimum number of gate levels actually required for dynamic hazards to occur, contradicting this claim?

A.
1
B.
2
C.
3
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: Dynamic hazards require multi-level circuits with at least 3 gate levels.

Step 1: Recall the definition from the topic introduction:

"Dynamic hazard: Common in multi-level logic circuits (3+ gate levels)"

Step 2: The student's claim: dynamic hazards in 2-level circuits

Step 3: The contradiction: dynamic hazards require at least 3 levels

Step 4: Minimum number of levels = 3

Answer: 3 (Option C)

Question 7 · Digital Logic MCQ

A combinational circuit exhibits momentary false outputs due to unequal propagation delays. Considering all possible hazard types (Static-1, Static-0, and Dynamic), what is the total number of distinct hazard categories?

A.
1
B.
2
C.
4
D.
3

Correct Answer:

Step-by-Step Solution

Key idea: This is a casework question asking you to count the distinct types of hazards.

Step 1: List all hazard types mentioned in the concept:

Static-1 hazard
Static-0 hazard
Dynamic hazard

Step 2: Count the total number of distinct categories.

Total = 3

Step 3: Match with the given options.

Answer: 3 (Option D)

Question 8 · Digital Logic MCQ

Assertion (A): To select one of 32 words in a register file, 5 address bits are required.

Reason (R): The memory size is 32 bytes, and since each byte requires 1 address bit, 32 bytes require 5 bits.

Which of the following is correct?

A.
Both A and R are true and R is the correct explanation of A
B.
A is true but R is false
C.
Both A and R are true but R is NOT the correct explanation of A
D.
A is false but R is true

Correct Answer:

Step-by-Step Solution

Key idea: This is an assertion-reason question testing the distinction between words and bytes in memory interfacing.

Step 1: Evaluate Assertion (A): To select 1 of 32 words, we need $n$ bits where $2^{n} = 32$ . Thus, $n = 5$ . Assertion A is True.

Step 2: Evaluate Reason (R): R states the size is 32 bytes and each byte requires 1 bit. This is a unit mismatch. Address bits select words (or bytes, depending on addressing), but the calculation "32 bytes require 5 bits because each byte requires 1 bit" is logically false. The correct reasoning is $2^{5} = 32$ unique addresses.

Step 3: Conclusion: A is true, but R is false.

Answer: Option B

Question 9 · Digital Logic MCQ

A signal can travel from input A to output Y through two paths. Path 1 has 3 gates each of delay 2 ns. Path 2 has 2 gates each of delay 4 ns. What is the maximum propagation delay from A to Y?

A.
6 ns
B.
8 ns
C.
10 ns
D.
14 ns

Correct Answer:

Step-by-Step Solution

Key idea: Calculate the total delay for each path, then identify the maximum.

Step 1: Calculate delay for Path 1:

Path 1 delay = 3 gates × 2 ns/gate = 6 ns

Step 2: Calculate delay for Path 2:

Path 2 delay = 2 gates × 4 ns/gate = 8 ns

Step 3: Identify the maximum delay:

Max(6 ns, 8 ns) = 8 ns

Answer: 8 ns (Option B)

Question 10 · Digital Logic MCQ

Consider the following two statements about a timing analysis circuit where input A transitions at t = 0 and feeds through a NOT gate (delay 2 ns) and an AND gate (delay 3 ns) to an OR gate (delay 2 ns):

Assertion (A): The output Y experiences a static-1 hazard.

Reason (R): The hazard occurs because the NOT gate path (2 ns) and AND gate path (3 ns) have unequal propagation delays.

Which of the following is correct?

A.
Both A and R are true and R is the correct explanation of A
B.
Both A and R are true but R is NOT the correct explanation of A
C.
A is true but R is false
D.
A is false but R is true

Correct Answer:

Step-by-Step Solution

Key idea: Analyze the timing to verify both the assertion and the reason.

Step 1: Trace the signal paths:

Path 1 (NOT): A changes at t=0, NOT output changes at t=2 ns
Path 2 (AND): A changes at t=0, AND output changes at t=3 ns

Step 2: Analyze OR gate output Y:

At t=0: Assume initial state Y=1
At t=2 ns: NOT output changes (1→0), AND output still 0, so Y=0
At t=3 ns: AND output changes (0→1), so Y=1
Y transitions: 1 → 0 → 1 (glitch)

Step 3: Verify Assertion (A):

Y should stay at 1 but glitches to 0 → Static-1 hazard ✓

Step 4: Verify Reason (R):

Unequal delays (2 ns vs 3 ns) cause the timing mismatch ✓
This unequal delay is why the glitch occurs ✓

Step 5: Check if R explains A:

Yes, the unequal delays directly cause the hazard

Answer: Both A and R are true, and R correctly explains A (Option A)

Try a question

Login to view the complete practice questions and solutions

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

Combinational Logic Circuits and Data Selectors Practice Questions for GATE CS

Chapter Roadmap: Combinational Logic Circuits and Data Selectors

Combinational Logic Circuits and Data Selectors

1. Logic Gate Circuit Analysis and Hazards

2. Decoders, Multiplexers and Address Selection

3. Boolean Function Realization Using Multiplexers

4. Cascaded Multiplexer and Decoder Circuits

Logic Gate Circuit Analysis and Hazards

Logic Gate Circuit Analysis and Hazards

What you will learn here

Combinational Logic Circuits and Data Selectors: Solved Questions with Step-by-Step Explanations (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

More practice questions in this unit