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GATE Electrical Engineering (EE) Test: Comparators Free Online Test 2026


MCQ Practice Test & Solutions: Test: Comparators (10 Questions)

You can prepare effectively for Electrical Engineering (EE) GATE Electrical Engineering (EE) Mock Test Series 2027 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Comparators". These 10 questions have been designed by the experts with the latest curriculum of Electrical Engineering (EE) 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 30 minutes
  • - Number of Questions: 10

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Test: Comparators - Question 1

Identify the 1 – bit comparator circuit 

Detailed Solution: Question 1

Comparator: 
A comparator is a combinational circuit that gives the output in terms of input signals.
It is used to compare numbers and represent their relationship with each other.
It compares the signals as A > B, A < B, and A = B.
1-bit comparator:
A 1-bit comparator circuit is shown below.

In this circuit, we can compare two input signals
The truth table for the 1-bit comparator is

The expression for A = B: A ⊙ B
The expression for A > B: A B̅ 
The expression for A < B: A̅ B

Test: Comparators - Question 2

A digital circuit which compares two numbers A3 A2 A1 A0, B3 B2 B1 B0 is shown in figure. To get output Y = 1, the correct pair of input numbers are

Detailed Solution: Question 2


Y = (A0 ⊕ B0) + (A1 ⊕ B1) + (A2 ⊕ B2) + (A3 ⊕ B3
Given that output is 1.
To get the output y = 1, all the inputs of OR gate should not be zero.
From the options, A3 A2 A1 A0 = 0010, B3 B2 B1 B0 = 1011 satisfies this condition.

Test: Comparators - Question 3

X = X1X0 and Y = Y1Y0 are 2-bit binary numbers. The Boolean function S that satisfies the condition “If X > Y, then S = 1”, in its minimized form, is

Detailed Solution: Question 3

x = x1 x0 y = y1 y0
The Boolean function S that satisfies condition,
If x > y, then s = 1.
We can represent the above condition through a truth table.

By using k-maps

Minimized form:
s = x11 + x1x00 + x010  

Test: Comparators - Question 4

Maximum conversion time in clock cycles for three types of 8 bit ADCs (i) Successive approximation, (ii) Dual slope and (iii) Parallel comparator are respectively

Detailed Solution: Question 4

Concept:
The conversion time of different types of n-bit ADC is shown :

  • From the above table, it is clear that the dual-slope is the slowest ADC and Flash Type is the fastest ADC.
  • The conversion type of Flash type is independent of the number of bits 

Application:
n = 8

  • Maximum conversion time in clock cycles of 8-bit Successive approximation type ADC  = n=  8
  • Maximum conversion time in clock cycles of 8 bit Dual slope type ADC = 2n+1 = 29 = 512
  • Maximum conversion time in clock cycles of 8 bit Successive approximation type ADC = 1

Test: Comparators - Question 5

The circuit shown in the figure is a –

Detailed Solution: Question 5

From the given circuit we can find the outputs expression as follows:

Now, we can observe that

So, the given circuit in question behaves like a comparator circuit. 

Test: Comparators - Question 6

Circuit for comparing 2 n-bit numbers has ____ entries in truth table

Detailed Solution: Question 6

In an n-bit, compare the n columns of bits in one binary number (let it be A) and n columns of bits of another number (let it be B).
For all possible values of bits in A and B truth table is taken for A > B, A < B and A = B.
So there are 2n inputs in the comparator.
For 2n inputs total possible combinations are 22n
So total number of entries is 22n.

Test: Comparators - Question 7

The given below logic circuit will work as

Detailed Solution: Question 7


By analysing the above truth table it is giving the expression for 1 bit comparator.

Test: Comparators - Question 8

The logic circuit of the figure is a

Detailed Solution: Question 8

Y = AB + A̅B̅
It represents a XNOR gate
The truth table of XNOR gate is


The output is high when both the inputs are equal.
It represents a 1-bit comparator (or) equality detector.

Test: Comparators - Question 9

The output Y of a 2-bit comparator is logic 1 whenever the 2-bit input A is greater than the 2-bit input B. The number of combinations for which the output is logic 1, is

Detailed Solution: Question 9

The only possible combinations are
A = 01 and B = 0 0
A = 10 and B = 00, 01
A = 11 and B = 00, 01, 10
So there are only 6 combinations
Tips and Tricks:

where n = 2 bit

Test: Comparators - Question 10

A 3 bit comparator circuit gives logic 1 output whenever the 3 bit input A is greater than the 3 bit input B. The output Boolean function y is represented by a K-map. The no. of 1 in the K-map is –

Detailed Solution: Question 10

Total no. of input combination of A & B are = 23 x 23 = 64
In 64 combination only in 8 combination A = B situation occurs. So the no. of combination for which A ≠ B is 64 - 8 = 56
Only of these 56 combinations only in 28 combinations A > B situation occur.
So, the no of 1 in K map will be 28

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