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GATE Computer Science Engineering(CSE) 2027 Test: Boolean Algebra & Logic


MCQ Practice Test & Solutions: Test: Boolean Algebra & Logic Gates- 2 (15 Questions)

You can prepare effectively for Computer Science Engineering (CSE) GATE Computer Science Engineering(CSE) 2027 Mock Test Series with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Boolean Algebra & Logic Gates- 2". These 15 questions have been designed by the experts with the latest curriculum of Computer Science Engineering (CSE) 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 45 minutes
  • - Number of Questions: 15

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Test: Boolean Algebra & Logic Gates- 2 - Question 1

The function

can be written as

Detailed Solution: Question 1

Test: Boolean Algebra & Logic Gates- 2 - Question 2

The Karnaugh map for the Boolean function F of 4 Boolean variables is given in Figure. A, B, C are don't care conditions. What values of A, B, C will result in the minimal expression?

Detailed Solution: Question 2

Test: Boolean Algebra & Logic Gates- 2 - Question 3

Let f(A, B) =   then simplified from of the function 

Detailed Solution: Question 3

Test: Boolean Algebra & Logic Gates- 2 - Question 4

Consider the circuit-shown below. Each of the control inputs, C0 through C3, must be tied to a constant, either ‘0’ or '1'.

What are the values of C0 through C3 that would cause F to be the exclusive OR of A and B?

Detailed Solution: Question 4

The output of

Test: Boolean Algebra & Logic Gates- 2 - Question 5

Write the minimized expression of the function of three variables that is 1 if the third variable is equal to the OR of the first two variable, 0 otherwise

Detailed Solution: Question 5

Constructing the table we have

Hence the required expression is

Test: Boolean Algebra & Logic Gates- 2 - Question 6

Which of the following functions implements the Karnaugh map shown below?

Detailed Solution: Question 6

Solving the given K-map we have

Hence (b) is correct option

Test: Boolean Algebra & Logic Gates- 2 - Question 7

The Boolean expression for the shaded area in the given Venn diagram is

Detailed Solution: Question 7

Hence, the required boolean expression is 


Hence (b) is the required option.

Test: Boolean Algebra & Logic Gates- 2 - Question 8

The Boolean expression is a simplified version of expression:

Then which of the following choice is correct:
1. Don’t care conditions don’t exist.
2. Don’t care conditions exist.
3. D (16, 18, 20, 23, 27, 29) is the set of don’t care conditions.
​4. D (16, 20, 22, 27, 29) is the set of don’t care conditions.

Detailed Solution: Question 8


Only (c) option satisfies the required condition.

Test: Boolean Algebra & Logic Gates- 2 - Question 9

The logical expression for K-Map shown above is:

Detailed Solution: Question 9


Test: Boolean Algebra & Logic Gates- 2 - Question 10

The Boolean expression for the shaded area in the Venn diagram is

Detailed Solution: Question 10

Test: Boolean Algebra & Logic Gates- 2 - Question 11

Consider, a four variable Boolean function, which contains half the number of minterms with odd number of 1 ’s. Then the Boolean can be realized with variables A, B, C, D as:

Detailed Solution: Question 11

Constructing the K-map we have


Hence the required output is

Test: Boolean Algebra & Logic Gates- 2 - Question 12

Consider

the Y is equivalent to:

Detailed Solution: Question 12

Test: Boolean Algebra & Logic Gates- 2 - Question 13

An X-Y flip-flop, whose characteristic table is given below, is to be implemented using a JK flip-flop. This can be done by making

Detailed Solution: Question 13


The JK fiip-flop can be implemented by making , K = x 

Which is true
Hence (d) is correct option.

Test: Boolean Algebra & Logic Gates- 2 - Question 14

For what logic gate, the output is complement of the input?

Detailed Solution: Question 14

The logic gate that produces the complement of the input is the NOT gate.

  • It inverts the input signal.
  • If the input is 1, the output is 0.
  • If the input is 0, the output is 1.

In essence, the NOT gate changes the state of its input.

Test: Boolean Algebra & Logic Gates- 2 - Question 15

The NAND can function as a NOT gate if

Detailed Solution: Question 15

NAND gate work as NOT if inputs are connected together i.e.

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