Test: Boolean Algebra & Logic Gates- 2 - Computer Science Engineering (CSE) MCQ

# Test: Boolean Algebra & Logic Gates- 2 - Computer Science Engineering (CSE) MCQ

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## 15 Questions MCQ Test GATE Computer Science Engineering(CSE) 2025 Mock Test Series - Test: Boolean Algebra & Logic Gates- 2

Test: Boolean Algebra & Logic Gates- 2 for Computer Science Engineering (CSE) 2024 is part of GATE Computer Science Engineering(CSE) 2025 Mock Test Series preparation. The Test: Boolean Algebra & Logic Gates- 2 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Test: Boolean Algebra & Logic Gates- 2 MCQs are made for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Boolean Algebra & Logic Gates- 2 below.
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Test: Boolean Algebra & Logic Gates- 2 - Question 1

### The function can be written as

Detailed Solution for Test: Boolean Algebra & Logic Gates- 2 - 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 for Test: Boolean Algebra & Logic Gates- 2 - Question 2

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

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

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

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

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

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

By using NOT gate output is complement of input

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

The NAND can function as a NOT gate if

Detailed Solution for Test: Boolean Algebra & Logic Gates- 2 - Question 15

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

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