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Test: Boolean Algebra & Logic Gates- 2

15 Questions MCQ Test Question Bank for GATE Computer Science Engineering | Test: Boolean Algebra & Logic Gates- 2

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This mock test of Test: Boolean Algebra & Logic Gates- 2 for Computer Science Engineering (CSE) helps you for every Computer Science Engineering (CSE) entrance exam. This contains 15 Multiple Choice Questions for Computer Science Engineering (CSE) Test: Boolean Algebra & Logic Gates- 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Boolean Algebra & Logic Gates- 2 quiz give you a good mix of easy questions and tough questions. Computer Science Engineering (CSE) students definitely take this Test: Boolean Algebra & Logic Gates- 2 exercise for a better result in the exam. You can find other Test: Boolean Algebra & Logic Gates- 2 extra questions, long questions & short questions for Computer Science Engineering (CSE) on EduRev as well by searching above.
QUESTION: 1

Solution:

QUESTION: 2

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

Solution:

From the given k-map we can observe that B = 1 always. Also if A is to be paired than it should be
1. If A is selected there is no need to take C.
Hence, A - 1, B = 1, C = 0 Hence option (d) is correct

QUESTION: 3

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

Solution:

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?

Solution:

The output of

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

Solution:

Constructing the table we have

Hence the required expression is

QUESTION: 6

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

Solution:

Solving the given K-map we have

Hence (b) is correct option

QUESTION: 7

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

Solution:

Hence, the required boolean expression is

Hence (b) is the required option.

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.

Solution:

Only (c) option satisfies the required condition.

QUESTION: 9

The logical expression for K-Map shown above is:

Solution:

QUESTION: 10

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

Solution:

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:

Solution:

Constructing the K-map we have

Hence the required output is

QUESTION: 12

Consider

the Y is equivalent to:

Solution:

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

Solution:

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

Which is true
Hence (d) is correct option.

QUESTION: 14

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

Solution:

By using NOT gate output is complement of input

QUESTION: 15

The NAND can function as a NOT gate if

Solution:

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