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Boolean Algebra And Logic Simplification Online Exam Quiz

Important questions about Boolean Algebra And Logic Simplification. Boolean Algebra And Logic Simplification MCQ questions with answers. Boolean Algebra And Logic Simplification exam questions and answers for students and interviews.

1. How many gates would be required to implement the following Boolean expression after simplification? XY + X(X + Z) + Y(X + Z)

Options

A : 1

B : 2

C : 4

D : 5

2. The commutative law of Boolean addition states that A + B = A × B .

Options

A : True

B : False

C :

D :

3. Determine the values of A, B, C, and D that make the sum term equal to zero.

Options

A : A = 1, B = 0, C = 0, D = 0

B : A = 1, B = 0, C = 1, D = 0

C : A = 0, B = 1, C = 0, D = 0

D : A = 1, B = 0, C = 1, D = 1

4. A Karnaugh map is a systematic way of reducing which type of expression?

Options

A : product-of-sums

B : exclusive NOR

C : sum-of-products

D : those with overbars

5. The Boolean expression is logically equivalent to what single gate?

Options

A : NAND

B : NOR

C : AND

D : OR

7. Which of the following is an important feature of the sum-of-products (SOP) form of expression?

Options

A : All logic circuits are reduced to nothing more than simple AND and OR gates.

B : The delay times are greatly reduced over other forms.

C : No signal must pass through more than two gates, not including inverters.

D : The maximum number of gates that any signal must pass through is reduced by a factor of two.

8. Which Boolean algebra property allows us to group operands in an expression in any order without affecting the results of the operation [for example, A + B = B + A ]?

Options

A : associative

B : commutative

C : Boolean

D : distributive

10. An AND gate with schematic "bubbles" on its inputs performs the same function as a(n)________ gate.

Options

A : NOT

B : OR

C : NOR

D : NAND

1. The truth table for the SOP expression has how many input combinations?

Options

A : 1

B : 2

C : 4

D : 8

3. When grouping cells within a K-map, the cells must be combined in groups of ________.

Options

A : 2s

B : 1, 2, 4, 8, etc.

C : 4s

D : 3s

4. What is the primary motivation for using Boolean algebra to simplify logic expressions?

Options

A : It may make it easier to understand the overall function of the circuit.

B : It may reduce the number of gates.

C : It may reduce the number of inputs required.

D : all of the above

9. Which statement below best describes a Karnaugh map?

Options

A : A Karnaugh map can be used to replace Boolean rules.

B : The Karnaugh map eliminates the need for using NAND and NOR gates.

C : Variable complements can be eliminated by using Karnaugh maps.

D : Karnaugh maps provide a cookbook approach to simplifying Boolean expressions.

10. Which of the examples below expresses the commutative law of multiplication?

Options

A : A + B = B + A

B : AB = B + A

C : AB = BA

D : AB = A × B

5. Occasionally, a particular logic expression will be of no consequence in the operation of a circuit, such as a BCD-to-decimal converter. These result in ________terms in the K-map and can be treated as either ________ or ________, in order to ________ the resulting term.

Options

A : don't care, 1s, 0s, simplify

B : spurious, ANDs, ORs, eliminate

C : duplicate, 1s, 0s, verify

D : spurious, 1s, 0s, simplify

6. A truth table for the SOP expression has how many input combinations?

Options

A : 1

B : 2

C : 4

D : 8

7. AC + ABC = AC

Options

A : True

B : False

C :

D :

9. Use Boolean algebra to find the most simplified SOP expression for F = ABD + CD + ACD + ABC + ABCD .

Options

A : F = ABD + ABC + CD

B : F = CD + AD

C : F = BC + AB

D : F = AC + AD

2. Which of the following combinations cannot be combined into K-map groups?

Options

A : corners in the same row

B : corners in the same column

C : diagonal

D : overlapping combinations

5. The expression W(X + YZ) can be converted to SOP form by applying which law?

Options

A : associative law

B : commutative law

C : distributive law

D : none of the above

6. One of De Morgan's theorems states that . Simply stated, this means that logically there is no difference between:

Options

A : a NOR and an AND gate with inverted inputs

B : a NAND and an OR gate with inverted inputs

C : an AND and a NOR gate with inverted inputs

D : a NOR and a NAND gate with inverted inputs

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