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Digital Electronics
Boolean Algebra

Practice questions from Boolean Algebra.

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Q#1 Boolean Algebra GATE EE 2025 (Set 1) MCQ +2 marks -0.66 marks

A Boolean function is given as

 

The simplified form of this function is represented by

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Q#2 Boolean Algebra GATE EE 2024 (Set 1) MCQ +1 mark -0.33 marks

Simplified form of the Boolean function

 Is

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Q#3 Boolean Algebra GATE EE 2019 (Set 1) MCQ +2 marks -0.66 marks

In the circuit shown below, X and Y are digital inputs, and Z is a digital output. The equivalent circuit is a

XNOR

NOR gate

XOR gate

NAND gate

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Q#4 Boolean Algebra GATE EE 2018 (Set 1) MCQ +1 mark -0.33 marks

In the logic circuit shown in the figure. Y is given by

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Q#5 Boolean Algebra GATE EE 2018 (Set 1) MCQ +2 marks -0.66 marks

Digital input signals A, B, C with A as the MSB and C as the LSB are used to realize the Boolean function

, Where to denote the minterm. In addition, F has a don't care for. The simplified expression for F is given by

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Q#6 Boolean Algebra GATE EE 2017 (Set 1) MCQ +1 mark -0.33 marks

The Boolean expression  simplifies to

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Q#7 Boolean Algebra GATE EE 2017 (Set 2) MCQ +1 mark -0.33 marks

For a 3-input logic circuit shown below, the output Z can be expressed as

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Q#8 Boolean Algebra GATE EE 2016 (Set 2) MCQ +2 marks -0.66 marks

The Boolean expression  simplifies to

1

a, b

0

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Q#9 Boolean Algebra GATE EE 2013 (Set 1) MCQ +1 mark -0.33 marks

A bulb in a staircase has two switches, one switch being at the ground floor and the other one at the first floor. The bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch. The logic of switching of the bulb resembles        

an AND gate

an OR gate

an XOR gate

a NAND gate

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Q#10 Boolean Algebra GATE EE 2011 (Set 1) MCQ +1 mark -0.33 marks

The output Y of the logic circuit given below is

1

0

X

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Q#11 Boolean Algebra GATE EE 2010 (Set 1) MCQ +2 marks -0.66 marks

The TTL circuit shown in the figure is fed with the waveform X (also shown). All gates have equal propagation delay of 10ns. The output Y of the circuit is

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Q#12 Boolean Algebra GATE EE 2009 (Set 1) MCQ +1 mark -0.33 marks

The complete set of only those Logic Gates designated as Universal Gates is

NOT, OR and AND Gates

XNOR, NOR and NAND Gates

NOR and NAND Gates

XOR, NOR and NAND Gates

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Q#13 Boolean Algebra GATE EE 2004 (Set 1) MCQ +2 marks -0.66 marks

The simplified form of the Boolean expression can be written as

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Q#14 Boolean Algebra GATE EE 2003 (Set 1) MCQ +2 marks -0.66 marks

The Boolean expression

can be simplified to

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Q#15 Boolean Algebra GATE EE 2002 (Set 1) MCQ +2 marks -0.66 marks

For the circuit shown in figure, the Boolean expression for the output Y in terms of inputs P, Q, R and S is

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Q#16 Boolean Algebra GATE EE 2001 (Set 1) MCQ +1 mark -0.33 marks

The output of a logic gate is “1” when all its inputs are at logic “0”.  The gate is either.

A NAND or an EX-OR gate

A NOR or an EX-OR gate

An AND or an EX-NOR gate

A NOR or an EX-NOR gate

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Q#17 Boolean Algebra GATE EE 1999 (Set 1) MCQ +2 marks -0.66 marks

The logic function  is the same as

None of (a), (b) (c)

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Q#18 Boolean Algebra GATE EE 1996 (Set 1) MCQ +1 mark -0.33 marks

The Boolean expression for the output of the logic circuit shown in figure is

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Q#19 Boolean Algebra GATE EE 1995 (Set 1) MCQ +2 marks -0.66 marks

The min-terms of a four variable Boolean function Y is given by  

Use Karnaugh map to minimize the function Y. Realize function Y employing only three-input NAND gates.

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