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Digital Electronics
Boolean Algebra

Practice questions from Boolean Algebra.

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Q#1 Boolean Algebra GATE EC 2025 (Set 1) MCQ +1 mark -0.33 marks

A 3-input majority logic gate has inputs , and . The output F of the gate is logic '' if two or more of the inputs are logic ''. The output  is logic '  ' if two or more of the inputs are logic '0'.

Which one of the following options is a Boolean expression of the output F ?

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

For the Boolean function

 

the essential prime implicants are________

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Q#3 Boolean Algebra GATE EC 2022 (Set 1) MSQ +1 mark -0 marks

Select the Boolean function(s) equivalent to  , where , and  are Boolean variables, and + denotes logical OR operation.

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Q#4 Boolean Algebra GATE EC 2022 (Set 1) MSQ +2 marks -0 marks

Consider a Boolean gate  where the output  is related to the inputs  and  as, , where + denotes logical OR operation. The Boolean inputs ' 0 ' and ' 1 ' are also available separately. Using instances of only  gates and inputs ' 0 ' and ' 1 ', (select the correct options)

NAND logic can be implemented

OR logic cannot be implemented

NOR logic can be implemented

AND logic cannot be implemented

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Q#5 Boolean Algebra GATE EC 2018 (Set 1) MCQ +1 mark -0.33 marks

A function  defined by three Boolean variables A, B and C when expressed as sum of products is given by

Whereandare the complements of the respective variables. The product of sums (PUS) form of the function F is

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Q#6 Boolean Algebra GATE EC 2018 (Set 1) NAT +2 marks -0 marks

The logic gates shown in the digital circuit below use strong pull-down NMOS transistors for LOW logic level at the outputs. When the pull-downs are off, high-value resistors set the output logic levels to HIGH (i.e. the pull-ups are weak). Note that some nodes are intentionally shorted to implement -wired logic". Such shorted nodes will be HIGH only if the outputs of all the gates whose outputs are shorted are HIGH.  

Untitled-13.png

The number of distinct values of  (out of the 16 possible values) that give Y =1 is _______

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Q#7 Boolean Algebra GATE EC 2017 (Set 1) MCQ +2 marks -0.66 marks

Which one of the following gives the simplified sum of products expression for the Boolean function are minterms corresponding to the inputs A, B and C with A as the MSB and C as the LSB?

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Q#8 Boolean Algebra GATE EC 2016 (Set 1) MCQ +1 mark -0.33 marks

The output of the combinational circuit given below is

Q

A+B+C

A(B+C)

B(C+A)

C(A+B)

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Q#9 Boolean Algebra GATE EC 2016 (Set 1) MCQ +1 mark -0.33 marks

The minimum number of 2-input NAND gates required to implement a 2-input XOR gate is

4

5

6

7

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Q#10 Boolean Algebra GATE EC 2016 (Set 1) MCQ +2 marks -0.66 marks

Following is the K-map of a Boolean function of five variables P, Q, R, S and X.  The minimum sum-of-product (SOP) expression for the function is

30.jpg

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Q#11 Boolean Algebra GATE EC 2015 (Set 1) MCQ +2 marks -0.66 marks

The Boolean expression  converted into the canonical product of sum (POS) form is

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Q#12 Boolean Algebra GATE EC 2015 (Set 1) NAT +2 marks -0 marks

All the logic gates shown in the figure have a propagation delay of 20 ns. Let A = C = 0 and B = 1 until time t = 0. At t = 0, all the inputs flip (i.e., A = C = 1 and B = 0) and remain in that state. For t > 0, output Z = 1 for a duration (in ns) of ______________.

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Q#13 Boolean Algebra GATE EC 2015 (Set 1) MCQ +2 marks -0.66 marks

A 3-input majority gate is defined by the logic function. Which one of the following gates is represented by the function?

3-input NAND gate

3-input XOR gate

3-input NOR gate

3-input XNOR gate

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Q#14 Boolean Algebra GATE EC 2015 (Set 2) MCQ +1 mark -0.33 marks

In the figure shown, the output Y is required to be. The gates G1 and G2 must be, respectively,

NOR, OR

OR, NAND

NAND, OR

AND, NAND

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Q#15 Boolean Algebra GATE EC 2015 (Set 2) MCQ +2 marks -0.66 marks

A function of Boolean variables X, Y and Z is expressed in terms of the min-terms as

Which one of the product of sums given below is equal to the function F(X, Y, Z)?

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Q#16 Boolean Algebra GATE EC 2014 (Set 1) MCQ +1 mark -0.33 marks

The Boolean expression  simplifies to

X

Y

XY

X+Y

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Q#17 Boolean Algebra GATE EC 2014 (Set 1) MCQ +2 marks -0.66 marks

The output F in the digital logic circuit shown in the figure is

Q40-1.jpg

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Q#18 Boolean Algebra GATE EC 2014 (Set 1) MCQ +2 marks -0.66 marks

Consider the Boolean function,  . Which one of the following is the complete set of essential prime implicants?

w,y,xz

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Q#19 Boolean Algebra GATE EC 2014 (Set 2) MCQ +1 mark -0.33 marks

For an n-variable Boolean function, the maximum number of prime implicants is

2(n-1)

n/2

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Q#20 Boolean Algebra GATE EC 2014 (Set 4) MCQ +1 mark -0.33 marks

In the circuit shown in the figure, if C=0, the expression for Y is

Q15-1.jpg

Y=A+B

Y=AB

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Q#21 Boolean Algebra GATE EC 2014 (Set 3) MCQ +2 marks -0.66 marks

A universal logic gate can implement any Boolean function by connecting sufficient number of them appropriately. Three gates are shown.

Which one of the following statements is TRUE?

Gate 1 is a universal gate

Gate 2 is a universal gate

Gate 3 is a universal gate

None of the gates shown is a universal gate

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Q#22 Boolean Algebra GATE EC 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#23 Boolean Algebra GATE EC 2012 (Set 1) MCQ +1 mark -0.33 marks

In the sum of products function f(X,Y,Z) = the prime implicant’s are

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Q#24 Boolean Algebra GATE EC 2011 (Set 1) MCQ +1 mark -0.33 marks

The output Y in the circuit below is always “1” when        

15.jpg

Two or more of the inputs P, Q, R are “0”

Two or more of the inputs P, Q, R are “1”

Any odd number of the inputs P, Q, R are “0”

Any odd number of the inputs P, Q, R are “1”

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Q#25 Boolean Algebra GATE EC 2010 (Set 1) MCQ +1 mark -0.33 marks

Match the logic gates in Column A with their equivalents in Column B.

P-2, Q-4, R-1, S-3

P-4, Q-2, R-1, S-3

P-2, Q-4, R-3, S-1

P-4, Q-2, R-3, S-1

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Q#26 Boolean Algebra GATE EC 2010 (Set 1) MCQ +1 mark -0.33 marks

For the output F to be 1 in the logic circuit shown, the input combination should be        

7.jpg

A = 1, B = 1,         C = 0

A = 1, B = 0,         C = 0

A = 0, B = 1,         C = 0

A = 0, B = 0,         C = 1

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Q#27 Boolean Algebra GATE EC 2009 (Set 1) MCQ +2 marks -0.66 marks

If X =1 in the logic equation , then

Y = Z

Z = 1

Z = 0

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Q#28 Boolean Algebra GATE EC 2009 (Set 1) MCQ +2 marks -0.66 marks

What are the minimum number of 2-to-1 multiplexers required to generate a 2-input AND gate and a 2-input Ex-OR gate?

1 and 2

1 and 3

1 and 1

2 and 2

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Q#29 Boolean Algebra GATE EC 2009 (Set 1) MCQ +2 marks -0.66 marks

Two products are sold from a vending machine, which has two push buttons and . When a button is pressed, the price of the corresponding product is displayed in a 7-segment display.

If no buttons are pressed, ‘0’ is displayed, signifying ‘Rs. 0’.

If only  is pressed, ‘2’ is displayed, signifying ‘Rs. 2’.

If only  is pressed, ‘5’ is displayed, signifying ‘Rs. 5’.

If both  and  are pressed, ‘E’ is displayed, signifying ‘Error’.

The names of the segments in the 7-segment display, and the glow of the display for ‘0’, ‘2’, ‘5’ and ‘E’, are shown below.

Consider

(i) Push button pressed /not pressed is equivalent to logic 1/0 respectively,

(ii) A segment glowing / not glowing in the display is equivalent to logic 1/0 respectively

If the segments a to g are considered as functions of  and , then which of the following is correct?        

,

,

,

,

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Q#30 Boolean Algebra GATE EC 2009 (Set 1) MCQ +2 marks -0.66 marks

Two products are sold from a vending machine, which has two push buttons and . When a button is pressed, the price of the corresponding product is displayed in a 7-segment display.

If no buttons are pressed, ‘0’ is displayed, signifying ‘Rs. 0’.

If only  is pressed, ‘2’ is displayed, signifying ‘Rs. 2’.

If only  is pressed, ‘5’ is displayed, signifying ‘Rs. 5’.

If both  and  are pressed, ‘E’ is displayed, signifying ‘Error’.

The names of the segments in the 7-segment display, and the glow of the display for ‘0’, ‘2’, ‘5’ and ‘E’, are shown below.

Consider

(i) Push button pressed /not pressed is equivalent to logic 1/0 respectively,

(ii) A segment glowing / not glowing in the display is equivalent to logic 1/0 respectively

What are the minimum numbers of NOT gates and 2-input OR gates required to design the logic of the driver for this 7-segment display?

3 NOT and 4 OR

2 NOT and 4 OR

1 NOT and 3 OR

2 NOT and 3 OR

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Q#31 Boolean Algebra GATE EC 2008 (Set 1) MCQ +2 marks -0.66 marks

Which of the following Boolean Expressions correctly represents the relation between P, Q, R and ?        

38.jpg

= (P OR Q) XOR R

= (P AND Q) XOR R

= (P NOR Q) XOR R        

= (P XOR Q) XOR R

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Q#32 Boolean Algebra GATE EC 2007 (Set 1) MCQ +1 mark -0.33 marks

The Boolean function Y = AB+CD is to be realized using only 2-input NAND gates. The minimum number of gates required is

2

3

4

5

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Q#33 Boolean Algebra GATE EC 2007 (Set 1) MCQ +2 marks -0.66 marks

The Boolean expression can be minimized to

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Q#34 Boolean Algebra GATE EC 2006 (Set 1) MCQ +1 mark -0.33 marks

The number of product terms in the minimized sum-of-product expression obtained through the following K-map is (where “d” denotes don’t care states)

1

0

0

1

0

d

0

0

0

0

d

1

1

0

0

1

2

3

4

5

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Q#35 Boolean Algebra GATE EC 2006 (Set 1) MCQ +2 marks -0.66 marks

The point P in the following figure is stuck at-1. The output f will be

A

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Q#36 Boolean Algebra GATE EC 2005 (Set 1) MCQ +2 marks -0.66 marks

The Boolean expression for the truth table shown is:

A

B

C

F

0

0

0

0

0

0

1

0

0

1

0

0

0

1

1

1

1

0

0

0

1

0

1

0

1

1

0

1

1

1

1

0

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Q#37 Boolean Algebra GATE EC 2004 (Set 1) MCQ +2 marks -0.66 marks

The Boolean expression  is equivalent to

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Q#38 Boolean Algebra GATE EC 2004 (Set 1) MCQ +2 marks -0.66 marks

A Boolean function f of two variables x and y is defined as follows:

;

Assuming complements of x and y are not available, a minimum cost solution for realizing f using only
2-input NOR gates and 2-input OR gates (each having unit cost) would have a total cost of

1 unit

4 unit

3 unit

2 unit

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Q#39 Boolean Algebra GATE EC 2003 (Set 1) MCQ +1 mark -0.33 marks

The number of distinct Boolean expression of 4 variables is

16

256

1024

65536

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Q#40 Boolean Algebra GATE EC 2003 (Set 1) MCQ +2 marks -0.66 marks

If the functions W, X, Y and Z are as follows

Then

W = Z,

W = Z, X = Y

W = Y

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Q#41 Boolean Algebra GATE EC 2003 (Set 1) MCQ +2 marks -0.66 marks

The circuit shown in figure converts        

BCD to binary code

Binary to excess - 3 code

Excess – 3 to Gray code

Gray to Binary code

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Q#42 Boolean Algebra GATE EC 2002 (Set 1) MCQ +1 mark -0.33 marks

If the input to the digital circuit (Figure) consisting of a cascade of 20 XOR-gates in X, then the output Y is equal to

0

1

X

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Q#43 Boolean Algebra GATE EC 2002 (Set 1) MCQ +2 marks -0.66 marks

he gates  and  in figure have propagation delays of 10nsec 20nsec respectively. If the input  makes an abrupt change from logic 0 to 1 at time
 then the output waveform  is        

17.jpg

18.jpg

19.jpg

20.jpg

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Q#44 Boolean Algebra GATE EC 2001 (Set 1) MSQ +2 marks -0 marks

For the digital block shown in Figure , the output  where  is MSB and  is LSB. Y is given in terms of minterms as and its complement is

(a) Enter the logical values in the given karnaugh map [Figure] for the output Y.

(b) Write down the expression for Y in sum-of products from using minimum number of terms

(c) Draw the circuit for the digital logic boxes using four 2-input NAND gates only for each of the boxes.

(a) Untitled-8.png

(b)

(c)Untitled-10.png

None of these

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Q#45 Boolean Algebra GATE EC 2000 (Set 1) MCQ +1 mark -0.33 marks

For the logic circuit shown in Figure, the required input condition (A, B, C) to make the output (X)=1 is        

8.jpg

1, 0, 1

0, 0, 1

1, 1, 1

0, 1, 1

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Q#46 Boolean Algebra GATE EC 2000 (Set 1) MCQ +2 marks -0.66 marks

For the logic circuit shown in figure, the simplified Boolean expression for the output Y is

A+B+C

A

B

0

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Q#47 Boolean Algebra GATE EC 2000 (Set 1) MSQ +2 marks -0 marks

The operating conditions (ON = 1, OFF = 0) of three pumps (x, y, z) are to be monitored.  implies that pump X is on. It is required that the indicator (LED) on the panel should glow when a majority of the pumps fail.

(a) Enter the logical values in the K-map in the format shown in figure. Derive the minimal Boolean sum of products expression whose output is zero when a majority of the pumps fail.

(b) The above expression is implemented using logic gates, and point P is the output of this circuit, as shown in figure. P is at 0 V when a majority of the pumps fails and is at 5 V otherwise. Design a circuit to drive the LED using this output. The current through the LED should be 10mA and the voltage drop across it is 1V. Assume that P can source or sink 10mA and a 5V supply is available.

(a)

(b) Untitled-7.png

(b) R= 400 ohm

(b) R=300 Ohm

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Q#48 Boolean Algebra GATE EC 1999 (Set 1) MCQ +1 mark -0.33 marks

The logical expression  is equivalent to

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Q#49 Boolean Algebra GATE EC 1999 (Set 1) MCQ +2 marks -0.66 marks

The minimized form of the logical expression  is

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Q#50 Boolean Algebra GATE EC 1999 (Set 1) MSQ +2 marks -0 marks

In a certain application, four inputs A, B, C, D (both true and complement forms available) are fed to logic circuit, producing an output F, which operates a relay. The relay turns on when  for the following states of the input  and . States ‘1000’ and ‘1001’ do not occur, and for the remaining states, the relay is off. Minimize F with the help of a Karnaugh map and Find the minimum number of 3 – input NAND gates required to realize this.

Minimum of 5,  3-input NAND gates are required

Minimum of 4,  3-input NAND gates are required

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Q#51 Boolean Algebra GATE EC 1998 (Set 1) MCQ +2 marks -0.66 marks

The minimum number of 2-input NAND gates required to implements the Boolean function, assuming that A, B and C are available, is

Two

Three

Five

Six

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Q#52 Boolean Algebra GATE EC 1998 (Set 1) MCQ +2 marks -0.66 marks

Two 2’s complement number having sign bits x and y are added and the sign bit of the result is z. Then, the occurrence of overflow is indicated by the Boolean function

x y z

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Q#53 Boolean Algebra GATE EC 1998 (Set 1) MCQ +2 marks -0.66 marks

For the identify , the dual form is

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Q#54 Boolean Algebra GATE EC 1998 (Set 1) MCQ +2 marks -0.66 marks

The K-map for a Boolean function is shown in Fig. 2.13. The number of essential prime implicants for this function is

9.jpg

4

5

6

8

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Q#55 Boolean Algebra GATE EC 1997 (Set 1) MCQ +1 mark -0.33 marks

The output of the logic gate in the figure is7.jpg

0

1

A

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Q#56 Boolean Algebra GATE EC 1997 (Set 1) MCQ +1 mark -0.33 marks

The Boolean function  is a reduced form of

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Q#57 Boolean Algebra GATE EC 1995 (Set 1) MCQ +1 mark -0.33 marks

The output of the circuit in the figure is equal to

2.jpg

0

1

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Q#58 Boolean Algebra GATE EC 1995 (Set 1) MCQ +1 mark -0.33 marks

The minimum number of NAND gates required to implement the Boolean function, is equal to

Zero

1

4

7

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Q#59 Boolean Algebra GATE EC 1994 (Set 1) MCQ +1 mark -0.33 marks

The output of a logic gate is ‘1’ when all its inputs are at logic ‘0’. Then gate is either

A NAND or an EX-OR gate

A NOR or an EX-NOR gate

AN OR or an EX-NOR gate

AN AND or an EX-OR gate

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Q#60 Boolean Algebra GATE EC 1993 (Set 1) MCQ +1 mark -0.33 marks

For the logic circuit shown in figure, the output Y is equal to

None

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Q#61 Boolean Algebra GATE EC 1993 (Set 1) MCQ +2 marks -0.66 marks

The truth table for the output Y in terms of three inputs A, B and C are given in table. Draw a logic circuit realization using only NOR gates.

A

0

1

0

1

0

1

0

1

B

0

0

1

1

0

0

1

1

C

0

0

0

0

1

1

1

1

Y

1

1

1

0

1

0

0

0

None of these

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Q#62 Boolean Algebra GATE EC 1993 (Set 1) MCQ +2 marks -0.66 marks

Boolean expression for the output of XNOR (Equivalent) logic gate with inputs A and B is:

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Q#63 Boolean Algebra GATE EC 1991 (Set 1) MCQ +2 marks -0.66 marks

The four variable function f is given in terms of min-terms as: .

Using the k-map minimize the function in the sum of products form.

None of these

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