The Boolean expression $$AC + B bar{C}$$ is equivalent to(GATE 2004 || EC || MCQ ||2 MARK)

$$ABC + bar{A}B bar{C} + AB bar{C} + A bar{B}C$$

$$ bar{B}C + AC + B bar{C} + bar{A}C bar{B}$$

$$AC + B bar{C} + bar{B}C + ABC$$

In the sum of products function f(X, Y, Z) = $$ sum (2, 3, 4, 5)$$the prime implicants are (GATE 2012 || EC || MCQ ||1 MARK)

$$ bar{X}Y, X bar{Y} bar{Z}, X bar{Y}Z$$

$$ bar{X}Y bar{Z}, bar{X}YZ, X bar{Y}$$

$$ bar{X} bar{Y} bar{Z}, bar{X}YZ, X bar{Y} bar{Z}, X bar{Y}Z$$

A function of Boolean variables, X, Y and Z, is expressed in terms of the min-terms as $$F(X, Y, Z) = sum (1, 2, 5, 6, 7).$$ Which one of the product of sums given below is Equal to the function F (X, Y, Z)?(GATE 2025 || EC || MCQ ||1 MARK)

$$(X + Y + Z)(X + bar{Y} + bar{Z})( bar{X} + Y + Z)$$

$$( bar{X} + bar{Y} + bar{Z})( bar{X} + Y + Z)(X + bar{Y} + bar{Z})$$

$$( bar{X} + bar{Y} + Z)( bar{X} + Y + bar{Z})(X + bar{Y} + Z)(X + Y + bar{Z})(X + Y + Z)$$

$$(X + Y + bar{Z})( bar{X} + Y + Z)( bar{X} + Y + bar{Z})( bar{X} + bar{Y} + Z)( bar{X} + bar{Y} + bar{Z})$$

The Boolean expression F(X, Y, Z) = $$ bar{X}{Y} bar{Z} + X bar{Y} bar{Z} + XY bar{Z} + XYZ.$$ converted into canonical product of sum (POS) form is ( GATE 2015 || EC || MCQ ||1 MARK)

$$(X + Y + Z)(X + Y + bar{Z})(X + bar{Y} + bar{Z})( bar{X} + Y + bar{Z})$$

$$(X + bar{Y} + Z)( bar{X} + Y + bar{Z})( bar{X} + bar{Y} + Z)( bar{X} + bar{Y} + bar{Z})$$

$$(X + Y + Z)( bar{X} + Y + bar{Z})(X + bar{Y} + Z)( bar{X} + bar{Y} + bar{Z})$$

$$(X + bar{Y} + bar{Z})( bar{X} + Y + Z)( bar{X} + bar{Y} + Z)(X + Y + Z)$$

The Boolean expression F(X, Y, Z) = $$ bar{X}YZ + X bar{Y} bar{Z} + XY bar{Z} + XYZ$$ converted into canonical product of sum (POS) form is (GATE 2015 || EC || MCQ ||1 MARK)

$$(X + Y + Z)(X + Y + bar{Z})(X + bar{Y} + bar{Z})( bar{X} + Y + bar{Z})$$

$$(X + bar{Y} + Z)( bar{X} + Y + bar{Z})( bar{X} + bar{Y} + Z)( bar{X} + bar{Y} + bar{Z})$$

$$(X + Y + Z)( bar{X} + Y + bar{Z})(X + bar{Y} + Z)( bar{X} + bar{Y} + bar{Z})$$

$$(X + bar{Y} + bar{Z})( bar{X} + Y + Z)( bar{X} + bar{Y} + Z)(X + Y + Z)$$

GATE EC|| DIGITAL LOGIC ||K MAP ||PYQS(2000-2025)

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Question 1

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

[Tex]W = R + \bar{P}Q + \bar{R}S.[/Tex]

[Tex]X = PQ\bar{R}\bar{S} + \bar{P}\bar{Q}\bar{R}\bar{S} + P\bar{Q}\bar{R}\bar{S}[/Tex]

[Tex]Y = RS + \overline{PR + P\bar{Q} + \bar{P}\bar{Q}}[/Tex]

[Tex]Z = R + S + \overline{PQ + \bar{P}\bar{Q}\bar{R} + P\bar{Q}\bar{S}}[/Tex]

then

( GATE 2003 || EC || MCQ ||1 MARK)

W = Z, X = [Tex]{\bar{Z}}[/Tex]
[Tex]W = \bar{Z}, \; X = Y[/Tex]
W = Z, X = Y
[Tex]W = Y = \bar{Z}[/Tex]

Question 2

The Boolean expression

[Tex]AC + B\bar{C}[/Tex]

is equivalent to

(GATE 2004 || EC || MCQ ||2 MARK)

[Tex]\bar{A}C + B\bar{C} + AC[/Tex]
[Tex]\bar{B}C + AC + B\bar{C} + \bar{A}C\bar{B}[/Tex]
[Tex]AC + B\bar{C} + \bar{B}C + ABC[/Tex]
[Tex]ABC + \bar{A}B\bar{C} + AB\bar{C} + A\bar{B}C[/Tex]

Question 3

The number of product terms in the minimized sum-of-products 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

( GATE 2006 || EC || MCQ ||1 MARK)

Question 4

In the sum of products function f(X, Y, Z) = [Tex]\sum (2, 3, 4, 5)[/Tex]

the prime implicants are

(GATE 2012 || EC || MCQ ||1 MARK)

[Tex]\bar{X}Y, X\bar{Y}[/Tex]
[Tex]\bar{X}Y, X\bar{Y}\bar{Z}, X\bar{Y}Z[/Tex]
[Tex]\bar{X}Y\bar{Z}, \bar{X}YZ, X\bar{Y}[/Tex]
[Tex]\bar{X}\bar{Y}\bar{Z}, \bar{X}YZ, X\bar{Y}\bar{Z}, X\bar{Y}Z[/Tex]

Question 5

In the sum of products function f(x, y, z) = Σ(2, 3, 4, 5) the prime implicants are

(GATE 2012 || EC || MCQ ||1 MARK)

X̅Y, XY̅
X̅Y, XY̅Z̅, XY̅Z
X̅YZ̅, X̅YZ, XY̅
X̅Y̅Z̅, X̅YZ, XY̅Z̅, XY̅Z

Question 6

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

[Tex]F(X, Y, Z) = \sum (1, 2, 5, 6, 7).[/Tex]

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

(GATE 2025 || EC || MCQ ||1 MARK)

[Tex](\bar{X} + \bar{Y} + \bar{Z})(\bar{X} + Y + Z)(X + \bar{Y} + \bar{Z})[/Tex]
[Tex](X + Y + Z)(X + \bar{Y} + \bar{Z})(\bar{X} + Y + Z)[/Tex]
[Tex](\bar{X} + \bar{Y} + Z)(\bar{X} + Y + \bar{Z})(X + \bar{Y} + Z)(X + Y + \bar{Z})(X + Y + Z)[/Tex]
[Tex](X + Y + \bar{Z})(\bar{X} + Y + Z)(\bar{X} + Y + \bar{Z})(\bar{X} + \bar{Y} + Z)(\bar{X} + \bar{Y} + \bar{Z})[/Tex]

Question 7

The Boolean expression

F(X, Y, Z) = [Tex]\bar{X}{Y}\bar{Z} + X\bar{Y}\bar{Z} + XY\bar{Z} + XYZ.[/Tex]

converted into canonical product of sum (POS) form is

( GATE 2015 || EC || MCQ ||1 MARK)

[Tex](X + Y + Z)(X + Y + \bar{Z})(X + \bar{Y} + \bar{Z})(\bar{X} + Y + \bar{Z})[/Tex]
[Tex](X + \bar{Y} + Z)(\bar{X} + Y + \bar{Z})(\bar{X} + \bar{Y} + Z)(\bar{X} + \bar{Y} + \bar{Z})[/Tex]
[Tex](X + Y + Z)(\bar{X} + Y + \bar{Z})(X + \bar{Y} + Z)(\bar{X} + \bar{Y} + \bar{Z})[/Tex]
[Tex](X + \bar{Y} + \bar{Z})(\bar{X} + Y + Z)(\bar{X} + \bar{Y} + Z)(X + Y + Z)[/Tex]

Question 8

The Boolean expression

F(X, Y, Z) = [Tex]\bar{X}YZ + X\bar{Y}\bar{Z} + XY\bar{Z} + XYZ[/Tex]

converted into canonical product of sum (POS) form is

(GATE 2015 || EC || MCQ ||1 MARK)

[Tex](X + Y + Z)(X + Y + \bar{Z})(X + \bar{Y} + \bar{Z})(\bar{X} + Y + \bar{Z})[/Tex]
[Tex](X + \bar{Y} + Z)(\bar{X} + Y + \bar{Z})(\bar{X} + \bar{Y} + Z)(\bar{X} + \bar{Y} + \bar{Z})[/Tex]
[Tex](X + Y + Z)(\bar{X} + Y + \bar{Z})(X + \bar{Y} + Z)(\bar{X} + \bar{Y} + \bar{Z})[/Tex]
[Tex](X + \bar{Y} + \bar{Z})(\bar{X} + Y + Z)(\bar{X} + \bar{Y} + Z)(X + Y + Z)[/Tex]

Question 9

The Boolean expression F(X, Y, Z) = X̅YZ̅ + XY̅Z̅ + XYZ̅ + XYZ converted into canonical product of sum (POS) form is

(GATE 2015 || EC || MCQ || 1 MARK)

(X + Y + Z)(X + Y + Z̅)(X + Y̅ + Z̅)(X̅ + Y + Z̅)
(X + Y̅ + Z)(X̅ + Y + Z̅)(X̅ + Y̅ + Z)(X̅ + Y̅ + Z̅)
(X + Y + Z)(X̅ + Y + Z̅)(X + Y̅ + Z)(X̅ + Y̅ + Z̅)
(X + Y̅ + Z̅)(X̅ + Y + Z)(X̅ + Y̅ + Z)(X + Y + Z)

Question 10

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

( GATE 2014 || EC || MCQ ||1 MARK)

2(n-1)
n/2
2ⁿ
2^n-1

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