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CBSE AISSCE 2017-2018 Marking Scheme for Computer Science
(2018-2019 Sub Code: 083 Paper Code: 91)

X . (X’+ Y)= X . Y
Verification:
X Y X’ X’+ Y X.(X’+Y) X.Y
0 0 1 1 0 0
0 1 1 1 0 0

1 0 0 0 0 0
1 1 0 1 1 1

(1 Mark for stating any one Absorption Law correctly)
(1 Mark for correctly verifying the stated Law using Truth Table)
(b) Draw the Logic Circuit of the following Boolean Expression: 2
​(U’+V).(V’+W’)
Ans

(Full 2 Marks for drawing the Logic Circuit for the expression correctly)
OR
(½ Mark for drawing Logic circuit for (U’ + V) correctly)
(½ Mark for drawing Logic circuit for (V’ + W’) correctly)
(c) Derive a Canonical POS expression for a Boolean function FN, represented by the 1
following truth table:
X Y Z FN(X,Y,Z)
0 0 0 1
0 0 1 1
0 1 0 0
0 1 1 0

1 0 0 1
1 0 1 0
1 1 0 0
1 1 1 1

Ans FN(X,Y,Z)= (X+Y’+Z).(X+Y’+Z’).(X’+Y+Z’).(X’+Y’+Z)
OR
FN(X,Y,Z)= ​∏​(2,3,5,6)

(1 Mark for correctly writing the POS form)
OR
(½ Mark for any two correct terms)
Note: Deduct ½ mark if wrong variable names are written in the expression

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