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CBSE AISSCE 2016-2017 Marking Scheme for Computer Science
(Sub Code: 083 Paper Code 91/1 Delhi)

(ii)
X Y Z Y.Z X+Y.Z (X+Y) (X+Z) (X+Y).(X+Z)
0 0 0 0 0 0 0 0

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

(1 Mark for stating any one Distributive Law correctly)
(1 Mark for correctly verifying any one Distributive Law using Truth Table)

(b) Draw the Logic Circuit of the following Boolean Expression using only NAND Gates: 2
X.Y + Y.Z

Ans

(Full 2 Marks for drawing the Logic Circuit for the expression correctly)
OR
(½ Mark for drawing Logic circuit for (X NAND Y) correctly)
(½ Mark for drawing Logic circuit for (Y NAND Z) correctly)

(c) Derive a Canonical SOP expression for a Boolean function F, represented by 1
the following truth table:

U V W F(U,V,W)
0 0 0 1
0 0 1 0

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

1 1 0 1
1 1 1 0

Ans F(U,V,W)= U’V’W’ + U’VW’ + U’VW + UVW’

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