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

(½ Mark for drawing Logic circuit for (A NOR B) correctly)
(½ Mark for drawing Logic circuit for (C NOR D) correctly)

c. Derive a Canonical POS expression for a Boolean function G, represented by 1
the following truth table:

X Y Z G(X,Y,Z)

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

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

Ans G(X,Y,Z)= (X+Y+Z).( X+Y+Z’).( X+Y’+Z’).(X’+Y’+Z)
OR

G(X,Y,Z)= ​∏​(0,1,3,6)
(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
d. Reduce the following Boolean expression to its simplest form using K-Map: 3
E(U,V,Z,W)= (2,3,6,8,9,10,11,12,13)Σ

Ans

OR

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