ISRO EC 2023 Q.25: solution and answer

ISRO 2023 · ECQ.25 · 1 mark · MCQ

Simplified Boolean function $f (A, B, C, D) = \sum (0, 2, 4, 5, 6, 7, 8, 10, 13, 15)$ in POS form is

(A)
$(B^{'} + D) (A^{'} + B^{'} + D)$
(B)
$(B + D^{'}) (A + B^{'} + D)$
(C)
$(B + D) (B^{'} + D^{'}) (A^{'} + B)$
(D)
$(B + D^{'}) (A^{'} + B^{'} + D)$

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Answer and solution

ANSWER

The POS form is $(B + D^{'}) (A^{'} + B^{'} + D)$ , which is option (d).

Official answer: DSource: ISRO's revised and final answer key of 16 February 2024

Four variables give 16 minterms. The function is 0 at the minterms not listed, so $f = \prod M (1, 3, 9, 11, 12, 14)$ .
In binary ( $A$ is the MSB) these zeros are $0001, 0011, 1001, 1011, 1100, 1110$ .
Group $1, 3, 9, 11$ : here $B = 0$ and $D = 1$ while $A$ and $C$ take every value, so $f^{'}$ contains $B^{'} D$ .
Group $12, 14$ : here $A = 1, B = 1, D = 0$ while $C$ varies, so $f^{'}$ contains $A B D^{'}$ .
So $f^{'} = B^{'} D + A B D^{'}$ .
Complement with De Morgan: $f = (B + D^{'}) (A^{'} + B^{'} + D)$ .
Check with minterm 5 ( $0101$ ): $B + D^{'} = 1$ and $A^{'} + B^{'} + D = 1$ , so $f = 1$ . Check with maxterm 12 ( $1100$ ): $A^{'} + B^{'} + D = 0$ , so $f = 0$ .
This is option (d). Option (b) has $A$ unbarred, which gives $f = 1$ at $1100$ because $B + D^{'} = 1$ and $A + B^{'} + D = 1$ , but $f (12)$ must be 0, so it fails.

KEY IDEA

For a POS answer, group the zeros of the function (maxterms) on the K-map, write $f^{'}$ as a sum of products, then complement it with De Morgan's theorem.

WATCH OUT FOR

Grouping the ones instead of the zeros, or forgetting to complement the variables when converting $f^{'}$ to POS, which turns $A^{'}$ into $A$ and gives option (b).

From the ISRO EC 2023 paper. Every question of this paper · The full paper (PDF, ISRO) · All 6 ISRO Boolean logic questions, with notes

ISRO EC 2023 Q.25: solution and answer

Answer and solution

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