GATE CSE 2015 Set 2 | Question: 37

Question

The number of min-terms after minimizing the following Boolean expression is _______.

$[D'+AB'+A'C+AC'D+A'C'D]'$

Comments

Directly apply the boolean algebra and get the right answer

min-term=1

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15 max terms

1 min terms 

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Can’t I directly do complement of all terms like this..

[D.AB’.AC’.A’CD.ACD]  = 0    {AS  A.A’ =0}

Answer 1

answer is 1

solve it we get 15 max terms

so no. of min terms are 1

Answer 2

on putting some element say A = 0 we solve the boolean expression we get ans = 0; so for all minterms where A=0 function is 0. On taking A=1 we solve the boolean expression and get function = BCD, on K-Map its easy to put value of BCD.

The end result of this gives us only one minterm = ABCD

hence, answer = 1

Answer 3

The minterm we get after minimizing this expression is ABCD.
So, number of minterms is 1.

Answer 4

https://gateoverflow.in/?qa=blob&qa_blobid=55204506286937687