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### 1 comment

It is the option B. Which includes min term $\sum m(3,5,6,7)$ whose binary equivalent is $011,101,110,111$. And in this you will find that no of 1's are more than the number of 0's.

## 2 Answers

B matches as correct option

In such questions we should take the help of TRUTH TABLE while solving. The question clearly says that circuit will only work if majority of the inputs are 1 so we'll choose only those inputs from the truth table where we find two or more than 1s and hence write their equivalent values in the form of literals and ultimately ORing them to get the final output.

Therefore, B is the right answer.

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