Digital Electronics - Boolean Algebra and Logic Simplification - Discussion

I agree. 'A' is not the correct answer. It should be B. We need one the sets of parentheses to be (0+0+0), which eliminates 'A' and 'D'.

Option 'C' is also eliminated because it has B and B' both equal to 0, which is impossible. This leaves Option 'B' as the correct response.

I think option B is correct and not option A. Because as option A gives the following (1+0+1) (0+1+0) whose output is 1 and not 0.

The answer A is correct. When A is true then the binary representation is 1 and when A is false (Algebra) then binary representation is 0.

Taking this into account, when faced with (A+B'+C) , it would suggest that the binary bracket would be (1+0+1).

Looking above, A is correct as (0 + 1 + 0) (1 + 0 + 1) is the binary representation of (A+B'+C) (A'+B+C') only the two brackets have been swapped around.

Its really easy when you look at it they've only changed the layout to (A'+B+C') (A'+B+C') essentially!

Answer A is correct.

Draw Truth tables.

"C" and "D" cannot be correct, because if A' = 1 then A cannot be 1 or B' = 0 , B cannot be 0 respectively.

"B" cannot be the answer because (1+1+1)(0+0+0) is a SOP and the question asks for POS.

"B" is the correct answer.

The output of "A" will be 1.

In "C" B and C bar are given the same value.

In "D" C and C bar are given the same value.

I think option "D" is the correct ans, since we know that any unbarred letter is 0 and barred letter is 1.

"C" and "B" both are correct. But not "A".

I think B should be ans. as 1.0 = 0 which we required. while in A 1.1 WHICH IS EQUAL TO 1. while we need 0.

Option "C" is the correct answer.