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Q.20 Consider, a four-variable Boolean function, which contains half the number of minterms with an odd number of $1's$. Then the Boolean can be realized with variables $A,B, C,D$ as:
(a) $A \oplus B \oplus C \oplus D$
(b) $A \oplus B \oplus C$
(c) $\mathrm{B} \oplus \mathrm{C}$
(d) $(A \oplus B)+(C \oplus D)$

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$\color{olive}{F = A_1 \oplus A_2 \dots \oplus A_{n-1} \oplus A_n}$ is true if it contains odd number of $1's$, which makes me say $\color{olive}{A \oplus B \oplus C \oplus D}$ is correct answer. But let's also see the other way

Question says function contains half of minterms (means $8$ minterms), with odd number of $1's$. Some minterms would be $\overline{A}BCD$, $ABC\overline{D}$, and so on. K-MAP looks like $\Rightarrow$



$\color{blue}{F(A,B,C,D) = \sum (1,2,4,7,8,11,13,14) = A \oplus B \oplus C \oplus D}$

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Odd no of 1s na   

1  2. 4. 7. 8. 11. 13. 14.

Now if u plot this values in kmap  u we will see that we cannot group any of of the minterms so this situation is called check mate so the and would be  A (+)B(+)C(+)D

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