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54 votes
54 votes

Let, $x_{1} ⊕ x_{2} ⊕ x_{3} ⊕ x_{4}= 0$ where $x_{1}, x_{2}, x_{3}, x_{4}$ are Boolean variables, and $⊕$ is the XOR operator.

Which one of the following must always be TRUE?

  1. $x_{1}x_{2}x_{3}x_{4} = 0$
  2. $x_{1}x_{3} + x_{2} = 0$
  3. $\bar{x}_{1} ⊕ \bar{x}_{3} = \bar{x}_{2} ⊕ \bar{x}_{4}$
  4. $x_{1} + x_{2} + x_{3} + x_{4} = 0$
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10 Answers

Best answer
81 votes
81 votes

Let $x_1 =1\; x_2=1\;x_3=1\;\text{and}\; x_4=1$

such that $x_1\oplus x_2\oplus x_3 \oplus x_4= 1\oplus 1\oplus 1\oplus1=0$

  1. $x_1x_2x_3x_4= 1.1.1.1=1$ , False
  2. $x_1x_3+x_2= 1.1+1=1$ , False
  3. is always True.
  4. $x_1+x_2+x_3+x_4= 1+1+1+1=1$ , False

Correct Answer: $C$

edited by
17 votes
17 votes

We know $\mathrm{A}\oplus \mathrm{B}=\mathrm{A}\mathrm{\bar{B}}+\mathrm{\bar{A}}\mathrm{B}$ and $\overline{\mathrm{A}\oplus \mathrm{B}}=\mathrm{A}\mathrm{B}+\mathrm{\bar{A}}\mathrm{\bar{B}}$

Now let's have some facts from this boolean equation.

  1. $\mathrm{A}\oplus \mathrm{A}=\mathrm{A}\mathrm{\bar{A}}+\mathrm{\bar{A}}\mathrm{A}=0$
  2. $\mathrm{B}\oplus \mathrm{A}=\mathrm{B}\mathrm{\bar{A}}+\mathrm{\bar{B}}\mathrm{A}=\mathrm{\bar{A}}\mathrm{B}+\mathrm{A}\mathrm{\bar{B}}=\mathrm{A}\mathrm{\bar{B}}+\mathrm{\bar{A}}\mathrm{B}=\mathrm{A}\oplus \mathrm{B}$
  3. $\mathrm{\bar{A}}\oplus \mathrm{\bar{B}}=\mathrm{\bar{A}}\mathrm{\bar{\bar{B}}}+\mathrm{\bar{\bar{A}}}\mathrm{\bar{B}}=\mathrm{\bar{A}}\mathrm{B}+\mathrm{A}\mathrm{\bar{B}}=\mathrm{A}\oplus \mathrm{B}$
  4. $(\mathrm{A}\oplus \mathrm{B})\oplus \mathrm{C}=(\mathrm{A}\mathrm{\bar{B}}+\mathrm{\bar{A}}\mathrm{B}) \mathrm{\bar{C}}+(\mathrm{A}\mathrm{B}+\mathrm{\bar{A}}\mathrm{\bar{B}})\mathrm{C}=\mathrm{A}\mathrm{\bar{B}}\mathrm{\bar{C}}+\mathrm{\bar{A}}\mathrm{B}\mathrm{\bar{C}}+\mathrm{A}\mathrm{B}\mathrm{C}+\mathrm{\bar{A}}\mathrm{\bar{B}}\mathrm{C}=\mathrm{A}(\mathrm{B}\mathrm{C}+\mathrm{\bar{B}}\mathrm{\bar{C}})+\mathrm{\bar{A}}(\mathrm{B}\mathrm{\bar{C}}+\mathrm{\bar{B}}\mathrm{C})=\mathrm{A}(\overline{\mathrm{B} \oplus \mathrm{C}})+\mathrm{\bar{A}}(\mathrm{B} \oplus \mathrm{C})=\mathrm{A}\oplus (\mathrm{B}\oplus \mathrm{C})$

 

Now

$\begin{align} x_1\oplus x_2\oplus x_3\oplus x_4&=0\\ \Rightarrow x_1\oplus (x_2\oplus x_3)\oplus x_4&=0 ~;~[\mathrm{Using~no(4)}]\\ \Rightarrow x_1 \oplus (x_3\oplus x_2)\oplus x_4&=0 ~;~[\mathrm{Using~no(2)}] \\ \Rightarrow (x_1 \oplus x_3)\oplus (x_2\oplus x_4)&=0 ~;~[\mathrm{Using~no(4)}] \\ \Rightarrow x_1 \oplus x_3&=x_2\oplus x_4 ~;~[\mathrm{Using~no(1)}] \\ \Rightarrow \bar{x_1} \oplus \bar{x_3}&=\bar{x_2}\oplus \bar{x_4} ~;~[\mathrm{Using~no(3)}] \end{align}$

 

So the correct answer is C as it must always be true.

11 votes
11 votes

Option C) x'1⊕x'3=x'2⊕x'4

It can be easily observed with Truth table method

x1 x2 x3 x4
x1⊕x2⊕x3⊕x4
0 0 0 0 0
0 0 0 1 1
0 0 1 0 1
0 0 1 1 0
0 1 0 0 1
0 1 0 1 0
0 1 1 0 0
0 1 1 1 1
1 0 0 0 1
1 0 0 1 0
1 0 1 0 0
1 0 1 1 1
1 1 0 0 0
1 1 0 1 1
1 1 1 0 1
1 1 1 1 0



 

2 votes
2 votes

we know that XOR gate is odd 1s detector. The output will be 1 only when number of 1's in the input is odd else it is 0

So lets consider the case when the output can be zero i.e. number of 1s is even

x1     x2      x3    x4

0       0        0      0

0       0        1      1

0       1        0      1

0       1        1      0

1       0        0      1

1       0        1      0   option B fails here

1       1        0      0

1       1        1      1   here option A and D fails

option C

$\overline{x_{1}} \oplus \overline{x_{3}}=\overline{x_{2}} \oplus \overline{x_{4} }$

=> $x_{1} \oplus x_{3}=x_{2} \oplus x_{4}$

option C holds for all the combinations

hence it is correct

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Answer:

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