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43 votes
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Consider the circuit in below figure which has a four bit binary number $b_3b_2b_1b_0$ as input and a five bit binary number, $d_4d_3d_2d_1d_0$ as output.

  1. Binary to Hex conversion

  2. Binary to BCD conversion

  3. Binary to Gray code conversion

  4. Binary to $radix-12$ conversion

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5 Answers

Best answer
28 votes
28 votes
Whenever, $b_{2}$ = $b_{3}$$=$ $1$, then only $ 0100$ i.e., $4$ is added to the given binary number. Let's write all possibilities for $b$.

$${\begin{array}{|c|c|c|c|}\hline
\bf{b_3}&    \bf{b_2}&  \bf{b_1}&\bf{ b_0} \\\hline
0&0&0&0 \\ 0&0&0&1\\    0&0&1&0 \\    0&0&1&1 \\    0&1&0&0 \\   0&1&0&1  \\   0& 1&1&0 \\   0&1&1&1\\   1&0&0&0\\   1&0&0&1\\    1&0&1&0 \\    1&0&1&1 \\   1&1&0&0  \\   1& 1&0&1 \\   1&1&1&0\\ 1&1&1&1 \\\hline
\end{array}}$$

Note that the last $4$ combinations ($1100$, $1101$, $1110$, $1111$) leads to $b_{3}$ and $b_2$ as $1$. So, in these combinations only $0100$ will be added.

$1100$ is $12$
$1101$ is $13$
$1110$ is $14$
$1111$ is $15$
in binary unsigned number system.

$1100 + 0100 = 10000$.
$1101 + 0100 = 10001$ and so on.
This is conversion to radix $12$.

Correct Answer: $D$
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87 votes
87 votes

Hope this helps.. I tried the possible way to explain this question. Please, find the attached screen shot.

20 votes
20 votes

Φ means nothing (we consider it as 0 or null)

So if we add number which is greater than or equal to 12 then it returns according to input

For example, 

For 0-11 it returns as it is

       12 it returns 0(0100+1100=0000)

       13 it returns 1(0100+1101=0001)

       14 it returns 2(0100+1110=0010)

       15 it returns 3(0100+1111=0011)

Answer is option D Binary to radix-12 conversion

9 votes
9 votes
when b3b2 is 11 restore circuit to 0000..
it counts from 0 to 11 i.e. Radix 12 no ..
circuit used for  binary to radix 12 conversion..
Answer:

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