1000 of 8421 binary coding system is equal to 1011 in ........ binary coding.

Question

Accepted Answer

The correct option is (2):
The question asks to find a binary coding system where the code 1011 is equivalent to the decimal value represented by 1000 in the 8421 system. The 8421 code is the standard weighted Binary Coded Decimal (BCD) code, where the four bits are assigned the weights 8, 4, 2, and 1, respectively. Therefore, the code $1000\_{8421}$ represents the decimal digit calculated as $(1 \times 8) + (0 \times 4) + (0 \times 2) + (0 \times 1)$, which equals **decimal 8**. We must now check which of the options assigns the value 8 to the binary pattern 1011. **Excess-3 (XS-3)** is an unweighted BCD code formed by adding $0011\_2$ (decimal 3) to the 8421 BCD code of the decimal digit. To find the decimal digit represented by $1011\_{Excess-3}$, we subtract $0011\_2$: $1011\_2 - 0011\_2 = 1000\_2$. Since $1000\_{8421}$ corresponds to decimal 8, the code $1011\_{Excess-3}$ also represents **decimal 8**. The equality holds, making Excess-3 the correct answer.

The incorrect option is (1): 5211.
The 5211 code is a weighted BCD code, where the weights for the bits are 5, 2, 1, and 1. To find the decimal value of the binary pattern 1011 in this system, we sum the weights corresponding to the '1' bits: $(1 \times 5) + (0 \times 2) + (1 \times 1) + (1 \times 1) = 5 + 1 + 1 = 7$. Since $1000\_{8421}$ is decimal 8 and $1011\_{5211}$ is decimal 7, they are not equivalent.

The incorrect option is (3): 2421.
The 2421 code is another weighted BCD code with positional weights 2, 4, 2, and 1. We calculate the decimal value of $1011\_{2421}$ as the sum of its weighted bits: $(1 \times 2) + (0 \times 4) + (1 \times 2) + (1 \times 1) = 2 + 2 + 1 = 5$. Since this code represents decimal 5, and not the required decimal 8, the 2421 code is incorrect.

The incorrect option is (4): 4221.
The 4221 code is a weighted BCD code that uses the weights 4, 2, 2, and 1. The decimal value for $1011\_{4221}$ is found by summing the weights of its '1' bits: $(1 \times 4) + (0 \times 2) + (1 \times 2) + (1 \times 1) = 4 + 2 + 1 = 7$. As this results in decimal 7, which is not equal to decimal 8, this option is incorrect.