SSC IMD SA 2017 – Question 102

If (11A1B)8 = (12C9)16, then what are the values of A and B? (C represents decimal 12)

• A = 5, B = 1
• A = 7, B = 5
• A = 5, B = 7
• A = 3 , B = 1

Detailed Explanation:

The correct answer is d) A = 3 , B = 1.

We are given: $$(11A1B)_8 = (12C9)_{16}$$ where $C = 12$ in decimal.

Step 1: Convert Hexadecimal $(12C9)_{16}$ to Binary
Each hexadecimal digit corresponds to a 4-bit binary sequence:

• $1_{16} = 0001_2$
• $2_{16} = 0010_2$
• $C_{16} = 12_{10} = 1100_2$
• $9_{16} = 1001_2$

Combining these: $$(12C9)_{16} = 0001\ 0010\ 1100\ 1001_2 = 1001011001001_2$$

Step 2: Group Binary Digits in 3s (from right to left) to Convert to Octal

• Rightmost 3 bits: $001_2 = 1_8$ (this represents $B \implies \mathbf{B = 1}$)
• Next 3 bits: $001_2 = 1_8$
• Next 3 bits: $011_2 = 3_8$ (this represents $A \implies \mathbf{A = 3}$)
• Next 3 bits: $001_2 = 1_8$
• Leftmost bits: $001_2 = 1_8$

Writing in octal: $$(11311)_8$$

Step 3: Match with $(11A1B)_8$
Comparing $(11A1B)_8$ with $(11311)_8$, we find:

• $A = 3$
• $B = 1$

Decimal Verification:

• $(12C9)_{16} = 1 \times 16^3 + 2 \times 16^2 + 12 \times 16^1 + 9 = 4096 + 512 + 192 + 9 = 4809_{10}$
• $(11311)_8 = 1 \times 8^4 + 1 \times 8^3 + 3 \times 8^2 + 1 \times 8^1 + 1 \times 8^0 = 4096 + 512 + 192 + 8 + 1 = 4809_{10}$

Both values are equal, verifying $A = 3$ and $B = 1$.