Question

Consider following numbers :
A. $(1101.01)_{2} = (13.25)_{10}$

B. $(13.28)_{10} = (1101.011)_{2}$

C. $(2\text{A})_{16} = (42)_{10}$

D. $(37)_{8} = (11010)_{2}$

Choose the correct answer from the options given below :

• A. A, B, C only
• B. B, D only
• C. C, D only
• D. A, C only

Hint:

For base-8 to base-2, convert each octal digit to 3 bits: $3 \to 011$, $7 \to 111$. So $37_8 = 011111_2 = 31_{10}$.

Answer 1

The Correct Option is D

Concept:
• Positional number systems allow values to be converted between bases by expanding the digits with their corresponding powers of the base.

Step 1: Verify Statement A
\((1101.01)_2 = (1 \cdot 2^3) + (1 \cdot 2^2) + (0 \cdot 2^1) + (1 \cdot 2^0) + (0 \cdot 2^{-1}) + (1 \cdot 2^{-2})\)
\(= 8 + 4 + 0 + 1 + 0 + 0.25 = 13.25_{10}\). Statement A is Correct.

Step 2: Verify Statement B
From Step 1, we see that \(13.25_{10} = (1101.01)_2\). The binary value \((1101.011)_2\) would equal \(13.25 + (1 \cdot 2^{-3}) = 13.25 + 0.125 = 13.375_{10}\). Statement B is Incorrect.

Step 3: Verify Statement C
\((2A)_{16} = (2 \cdot 16^1) + (A \cdot 16^0)\). Since \(A = 10\):
\(= 32 + 10 = 42_{10}\). Statement C is Correct.

Step 4: Verify Statement D
\((37)_8 = (3 \cdot 8^1) + (7 \cdot 8^0) = 24 + 7 = 31_{10}\).
Binary \((11010)_2 = (1 \cdot 2^4) + (1 \cdot 2^3) + (0 \cdot 2^2) + (1 \cdot 2^1) + (0 \cdot 2^0) = 16 + 8 + 0 + 2 + 0 = 26_{10}\).
Since \(31 \neq 26\), Statement D is Incorrect.