Question

Subtract (29.A)\(_{16}\) from (4F.B)\(_{16}\)

• A. (26.1)\(_{16}\)
• B. (26.A)\(_{16}\)
• C. (4F.A)\(_{16}\)
• D. (16.1)\(_{16}\)

Hint:

When performing hexadecimal arithmetic, remember the decimal equivalents: A=10, B=11, C=12, D=13, E=14, F=15. For subtraction, if you need to borrow, you borrow 16 from the column to the left. In this case, no borrowing was needed.

Answer 1

The Correct Option is A

Step 1: Prepare the hexadecimal subtraction. The problem is: \[ \begin{array}{@{}c@{\,}c@{}c@{}c} & 4 & F & . & B \\ - & 2 & 9 & .& A \\ \hline \end{array} \]
Step 2: Subtract the fractional digits. For the rightmost column, B - A converts to \(11 - 10 = 1\) in decimal. The result is \(1_{16}\). \[ \begin{array}{@{}c@{\,}c@{}c@{}c} & 4 & F & . & B \\ - & 2 & 9 & . & A \\ \hline & & & . & 1 \\ \end{array} \]
Step 3: Subtract the integer digits from right to left. First, the units column: F - 9 is \(15 - 9 = 6\) in decimal. The result is \(6_{16}\). \[ \begin{array}{@{}c@{\,}c@{}c@{}c} & 4 & F & . & B \\ - & 2 & 9 & . & A \\ \hline & & 6 & . & 1 \\ \end{array} \] Next, the 16's column: 4 - 2 is \(4 - 2 = 2\) in decimal. The result is \(2_{16}\). \[ \begin{array}{@{}c@{\,}c@{}c@{}c} & 4 & F & . & B \\ - & 2 & 9 & . & A \\ \hline & 2 & 6 & . & 1 \\ \end{array} \]
Step 4: Combine the results. The final result is \((26.1)_{16}\).