$A)\ 0.\dfrac{1}{2^1}=0.5$

$B)\ 0.\dfrac{1}{2^2}=0.25\ \checkmark$

$C)\ 0.\dfrac{7}{2^5}=0.21875$

$B)\ 0.\dfrac{1}{2^2}=0.25\ \checkmark$

$C)\ 0.\dfrac{7}{2^5}=0.21875$

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Best answer

**First Multiplication Iteration**

Multiply $0.25$ by $2$

$\begin{array}{c c c}0.25 \ast 2 = 0.50\;\text{(Product)} & \text{Fractional part} = 0.50 & \text{Carry} = 0 \textbf{ (MSB)} \end{array}$

**Second Multiplication Iteration**

Multiply $0.50$ by $2$

$\begin{array}{c c c}0.50 \ast 2 = 1.00\;\text{(Product)} & \text{Fractional part} = 1.00 & \text{Carry} = 1 \textbf{ (LSB)} \end{array}$

The fractional part in the $2$nd iteration becomes zero and hence we stop the multiplication iteration.

Carry from the $1$st multiplication iteration becomes **MSB **and carry from $2$nd iteration becomes **LSB**.

So the result is $0.01$

Correct Answer: B.

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