GO Classes DPP | GATE CS | Digital Logic | Base Arithmetic

Answer 1

Here is the explanation Note:- No need to worry about heavy calculation because we have calculator in the exam 

Answer 2

Since the original equation is an addition problem:

$$(4x\mathrm{D})_{16} + (2\mathrm{B}7)_{16} = (744)_{16}$$

We can rearrange it by isolating the unknown term through subtraction:

$$(4x\mathrm{D})_{16} = (744)_{16} - (2\mathrm{B}7)_{16}$$

Here is how that exact step-by-step subtraction looks in base $16$ (where we borrow $16$ instead of $10$ or $8$):

$$\begin{array}{r@{\quad}l} (7\,\,4\,\,4)_{16} \\ -\quad (2\,\,\mathrm{B}\,7)_{16} \\ \hline \phantom{+}(4\,x\,\mathrm{D)}_{16} \end{array}$$

• $\mathbf{16^0}$ place (Units): $4 - 7$. We cannot do this, so we borrow $1$ from the $16^1$ column.

  • The $4$ in the middle column becomes $3$.

  • New value: $(4 + 16) - 7 = 20 - 7 = 13$, which is $\mathrm{D}$ in hex.

• $\mathbf{16^1}$ place: The middle $4$ became $3$ after the borrow. Now we have $3 - \mathrm{B}$ (remembering that $\mathrm{B} = 11$). We cannot do this, so we borrow $1$ from the $16^2$ column.

  • The $7$ in the leftmost column becomes $6$.

  • New value: $(3 + 16) - 11 = 19 - 11 = \mathbf{8}$.

• $\mathbf{16^2}$ place: The $7$ became $6$ after the borrow.

  • New value: $6 - 2 = \mathbf{4}$.

• Putting the columns together, the subtraction gives us:

$$\begin{array}{r@{\quad}l} (7\,\,4\,\,4)_{16} \\ -\quad (2\,\,\mathrm{B}\,7)_{16} \\ \hline \phantom{+}(4\,\,8\,\mathrm{D)}_{16} \end{array}$$

• Comparing $(48\mathrm{D})_{16}$ directly to the term $(4x\mathrm{D})_{16}$, we get:

$$\boxed{x = 8}$$

 

 

 

$\underline{\textbf{Alternative Solution :}}$

To find the hexadecimal digit $x$, we can perform column-by-column addition from right to left, just like standard decimal addition, but using base-$16$.

The addition is set up as follows:

$$\begin{array}{r@{\quad}l} \phantom{+}4\,\,x\,\mathrm{D}_{16} \\ +\quad 2\,\,\mathrm{B}\,\,7_{16} \\ \hline \phantom{+}7\,\,4\,\,4_{16} \end{array}$$

Step $1:$ Rightmost Column (Units place)

We add the digits in the first column on the right:

$$\mathrm{D}_{16} + 7_{16}$$

In decimal, $\mathrm{D}_{16} = 13_{10}$.

$$13_{10} + 7_{10} = 20_{10}$$

To convert $20_{10}$ back to hexadecimal:

$$20_{10} = 1 \times 16 + 4 = 14_{16}$$

We write down the $4$ and carry over $1$ to the next column. This matches the $4$ in the units place of the sum $(744)_{16}$.

Step $2:$ Middle Column (Sixteens place)

We add the carry and the digits in the middle column:

$$\text{Carry} + x + \mathrm{B}_{16}$$

In decimal, $\mathrm{B}_{16} = 11_{10}$ and the carry is $1$.

$$1 + x + 11 = 12 + x$$

The result in this column is $4$, which means the sum $12 + x$ must end in a $4$ in hexadecimal notation. Since $x$ is a single hexadecimal digit ($0$ to $15$), $12 + x$ must equal $20_{10}$ (which is $14_{16}$, giving a digit of $4$ and a carry of $1$).

Setting up the equation:

$$12 + x = 20$$

$$x = 20 - 12$$

$$x = 8$$

Step $3:$ Leftmost Column (Two-hundred-fiftysixes place)

Let's verify by adding the leftmost column with the carry of $1$:

$$\text{Carry} + 4 + 2 = 1 + 4 + 2 = 7$$

This perfectly matches the $7$ in the sum $(744)_{16}$.

Thus, the digit $x$ is $\mathbf{8}$.

Comments

A Note on Approach: Method $1$ is the smarter way to solve this in a test because it eliminates the guesswork of finding a missing middle variable. However, we have included Method $2$ below because it’s great for visualizing how hexadecimal carries work and gives you a second way to verify your answer.

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sir your techniques teache me a lot to learn

Answer 3

Answer is x = 8

Answer 4

Answers is X = 8