Assume that a 12-bit Hamming codeword consisting of 8-bit data and 4 check bits is $d_8 d_7 d_6 d_5 c_8 d_4 d_3 d_2 c_4 d_1 c_2 c_1$, where the data bits and the check bits are given in the following tables. Which one of the following choices gives the correct values of $x$ and $y$? 
Step 1: Understand the Hamming code structure.
In a $(12,8)$ Hamming code, parity (check) bits are placed at positions that are powers of two: $1$, $2$, $4$, and $8$. Each check bit ensures even parity over a specific set of bit positions.
Step 2: Determine the value of $x$.
Using the parity equations for the relevant check bits that include $d_5$, and enforcing even parity, the value of $x$ is obtained as $0$.
Step 3: Determine the value of $y$.
Similarly, applying the parity condition for check bit $c_8$, which covers positions including $d_8$, $d_7$, $d_6$, and $d_5$, and substituting known values, the parity requirement gives $y = 0$.
Step 4: Conclusion.
Both unknowns satisfy the parity conditions when $x = 0$ and $y = 0$.
If \( x \) and \( y \) are two decimal digits and \( (0.1101)_2 = (0.8xy5)_{10} \), the decimal value of \( x + y \) is \(\underline{\hspace{2cm}}\).
The format of the single-precision floating-point representation of a real number as per the IEEE 754 standard is as follows:
\[ \begin{array}{|c|c|c|} \hline \text{sign} & \text{exponent} & \text{mantissa} \\ \hline \end{array}\] Which one of the following choices is correct with respect to the smallest normalized positive number represented using the standard?

Consider the following code:
int a;
int arr[] = {30, 50, 10};
int *ptr = arr[10] + 1;
a = *ptr;
(*ptr)++;
ptr = ptr + 1;
printf("%d", a + arr[1] + *ptr);
In the diagram, the lines QR and ST are parallel to each other. The shortest distance between these two lines is half the shortest distance between the point P and the line QR. What is the ratio of the area of the triangle PST to the area of the trapezium SQRT?
Note: The figure shown is representative

Consider the following process information for Shortest Remaining Time First (SRTF) scheduling:
\[ \begin{array}{|c|c|c|} \hline \textbf{Process} & \textbf{Arrival Time (AT)} & \textbf{Burst Time (BT)} \\ \hline P1 & 0 & 10 \\ P2 & 1 & 13 \\ P3 & 2 & 6 \\ P4 & 8 & 9 \\ \hline \end{array} \]Find the turnaround time for each process.