A grouped frequency distribution has a mean mark of 24.75 and a total of 40 students. The marks and frequencies are given in the table. Find the values of p and k.

The midpoint of a class interval is calculated as $\frac{\text{lower limit} + \text{upper limit}}{2}$.

Marks	Frequency ($f$)	Midpoint ($x$)	$fx$
11-15	$p$	$\frac{11+15}{2} = 13$	$13p$
16-20	4	$\frac{16+20}{2} = 18$	$4 \times 18 = 72$
21-25	11	$\frac{21+25}{2} = 23$	$11 \times 23 = 253$
26-30	$k$	$\frac{26+30}{2} = 28$	$28k$
31-35	6	$\frac{31+35}{2} = 33$	$6 \times 33 = 198$
36-40	1	$\frac{36+40}{2} = 38$	$1 \times 38 = 38$

The total number of students is given as 40. The sum of the frequencies must equal 40.

$$p + 4 + 11 + k + 6 + 1 = 40$$ $$p + k + 22 = 40$$ $$p + k = 40 - 22$$ $$p + k = 18 \quad (\text{Equation 1})$$

The mean mark is given as 24.75. The formula for the mean of a grouped frequency distribution is $\text{Mean} = \frac{\sum fx}{\sum f}$.
First, calculate $\sum fx$:

$$\sum fx = 13p + 72 + 253 + 28k + 198 + 38$$ $$\sum fx = 13p + 28k + 561$$

Now, substitute this into the mean formula along with $\sum f = 40$ and $\text{Mean} = 24.75$:

$$24.75 = \frac{13p + 28k + 561}{40}$$

Multiply both sides by 40:

$$24.75 \times 40 = 13p + 28k + 561$$ $$990 = 13p + 28k + 561$$ $$13p + 28k = 990 - 561$$ $$13p + 28k = 429 \quad (\text{Equation 2})$$

From Equation 1, express $p$ in terms of $k$:

$$p = 18 - k$$

Substitute this expression for $p$ into Equation 2:

$$13(18 - k) + 28k = 429$$ $$234 - 13k + 28k = 429$$ $$234 + 15k = 429$$ $$15k = 429 - 234$$ $$15k = 195$$ $$k = \frac{195}{15}$$ $$k = 13$$

Now substitute the value of $k$ back into Equation 1 to find $p$:

$$p + 13 = 18$$ $$p = 18 - 13$$ $$p = 5$$

The values are $p=5$ and $k=13$.

The final answer is $\mathbf{p=5, k=13}$.