Let's break down the given problem step by step.

Let:

• $G$ = number of girls
• $B$ = number of boys
• $x_g$ = average marks of girls
• $x_b$ = average marks of boys

According to the problem:

1. The average marks of the girls are equal to the number of boys: $x_g = B$

2. The average marks of the boys are equal to the number of girls: $x_b = G$

3. The class average is 4 less than the average of the boys' and girls' average marks: $\text{Class average} = \frac{x_g + x_b}{2} - 4$ Substituting $x_g = B$ and $x_b = G$ from the first two conditions: $\text{Class average} = \frac{B + G}{2} - 4$

4. The class average is also the weighted average of the boys' and girls' averages: $\text{Class average} = \frac{B \cdot x_b + G \cdot x_g}{B + G}$ Substituting $x_b = G$ and $x_g = B$: $\text{Class average} = \frac{B \cdot G + G \cdot B}{B + G} = \frac{2BG}{B + G}$

Now we have two expressions for the class average:

1. $\frac{B + G}{2} - 4$
2. $\frac{2BG}{B + G}$

Equating these two expressions: $\frac{B + G}{2} - 4 = \frac{2BG}{B + G}$ Multiply through by 2 to eliminate fractions: $B + G - 8 = \frac{4BG}{B + G}$ Multiply both sides by $B + G$ to clear the denominator: $(B + G)(B + G - 8) = 4BG$ Expanding both sides: $(B + G)^2 - 8(B + G) = 4BG$ Simplifying: $B^2 + G^2 + 2BG - 8B - 8G = 4BG$ Rearranging: $B^2 + G^2 - 8B - 8G - 2BG = 0$ This is a quadratic equation in terms of $B$ and $G$, which we can solve using trial and error with the given options.

Testing the options:

We are asked to find the number of students, which means $B + G$. Let's test the values given.

Option 1: $B + G = 24$

If $B + G = 24$, then substituting into the equation $B^2 + G^2 - 8B - 8G - 2BG = 0$ does not satisfy the equation, so this is not a solution.

Option 2: $B + G = 48$

Similar checks for $B + G = 48$ do not satisfy the equation.

Option 3: $B + G = 50$

Testing $B + G = 50$ works, as it satisfies the given equation $B^2 + G^2 - 8B - 8G - 2BG = 0$.

Thus, the correct answer is: $\boxed{50}$

Would you like further details or explanations?

Here are 5 related questions you can try:

1. How would the solution change if the class average was 5 less instead of 4?
2. Can you derive a similar formula for different class average conditions?
3. What happens if there is only one student (either a boy or girl) in the class?
4. How does changing the total number of students affect the solution?
5. Could there be a solution with a fractional number of students?

Tip: In algebraic problems involving averages, always focus on representing the quantities clearly in terms of variables and setting up the right equations to avoid confusion.