Context: The average marks of the girls in a class areequal to the number of boys and the averagemarks of boys is equal to the number of girls.If the class average is 4 less than the averageof the boys’ average marks and the girls’average marks, which of the following couldbe the number of students in the class?(1) 24 (2) 48(3) 50 (4) 64
Answer question

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.

Context: The average marks of the girls in a class are equal to the number of boys and the average marks of boys is equal to the number of girls. If the class average is 4 less than the average of the boys’ average marks and the girls’ average marks, which of the following could be the number of students in the class? (1) 24 (2) 48 (3) 50 (4) 64

Explore how to determine the number of students in a class based on given conditions relating to averages and quadratic equations. This problem involves algebraic concepts like setting up equations for class average, solving quadratic equations, and working with weighted averages. Suitable for Grades 10-12.

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