1. Key Concepts and Formulas

• Weighted Average (Combined Average): When combining two groups with different numbers of members and individual averages, the overall average is not a simple arithmetic mean of the individual averages. Instead, each group's average is "weighted" by its size. The formula for the combined average ($A_{\text{combined}}$) of two groups is: $A_{\text{combined}} = \frac{n_1 A_1 + n_2 A_2}{n_1 + n_2}$ where $n_1$ and $n_2$ are the number of members in Group 1 and Group 2, respectively, and $A_1$ and $A_2$ are their respective average values.
• Percentage Calculation: To find the percentage of a part within a whole, we use the formula: $\text{Percentage} = \frac{\text{Part}}{\text{Whole}} \times 100\%$
• The Allegation Rule (Mixture Rule): This is a quick and visual method often used for problems involving weighted averages or mixtures. It helps to find the ratio of quantities of two components when their individual average values and the combined average value are known.

2. Step-by-Step Solution

Let's apply the weighted average concept to solve the problem.

• Step 1: Define Variables and State Given Information. To set up our equation, we first define variables for the unknown quantities and list the known values from the problem statement.

  • Let $x$ be the number of boys in the class.
  • Let $y$ be the number of girls in the class.
  • Average marks of boys ($A_{\text{boys}}$) = 52
  • Average marks of girls ($A_{\text{girls}}$) = 42
  • Combined average marks of boys and girls ($A_{\text{combined}}$) = 50

• Step 2: Formulate the Weighted Average Equation. We substitute the defined variables and given average marks into the combined average formula. Here, $n_1 = x$, $A_1 = 52$, $n_2 = y$, and $A_2 = 42$. The combined average is given as 50. The total marks for boys is $x \times 52$. The total marks for girls is $y \times 42$. The total number of students is $x+y$. The total marks for the class is $52x + 42y$. So, the equation becomes: $50 = \frac{52x + 42y}{x + y}$ This equation establishes a relationship between the number of boys ($x$) and the number of girls ($y$) based on their average marks and the overall class average.

• Step 3: Solve the Equation to Find the Ratio of Boys to Girls. Now, we algebraically manipulate the equation to find the ratio of $x$ to $y$.

  • Multiply both sides by the denominator $(x+y)$ to eliminate the fraction and simplify the equation: $50(x + y) = 52x + 42y$
  • Distribute the 50 on the left side: $50x + 50y = 52x + 42y$
  • Group like terms by moving all terms involving $x$ to one side and all terms involving $y$ to the other side. This is done by subtracting $50x$ from both sides and $42y$ from both sides: $50y - 42y = 52x - 50x$
  • Simplify both sides of the equation: $8y = 2x$
  • Divide both sides by 2 to find the simplest relationship between $x$ and $y$: $4y = x$ This result tells us that the number of boys ($x$) is four times the number of girls ($y$).

• Step 4: Calculate the Percentage of Boys. With the relationship $x = 4y$, we can now determine the percentage of boys in the class.

  • Calculate the total number of students: Total students = Number of boys + Number of girls Total students = $x + y$ Substitute $x = 4y$: Total students = $4y + y = 5y$
  • Calculate the percentage of boys: Percentage of boys = $\frac{\text{Number of boys}}{\text{Total students}} \times 100\%$ Substitute the expressions in terms of $y$: Percentage of boys = $\frac{4y}{5y} \times 100\%$ The variable $y$ cancels out, as expected, since the percentage depends only on the ratio: Percentage of boys = $\frac{4}{5} \times 100\%$ Percentage of boys = $0.8 \times 100\%$ Percentage of boys = $80\%$

3. Common Mistakes & Tips

• Common Mistake: Simple Average: A frequent error is to simply average the individual averages $(52 + 42)/2 = 47$. This is incorrect because it implicitly assumes an equal number of boys and girls, which is not true in this problem (or generally, unless stated). Always use the weighted average formula.
• Tip: Interpret the Combined Average: Notice that the combined average (50) is closer to the boys' average (52) than to the girls' average (42). This intuitively tells us that there must be more boys than girls in the class, which our calculation ($x=4y$) confirms.
• Tip: Use the Allegation Rule for Speed: For competitive exams, the Allegation Rule provides a quick way to solve such problems.
  1. Place individual averages (52 for boys, 42 for girls) on the left and right.
  2. Place the combined average (50) in the middle.
  3. Find the absolute differences diagonally: $$\begin{array}{ccc} \text{Boys} & & \text{Girls} \\ 52 & & 42 \\ & \searrow & \swarrow \\ & 50 & \\ & \swarrow & \searrow \\ (50-42) & : & (52-50) \\ 8 & : & 2 \end{array}$$
  The ratio of the number of boys to the number of girls is $8:2$, which simplifies to $4:1$. This means for every 4 boys, there is 1 girl. If there are $4k$ boys and $1k$ girls, the total is $5k$. The percentage of boys is $\frac{4k}{5k} \times 100\% = 80\%$.

4. Summary

This problem effectively demonstrates the application of the weighted average concept. By setting up an equation using the number of boys, girls, and their respective average marks, along with the combined average, we were able to establish a ratio between the number of boys and girls. The calculation revealed that the number of boys is four times the number of girls, leading to a final percentage of $80\%$ for boys in the class. The Allegation Rule offers a valuable shortcut for solving such problems efficiently.

The final answer is \boxed{80} which corresponds to option (A).