Solving the Average Score Problem for Boys and Girls

This problem involves calculating the average score of girls in a class, given the overall class average, the ratio of boys to girls, and the average score of the boys. We will use the concept of weighted averages to solve this.

Understanding the Given Information

• Overall average score of the class (boys and girls): \(p\)
• Ratio of boys to girls: 3 ∶ 1
• Average score of boys: \((p + 1)\)

We need to find the average score of the girls.

Setting Up the Problem with Variables

Let's assume the number of girls in the class is \(n\). Since the ratio of boys to girls is 3 ∶ 1, the number of boys will be \(3n\).

• Number of girls = \(n\)
• Number of boys = \(3n\)
• Total number of students = Number of boys + Number of girls = \(3n + n = 4n\)

The average score is calculated as:

Average Score = \(\frac{\text{Total Score}}{\text{Total Number of Students}}\)

Therefore, the total score for any group is:

Total Score = Average Score \(\times\) Number of Students

Calculating Total Scores

Using the formula above, let's calculate the total scores for each group and the entire class:

• Total score of boys = Average score of boys \(\times\) Number of boys
• Total score of boys = \((p + 1) \times 3n\)
• Total score of girls = Average score of girls \(\times\) Number of girls
• Let the average score of girls be \(g\).
• Total score of girls = \(g \times n\)
• Total score of the class = Overall average score \(\times\) Total number of students
• Total score of the class = \(p \times 4n\)

Using the Weighted Average Principle

The total score of the class is the sum of the total score of boys and the total score of girls.

Total score of the class = Total score of boys + Total score of girls

\(p \times 4n = (p + 1) \times 3n + g \times n\)

Solving for the Average Score of Girls (g)

Now, we need to solve the equation for \(g\). Notice that \(n\) is present in every term. Assuming there is at least one girl (\(n > 0\)), we can divide the entire equation by \(n\):

\(4pn = (3p + 3)n + gn\)

Divide by \(n\) (since \(n \ne 0\)):

\(4p = 3p + 3 + g\)

Now, isolate \(g\) by subtracting \(3p\) and \(3\) from both sides of the equation:

\(4p - 3p - 3 = g\)

\(p - 3 = g\)

So, the average score of the girls is \((p - 3)\).

Verification (Optional but Recommended)

Let's check if this result makes sense. If the boys have an average score higher than the class average \((p+1 > p)\), and they form a larger portion of the class (3 out of 4 parts), then the girls must have an average score lower than the class average to bring the overall average down. The result \((p-3)\) is indeed lower than \(p\).

Summary of Steps

• Assigned variables based on the given ratio.
• Used the relationship: Total Score = Average Score \(\times\) Number of Students.
• Set up an equation based on the fact that the total class score is the sum of boys' and girls' total scores.
• Solved the equation for the unknown average score of girls.

The average score of the girls is \((p - 3)\).

Revision Table: Key Concepts

Additional Information: Weighted Averages Explained

A weighted average is useful when different parts of a dataset contribute unevenly to the total. In a class average calculation, the number of students in each group (like boys or girls) is the 'weight'. If a group has more students, its average score has a greater impact on the overall class average.

Consider our problem:

• Boys make up \(\frac{3n}{4n} = \frac{3}{4}\) of the class.
• Girls make up \(\frac{n}{4n} = \frac{1}{4}\) of the class.

The overall average \(p\) is a weighted average of the boys' average \((p+1)\) and the girls' average \((g)\):

\(p = \left(\frac{3}{4}\right) \times (p+1) + \left(\frac{1}{4}\right) \times g\)

To solve for \(g\), we can multiply the entire equation by 4 to remove the fractions:

\(4p = 3(p+1) + 1g\)

\(4p = 3p + 3 + g\)

\(4p - 3p - 3 = g\)

\(p - 3 = g\)

This confirms the result obtained earlier using the total score method. Both methods rely on the principle of weighted averages but approach the calculation slightly differently.