Understanding the Average Score Problem

This problem involves calculating the average score in mathematics for two sections of a class, given the number of students in each section, the overall average score, and the relationship between the average scores of the two sections.

We are given:

• Number of students in Section A ($N_A$): 50
• Number of students in Section B ($N_B$): 62
• Overall average score ($A_{Total}$): 75
• Average score in Section A ($A_A$) is 20% more than the average score in Section B ($A_B$).

We need to find the average score of students in Section A ($A_A$), rounded to one decimal place.

Setting Up the Equations

First, let's determine the total number of students:

$\text{Total Students} = N_A + N_B$

$\text{Total Students} = 50 + 62 = 112$

The total score of all students in the class can be calculated using the overall average:

$\text{Total Score} = \text{Total Students} \times A_{Total}$

$\text{Total Score} = 112 \times 75$

$\text{Total Score} = 8400$

The total score is also the sum of the total scores from Section A and Section B. The total score for each section is the number of students in that section multiplied by their average score:

• Total score for Section A = $N_A \times A_A = 50 \times A_A$
• Total score for Section B = $N_B \times A_B = 62 \times A_B$

So, the equation for the total score is:

$50 A_A + 62 A_B = 8400$

We are also told that the average score in Section A is 20% more than the average score in Section B. This can be written as:

$A_A = A_B + 20\% \text{ of } A_B$

$A_A = A_B + 0.20 A_B$

$A_A = 1.20 A_B$

From this equation, we can express $A_B$ in terms of $A_A$:

$A_B = \frac{A_A}{1.20}$

Solving for the Average Score in Section A

Now we can substitute the expression for $A_B$ into the total score equation:

$50 A_A + 62 \left(\frac{A_A}{1.20}\right) = 8400$

Let's simplify the coefficient of $A_A$ in the second term:

$\frac{62}{1.20} = \frac{620}{12} = \frac{310}{6} = \frac{155}{3}$

So the equation becomes:

$50 A_A + \frac{155}{3} A_A = 8400$

Combine the terms with $A_A$:

$\left(50 + \frac{155}{3}\right) A_A = 8400$

Find a common denominator for the terms in the parenthesis:

$\left(\frac{50 \times 3}{3} + \frac{155}{3}\right) A_A = 8400$

$\left(\frac{150}{3} + \frac{155}{3}\right) A_A = 8400$

$\left(\frac{150 + 155}{3}\right) A_A = 8400$

$\frac{305}{3} A_A = 8400$

Now, solve for $A_A$ by multiplying both sides by $\frac{3}{305}$:

$A_A = \frac{8400 \times 3}{305}$

$A_A = \frac{25200}{305}$

Now, let's perform the division:

$A_A \approx 82.62295...$

Rounding the Average Score

The question asks for the average score of students in Section A (correct to one decimal place).

$A_A \approx 82.6$

Let's check if this answer is among the options.

Verification (Optional)

If $A_A \approx 82.6$, then $A_B = \frac{A_A}{1.20} \approx \frac{82.6}{1.20} \approx 68.833...$

Total Score = $(50 \times 82.6) + (62 \times 68.833...)$

Total Score $\approx 4130 + 4267.666... = 8397.666...$

Expected Total Score = $112 \times 75 = 8400$.

The slight difference is due to rounding $A_B$. Using the exact fraction for $A_A$ would yield the exact total score.

Using $A_A = \frac{25200}{305}$: $A_B = \frac{1}{1.20} \times \frac{25200}{305} = \frac{10}{12} \times \frac{25200}{305} = \frac{5}{6} \times \frac{25200}{305} = \frac{5 \times 4200}{305} = \frac{21000}{305}$

Total Score = $50 \times \frac{25200}{305} + 62 \times \frac{21000}{305} = \frac{1260000}{305} + \frac{1302000}{305} = \frac{2562000}{305} = 8400$. This confirms our calculation for $A_A$ is correct before rounding.

Rounding $\frac{25200}{305}$ to one decimal place gives 82.6.

Final Answer

The average score of students in section A, correct to one decimal place, is 82.6.

Revision Table: Key Concepts

Additional Information on Averages and Percentages

Understanding how to work with averages and percentages is crucial for many types of problems. An average gives a central value for a set of numbers. When combining groups with different averages and sizes, the overall average is influenced more by the larger groups – this is the concept of a weighted average.

In this problem, the overall average score (75) is a weighted average of the scores from Section A and Section B. Since Section B has more students (62) than Section A (50), the overall average will be closer to the average score of Section B than to the average score of Section A.

The relationship between the average scores was given as a percentage increase. An increase of 20% means the new value is 100% of the original value plus 20% of the original value, totaling 120% of the original value. Converting the percentage to a decimal (20% = 0.20) allows us to express this relationship mathematically as $A_A = A_B + 0.20 A_B = 1.20 A_B$. Similarly, if a value were 20% less, the relationship would be $A_A = A_B - 0.20 A_B = 0.80 A_B$. These concepts are fundamental in quantitative problems.