Solving a Classroom Marking Error Through Mathematical Analysis

Have you ever encountered a situation where a classroom's average marks mysteriously increased due to a simple yet significant error in marking? In this article, we will walk through a real-world example to illustrate how you can solve such a problem using basic algebra and your mathematical skills. This will not only enhance your problem-solving abilities but also help improve accuracy in your assessments.

Understanding the Problem

Imagine that a student's marks were entered as 85 instead of the correct 75. As a result, the average marks for the class increased by one-third (1/3). The question is: How many pupils are in the class?

Setting Up the Equation

To solve this problem, we need to establish a series of equations based on the information provided. Let's break it down into steps for clarity.

Step 1: Calculate the Original Average Marks

Let:

n number of pupils in the class, S original sum of the marks of the class.

The original average marks for the class are given by:

Original Average S/n

Step 2: Calculate the New Average After the Error

When the pupil's marks were wrongly entered as 85 instead of 75, the new sum of the marks becomes:

S 85 - 75 S 10

The new average marks for the class are then:

New Average (S 10)/n

Step 3: Set Up the Equation for the Increase in Average

According to the problem, the average increases by one-third (1/3) of the original average. This can be expressed as:

New Average Original Average (1/3) * Original Average

This expression can be further simplified to:

(S 10)/n (S/n) (1/3) * (S/n)

Step 4: Simplify the Equation

Combining the terms on the right side gives:

(S 10)/n (S S/3)/n (4S/3)/n

Step 5: Cross-Multiply to Eliminate the Fractions

Cross-multiplying yields:

3(S 10) 4S

Step 6: Expand and Rearrange the Equation

Expanding the left side gives:

3S 30 4S

Rearranging terms results in:

30 4S - 3S

30 S

Step 7: Find the Original Average

Now that we have S 30, we can calculate the original average:

Original Average S/n 30/n

Step 8: Use the New Average to Find n

Substituting S back into the equation for the new average:

New Average (S 10)/n (30 10)/n 40/n

We also have:

New Average (4S)/3n (4 * 30)/3n 120/3n 40/n

Step 9: Solve for n

Since both expressions for the new average are equal:

40/n 40/n

This confirms our calculations but to find the exact number of pupils, we can use the increase in average:

(30/n - 10/3n) 40/n

Solving:

30 - 10/3 40 implies (90 - 10)/3 40 implies 100/3 40 implies n 3

Hence, the number of pupils in the class is:

boxed{3}

This example not only highlights the importance of accuracy in marking but also demonstrates the application of algebra in real-world scenarios. Practicing such problems can significantly improve your analytical skills and ensure accuracy in your educational assessments.