Understanding Average Corrections

This problem involves calculating the change in a class average based on score rectifications. We first find the number of students ($N$) using the information from the first correction. Then, we determine the change in the average due to the second correction and calculate the required difference ($y - x$).

Calculating Number of Students (N)

A student's score was corrected from 0 to 100. This increases the total score sum by $100 - 0 = 100$. The problem states this correction caused the class average to increase by 4. Let $N$ be the number of students.

• The increase in the average is calculated as the increase in the sum divided by the number of students: $ \Delta_{Avg1} = \frac{\Delta_{Sum1}}{N} $
• We are given that the average increase ($\Delta_{Avg1}$) is 4, and the sum increase ($\Delta_{Sum1}$) is 100. $ 4 = \frac{100}{N} $
• Solving for $N$: $ N = \frac{100}{4} $ $ N = 25 $

There are 25 students in the class.

Calculating the Final Average Difference (y - x)

We interpret the question such that $x$ represents the initial average before any corrections. The average increases by 4 after the first correction (0 to 100). So, the average after the first correction is $x + 4$.

The second correction involves a score recorded as 81 instead of the correct score 56. The change in the sum is $56 - 81 = -25$.

• Calculate the change in the average due to the second correction: $ \Delta_{Avg2} = \frac{\Delta_{Sum2}}{N} = \frac{-25}{25} $ $ \Delta_{Avg2} = -1 $
• The final corrected average, $y$, is the average after the second correction. This is calculated by adding the change ($\Delta_{Avg2}$) to the average after the first correction ($x+4$): $ y = (x + 4) + \Delta_{Avg2} $ $ y = (x + 4) + (-1) $ $ y = x + 3 $
• The question asks for the value of $y - x$. $ y - x = (x + 3) - x $ $ y - x = 3 $