The class average x in a test increases by 4 when the score of a student is rectified, whose corrected score is 100 instead of 0. Later, the score of…

The correct option is (b) - 3.

Solution

Step 1: Determine the number of students (n).
Let the number of students in the class be n.
The first correction involves changing a score from 0 to 100. This is an increase of 100 marks.
We are told this increase causes the class average x to increase by 4.
Formula for change in average: Total Change / Number of Students = Increase in Average
$100 / n = 4$100/n=4
$n = 100 / 4 = 25$n=100/4=25 students.

Step 2: Calculate the effect of the second correction.
The second correction involves changing a score from 81 to 56.
This is a decrease in the total marks.
Change in marks = $56 - 81 = -25$56−81=−25 marks.

Step 3: Find the final average y and the difference y - x.
Let the initial total marks be T. Then $x = T / 25$x=T/25.
After the first correction (0 to 100), the new average is $x + 4$x+4.
After the second correction (-25 marks), the final total marks become: (Initial Total + 100 - 25).
Final Average y = (Initial Total + 75) / 25
y = (Initial Total / 25) + (75 / 25)
$y = x + 3$y=x+3
Therefore, $y - x = 3$y−x=3.

Why the other options are incorrect

• Option (a) - 2: This would be the result if the second correction resulted in a decrease of 50 marks instead of 25, or if the number of students was calculated incorrectly.
• Option (c) - 5: This value might be reached if a student incorrectly added the second change ($81 - 56 = 25$81−56=25) to the average instead of subtracting it from the total, or miscalculated the class size.
• Option (d) - 6: This value does not align with the calculated class size of 25; it would require a different net change in marks or a different number of students.

Key Concept

The relationship between the total sum, the number of observations, and the average, specifically how a net change in the sum affects the average across a fixed number of entities.