A 100 mark examination was administered to a class of 50 students. Despite only integer marks being given, the everge score of the class was 47.5. Then , the maximum number of students who could get marks more than the class average is

Options

A. 25
B. 35
C. 45
D. 49

Explanation

To solve the problem step by step, we need to analyze the given information and derive the maximum number of students who could have scored more than the average score of the class.

### Step 1: Calculate the total marks obtained by the class.
The average score of the class is given as 47.5, and there are 50 students. The total marks can be calculated using the formula:

\[ \text{Total Marks} = \text{Average} \times \text{Number of Students} \]

Substituting the values:

\[ \text{Total Marks} = 47.5 \times 50 = 2375 \]

### Step 2: Determine the implications of the average score.
Since the average score is 47.5, this means that some students scored below this average while others scored above it. To maximize the number of students scoring above the average, we need to consider how the remaining students can score.

### Step 3: Assume a scenario for maximum students scoring above average.
Let’s assume that \( x \) students scored more than 47.5 and the remaining \( 50 - x \) students scored the minimum possible marks, which is 0 (since marks are integers).

If \( x \) students scored just above the average, we can assume they scored 48 marks each (the smallest integer above 47.5). Therefore, the total marks contributed by these \( x \) students would be:

\[ \text{Marks from } x \text{ students} = 48x \]

The remaining \( 50 - x \) students scored 0 marks, contributing:

\[ \text{Marks from } (50 - x) \text{ students} = 0 \]

### Step 4: Set up the equation for total marks.
The total marks from all students must equal the total marks calculated earlier (2375):

\[ 48x + 0(50 - x) = 2375 \]

This simplifies to:

\[ 48x = 2375 \]

### Step 5: Solve for \( x \).
Now, we can solve for \( x \):

\[ x = \frac{2375}{48} \approx 49.79 \]

Since \( x \) must be an integer (as it represents the number of students), we take the maximum integer value less than or equal to 49.79, which is 49.

### Conclusion:
Thus, the maximum number of students who could score more than the average of 47.5 is:

\[ \boxed{49} \]