Question

In a class exam, Ramya's average mark was 90 per paper. If she had obtained 4 more marks in Maths paper and 20 more marks in Physics paper, then her average per paper would have been 94. How many papers were there in the exam?

• A. 6
• B. 8
• C. 9
• D. 7

Hint:

For average-based questions involving extra marks: \[ \text{Number of Items} = \frac{\text{Change in Total Sum}}{\text{Change in Average}} \] Here, \[ \frac{24}{94-90} = \frac{24}{4} = 6 \] This shortcut can solve many average problems within a few seconds.

Answer 1

The Correct Option is A

Concept: The average of a set of observations is defined as: \[ \text{Average} = \frac{\text{Total Sum of Observations}}{\text{Number of Observations}} \] From this formula: \[ \text{Total Sum} = \text{Average} \times \text{Number of Observations} \] Whenever some marks are added or removed, the total sum changes accordingly, causing a change in the average.

Step 1: Assume the number of papers. Let the number of papers be: \[ n \] Given that Ramya's average mark was 90 per paper. Therefore, the total marks obtained originally are: \[ 90n \]

Step 2: Calculate the increase in total marks. According to the question:
• Additional marks in Mathematics = 4
• Additional marks in Physics = 20 Hence, total additional marks: \[ 4 + 20 = 24 \] Therefore, the new total marks become: \[ 90n + 24 \]

Step 3: Use the new average. After adding these marks, the average becomes: \[ 94 \] Since the number of papers remains unchanged, \[ \text{New Total Marks} = 94n \] Therefore: \[ 94n = 90n + 24 \]

Step 4: Solve the equation. Subtract \(90n\) from both sides: \[ 94n - 90n = 24 \] \[ 4n = 24 \] Dividing both sides by 4: \[ n = \frac{24}{4} \] \[ n = 6 \]

Step 5: Verification. Original total marks: \[ 90 \times 6 = 540 \] Additional marks: \[ 24 \] New total marks: \[ 540 + 24 = 564 \] New average: \[ \frac{564}{6} = 94 \] The condition is satisfied perfectly. Therefore, the number of papers is: \[ \boxed{6} \] Hence, the correct answer is: \[ \boxed{\text{Option (A)}} \]