Question:medium

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?

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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.
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The Correct Option is A

Solution and Explanation


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: \[ {6} \] Hence, the correct answer is: \[ {\text{Option (A)}} \]
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