In a certain examination, the average marks of an examinee is 64 per paper. If he had obtained 18 more marks for his Mathematics paper and 4 more marks for his English paper, his average per paper would have been 66. How many papers were there in the examination?

Options

A. 11
B. 13
C. 9
D. none of the above

Answer

To solve the problem step by step, we will use the information given in the question.

### Step 1: Define Variables
Let \( n \) be the number of papers in the examination.

### Step 2: Calculate Total Marks
The average marks per paper is given as 64. Therefore, the total marks obtained by the examinee can be expressed as:
\[ \text{Total Marks} = \text{Average} \times \text{Number of Papers} = 64n \]

### Step 3: Adjust for Increased Marks
If the examinee had obtained 18 more marks in Mathematics and 4 more marks in English, the new total marks would be:
\[ \text{New Total Marks} = 64n + 18 + 4 = 64n + 22 \]

### Step 4: New Average Calculation
With the increased marks, the average becomes 66. Therefore, we can set up the equation:
\[ \frac{64n + 22}{n} = 66 \]

### Step 5: Clear the Fraction
To eliminate the fraction, multiply both sides by \( n \):
\[ 64n + 22 = 66n \]

### Step 6: Rearrange the Equation
Rearranging the equation gives:
\[ 66n - 64n = 22 \]
\[ 2n = 22 \]

### Step 7: Solve for \( n \)
Now, divide both sides by 2:
\[ n = \frac{22}{2} = 11 \]

### Conclusion
The number of papers in the examination is \( n = 11 \).

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