Calculating the Number of Papers in an Examination

This problem involves understanding how changes in total marks affect the average marks per paper. We are given the initial average, the new average after adding marks to specific papers, and the amount of marks added. We need to find the total number of papers in the examination.

Setting up the Problem

Let's define the terms:

• Let N be the total number of papers in the examination.
• Let S be the initial total marks obtained by the student in all papers.

According to the problem, the initial average marks per paper were 71.

$$\text{Initial Average} = \frac{\text{Initial Total Marks}}{\text{Number of Papers}}$$

$$71 = \frac{S}{N}$$

From this, we can express the initial total marks S in terms of N:

$$S = 71N \quad (*)$$

Understanding the Change in Marks

The student obtained additional marks in three specific papers:

• Science: 35 more marks
• History: 11 more marks
• Computer Science: 4 more marks

The total increase in marks is the sum of these additional marks:

$$\text{Total Increase in Marks} = 35 + 11 + 4$$

$$\text{Total Increase in Marks} = 50$$

The new total marks obtained by the student will be the initial total marks plus this increase:

$$\text{New Total Marks} = S + 50$$

Calculating the New Average

The problem states that with the additional marks, the new average marks per paper would have been 76. The number of papers (N) remains the same.

$$\text{New Average} = \frac{\text{New Total Marks}}{\text{Number of Papers}}$$

$$76 = \frac{S + 50}{N}$$

Solving for the Number of Papers (N)

Now we have an equation with S and N. We can substitute the expression for S from equation $(*)$ into this new average equation:

$$76 = \frac{71N + 50}{N}$$

To solve for N, multiply both sides of the equation by N:

$$76N = 71N + 50$$

Now, gather the terms involving N on one side of the equation by subtracting 71N from both sides:

$$76N - 71N = 50$$

Simplify the left side:

$$5N = 50$$

Finally, divide both sides by 5 to find the value of N:

$$N = \frac{50}{5}$$

$$N = 10$$

So, there were 10 papers in the examination.

Verification

Let's check if this answer is consistent with the given information.

• Number of papers (N) = 10
• Initial Average = 71
• Initial Total Marks (S) = $71 \times 10 = 710$
• Total Increase in Marks = 50
• New Total Marks = $710 + 50 = 760$
• New Average = $\frac{760}{10} = 76$

This matches the new average given in the problem (76), so our calculated number of papers is correct.

The problem essentially tells us that a total increase of 50 marks caused the average to increase by $76 - 71 = 5$ marks per paper. If an increase of 50 total marks leads to an increase of 5 marks *per paper*, then the total number of papers must be $\frac{\text{Total Increase in Marks}}{\text{Increase in Average per Paper}} = \frac{50}{5} = 10$. This confirms our detailed algebraic solution.

Revision Table: Average Marks and Examination Papers

Additional Information: Concepts related to Averages

Understanding averages is fundamental in many quantitative problems. Here are a few related concepts:

• Weighted Average: Used when different values contribute differently to the total (e.g., different papers having different maximum marks).
• Median: The middle value in a sorted dataset. Unlike the average, it is not affected by extreme outliers.
• Mode: The value that appears most frequently in a dataset.
• Sum of Deviations from Mean: The sum of the differences between each value and the mean (average) is always zero. This property is often useful in proofs and more advanced problems.

Problems involving averages in examinations often test your ability to set up equations based on the definition of the average and how changes in total values or the number of items affect the average.