In a History examination, the average mark for the entire class was 80. Ten percent of the students scored 35 marks each and twenty percent of the students scored 90 marks each. What was the average mark of the remaining students in the class, correct to one decimal place?

Introduction:
This question checks understanding of weighted averages. The class average is given, and the performance of two specific groups is known. The task is to determine the average of the remaining group of students so that the overall average condition is satisfied. Such problems are common in aptitude tests to assess the candidate s skill in handling averages and percentages simultaneously.

Given Data / Assumptions:
Average mark of the entire class is 80.
10% of the students scored 35 marks each.
20% of the students scored 90 marks each.
Remaining 70% of students scored an unknown average R marks.
We assume the total number of students is N and use percentages to represent their counts.

Concept / Approach:
The total marks for the class can be expressed as (average mark) * (number of students). We can also express total marks as the sum of marks scored by each group: low scoring group, high scoring group, and remaining students. By equating these two expressions for total marks, we can solve for the unknown average R of the remaining group. This is a direct application of the concept of weighted average.

Step-by-Step Solution:
Let total number of students be N. Total marks for the entire class = 80 * N. Number of students scoring 35 marks each = 10% of N = 0.10 * N. Marks contributed by this group = 0.10 * N * 35 = 3.5 * N. Number of students scoring 90 marks each = 20% of N = 0.20 * N. Marks contributed by this group = 0.20 * N * 90 = 18 * N. Remaining students = 70% of N = 0.70 * N with average R marks. Marks contributed by remaining students = 0.70 * N * R. Total marks by groups = 3.5 * N + 18 * N + 0.70 * N * R. This must equal total marks 80 * N. So 3.5 * N + 18 * N + 0.70 * N * R = 80 * N. Combine constants: (3.5 + 18) * N = 21.5 * N. So 21.5 * N + 0.70 * N * R = 80 * N. Divide both sides by N (N is positive): 21.5 + 0.70 * R = 80. Then 0.70 * R = 80 - 21.5 = 58.5. Hence R = 58.5 / 0.70 = 83.5714 approximately. Correct to one decimal place this is about 83.6.
Verification / Alternative check:
We can use a concrete value for N, for example N = 100. Then 10 students score 35 each, 20 score 90 each, and 70 score R each. Total marks from known groups = 10 * 35 + 20 * 90 = 350 + 1800 = 2150. Total marks for 100 students at average 80 is 100 * 80 = 8000. So remaining group must contribute 8000 - 2150 = 5850 marks. The average for 70 students is 5850 / 70 = 83.5714, which again rounds to 83.6. This confirms the result.

Why Other Options Are Wrong:
Values like 75, 80, 82.5, and 85 do not satisfy the weighted average equation when substituted for R. If we plug any of them into the calculation, the resulting overall average would differ from the given 80. Only about 83.6 ensures that the combined average of all three groups is exactly 80.

Common Pitfalls:
Common mistakes include averaging the three given numbers directly without weighting by the number of students, or forgetting that 10%, 20%, and 70% apply to counts of students, not to marks. Another pitfall is rounding too early in the calculation. It is best to carry exact decimal values to the end and then round only when presenting the final answer.

Final Answer:
The average mark of the remaining students is approximately 83.6.