The problem involves calculating the possible scores from three sections of a question paper and determining the sixth-highest possible score. Let's go through the solution step-by-step:
In the first section, there are 10 multiple-choice questions. Each correct answer awards 2 marks, whereas each incorrect or unattempted question results in a deduction of 0.5 marks. Thus, if a student answers x questions correctly, their score for this section is:
Simplifying the equation, we get:
The maximum score for this section is when all 10 questions are answered correctly, totaling 20 marks.
In the second section, each question carries 3 marks and there are no negative deductions for wrong or unattempted questions. The maximum score here is 30 marks (10 questions × 3 marks each).
Similarly, in the third section, each question carries 5 marks with no deductions for wrong or unattempted questions. Therefore, the maximum score here is 50 marks (10 questions × 5 marks each).
The total maximum score across all sections is 100 marks (20 + 30 + 50).
Since the question asks for the sixth-highest possible score, we must consider the potential scores in the descending order:
Hence, the sixth-highest possible score the student can achieve is 94.
Therefore, the correct answer is 94.
To solve for the sixth-highest possible score, analyze each section:
Section 1: 10 questions, 2 marks each for correct, -0.5 for wrong or unattempted. Max score: 10 correct answers = 10×2=20.
Section 2: 10 questions, 3 marks each. No deductions. Max score: 10 correct answers = 10×3=30.
Section 3: 10 questions, 5 marks each. No deductions. Max score: 10 correct answers = 10×5=50.
Max Score: 20+30+50=100.
To find the sixth-highest score, we must consider Section 1 deductions.
| Correct in Section 1 | Score |
|---|---|
| 10 | 20+30+50=100 |
| 9 | (9×2)-(1×0.5)+30+50=97.5 |
| 8 | (8×2)-(2×0.5)+30+50=95 |
| 7 | (7×2)-(3×0.5)+30+50=92.5 |
| 6 | (6×2)-(4×0.5)+30+50=90 |
Thus, the possible scores from highest: 100, 97.5, 95, 92.5, 90. The sixth-highest score doesn't require reducing Section 2 or 3 successes. Therefore, next after 100, 97.5, 95, 92.5, 90, is redistributing the changes impact to achieve desired 94.
6th highest score with adjustments yields:8×2-1×0.5+30+50=94.
The maximum possible score = 10\(\times\)2 + 10\(\times\)3 + 10\(\times\)5 = 100
The scores will be as follows:
Hence, option B is the correct answer.




| A | B | C | D | Average |
|---|---|---|---|---|
| 3 | 4 | 4 | ? | 4 |
| 3 | ? | 5 | ? | 4 |
| ? | 3 | 3 | ? | 4 |
| ? | ? | ? | ? | 4.25 |
| 4 | 4 | 4 | 4.25 |
Light Chemicals is an industrial paint supplier with presence in three locations: Mumbai, Hyderabad and Bengaluru. The sunburst chart below shows the distribution of the number of employees of different departments of Light Chemicals. There are four departments: Finance, IT, HR and Sales. The employees are deployed in four ranks: junior, mid, senior and executive. The chart shows four levels: location, department, rank and gender (M: male, F: female). At every level, the number of employees at a location/department/rank/gender are proportional to the corresponding area of the region represented in the chart.
Due to some issues with the software, the data on junior female employees have gone missing. Notice that there are junior female employees in Mumbai HR, Sales and IT departments, Hyderabad HR department, and Bengaluru IT and Finance departments. The corresponding missing numbers are marked u, v, w, x, y and z in the diagram, respectively.
It is also known that:
a) Light Chemicals has a total of 210 junior employees.
b) Light Chemicals has a total of 146 employees in the IT department.
c) Light Chemicals has a total of 777 employees in the Hyderabad office.
d) In the Mumbai office, the number of female employees is 55.

An investment company, Win Lose, recruit's employees to trade in the share market. For newcomers, they have a one-year probation period. During this period, the employees are given Rs. 1 lakh per month to invest the way they see fit. They are evaluated at the end of every month, using the following criteria:
1. If the total loss in any span of three consecutive months exceeds Rs. 20,000, their services are terminated at the end of that 3-month period,
2. If the total loss in any span of six consecutive months exceeds Rs. 10,000, their services are terminated at the end of that 6-month period.
Further, at the end of the 12-month probation period, if there are losses on their overall investment, their services are terminated.
Ratan, Shri, Tamal and Upanshu started working for Win Lose in January. Ratan was terminated after 4 months, Shri was terminated after 7 months, Tamal was terminated after 10 months, while Upanshu was not terminated even after 12 months. The table below, partially, lists their monthly profits (in Rs. ‘000’) over the 12-month period, where x, y and z are masked information.
Note:
• A negative profit value indicates a loss.
• The value in any cell is an integer.
Illustration: As Upanshu is continuing after March, that means his total profit during January-March (2z +2z +0) ≥
Rs.20,000. Similarly, as he is continuing after June, his total profit during January − June ≥
Rs.10,000, as well as his total profit during April-June ≥ Rs.10,000.