A Tale on Problem Solving

A Tale on Problem Solving

S Suudharshan Vaidhya

Let’s say an exam is being conducted in the following manner. It is held in four classrooms, with 36 students and 2 invigilators in each classroom. With an actual exam duration of 60 minutes, the interesting situation that happens is, in three of the four classrooms, the invigilators give 70 minutes for writing the exam, and in one classroom, the invigilators give only 60 minutes for writing the exam. Now, naturally, after the completion of the exam, the students of the one classroom are of the opinion that some compensatory marks mechanism is required for them due to the reduced time given. In this situation, what would be a good solution? I thought of three ideas:

Let the marks of the students of that particular classroom be M1, M2, M3, …., M36. Let the marks of the remaining students be M37, M38, M39, …., M144.

Method 1:?

Firstly, we have to see, on average, what marks did the students who had the full time to write the exam got.

• Let the median(M37, M38, …, M144) = A2

Now, since the issue here is of extra time, the next step would be to observe what marks (per-minute) did those students get.

• The “per-minute marks” of those students = A2/70

As they got 10 minutes extra (relatively), then what would this value be for 10 minutes:

• The marks gained by the students (who got full time to write the exam) in 10 minutes, on an average = (A2/70)*10 = E (say)

Now, one can simply add this value to each of M1, M2, …, M36. But that would ignore the variation in marks present in those who wrote the exam for 60 minutes. Hence, a better approach would be to weight this value according to the marks a student got, and then add it:

• Let the median(M1, M2, …, M36) = A1

M1* = M1 + ((M1/A1)*E)?

M2* = M2 + ((M2/A1)*E)

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M36* = M36 + ((M36/A1)*E)

Therefore, through this method, we have got compensatory modified marks for the students who got less time.

But, maybe, one can also see it from the professor’s perspective. Let’s go to Method 2.

Method 2:

In the view of the professor, the students’ per-minute marks as such don’t matter, because the ideal way of attempting it would be according to the professor who set it.

So, it will be as follows:

• The actual duration of the exam = 70 minutes
• Let the total marks of the exam = T

Hence, with respect to the professor’s expectation, the “per-minute mark” value will be:

• per-minute marks = T/70?
• This value for 10 minutes = (T/70)*10 = E’ (say)

Then, one can follow the same last step of the previous method:

M1* = M1 + ((M1/A1)*E’)?

M2* = M2 + ((M2/A1)*E’)

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M36* = M36 + ((M36/A1)*E’)

This method has the difference that it ignores the performance of the students who got the full 70 minutes, rather it takes the professor’s time allocation perspective alone, and accordingly modifies the marks of those who got less time.

However, one might argue after seeing these two methods, why aren’t we taking into account the performance of those who got less time, and only using those values as weights? Well, let’s move to Method 3.

Method 3:

One can say that, if each classroom’s students knew at the start that they were going to get so-and-so amount of time, then from the start of the exam itself, students 1-36 knew they were writing for a 60 minutes exam, and the others for 70 minutes. Hence, it would seem logical to take the performances of both categories into account when observing the marks on average, as the per-minute mark of both categories would tend to be similar.

Solving it as follows:

• Let the median(M1, M2, …, M36) = A1
• Per-minute mark(M1, M2, …, M36) = A1/60 = PM1
• Let the median(M37, M38, …, M144) = A2
• Per-minute mark(M37, M38, …, M144) = A2/70 = PM2
• Average per-minute mark = (PM1 + PM2)/2 = avg(PM)
• Average marks gained in 10 minutes = 10*avg(PM) = E’’

The last step slightly changes here, as the per-minute marks of those who got 60 minutes are already taken into account, i.e., the variation in M1, …, M36 is already accounted for, one can just simply add this value to each of M1, …, M36

M1* = M1 + E’’

M2* = M2 + E’’

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M36* = M36 + E’’

Hence, we get the modified marks by accounting for the per-minute marks of both categories in the initial stage itself, through this method.

What is the aim of this article?

Well, hypothetically, what do you think would have happened in the real world, had this situation come in a real exam? One possible way would have been to handle it on a case-by-case basis, asking each of those students in the less time category to explain to the professor how they would have utilized the extra time effectively, and then modifying the marks depending on the communication. Now, the problems with this approach are that it is highly subjective, it would be difficult to discern what exactly would be the quantum of change in marks, and ex-post explanations can vary a lot with what could have actually happened in the exam.?

Of course, the three methods presented above are also debatable. One might agree with some of them, disagree with some, and there will exist other variations and different types of solutions as well. However, the main advantage is that the three methods bring about analytical rigor which results in robustness in the way of problem-solving. The logical foundations back the statistical execution, and one could argue that it puts forward a degree of fairness to the situation objectively, just through application of basic statistics.

Therefore, through this article, it is interesting to observe how a problem can be solved to ensure “fairness” in a situation, and how common things such as basic statistical concepts can be applied creatively to solve important scenarios that can occur in real life.