There are only three female students - Amala, Koli and Rini - and only three male students - Biman, Mathew and Shyamal - in a course. The course has two evaluation components, a project and a test. The aggregate score in the course is a weighted average of the two components, with the weights being positive and adding to 1 .

The projects are done in groups of two, with each group consisting of a female and a male student. Both the group members obtain the same score in the project.

The following additional facts are known about the scores in the project and the test.

1. The minimum, maximum and the average of both project and test scores were identical – 40, 80 and 60 , respectively.

2. The test scores of the students were all multiples of 10 ; four of them were distinct and the remaining two were equal to the average test scores.

3. Amala's score in the project was double that of Koli in the same, but Koli scored 20 more than Amala in the test. Yet Amala had the highest aggregate score.

4. Shyamal scored the second highest in the test. He scored two more than Koli, but two less than Amala in the aggregate.

5. Biman scored the second lowest in the test and the lowest in the aggregate.

6. Mathew scored more than Rini in the project, but less than her in the test.

What was the maximum aggregate score obtained by the students?

To find the maximum aggregate score obtained by the students, we can break down the information given about their project and test scores step-by-step.

Step 1: Analyzing the Project Scores
There are three female students (Amala, Koli, Rini) and three male students (Biman, Mathew, Shyamal) paired up into three groups.
Let the project scores of the three groups be $p_1$, $p_2$, and $p_3$.
The minimum project score is 40, the maximum is 80, and the average is 60.
Since there are 6 students in total:
$\frac{2(p_1+p_2+p_3)}{6}=60\Rightarrow p_1+p_2+p_3=180$
From fact 3, Amala's project score was double that of Koli's project score:
$P\left(A\right)=2P\left(K\right)$
Since the minimum score is 40 and maximum is 80, we must have:
$P\left(K\right)=40$ and $P\left(A\right)=80$
This gives the third project score (for Rini) as:
$P\left(R\right)=180-(40+80)=60$
Mathew scored more than Rini in the project, so $P\left(M\right)>60\Rightarrow P\left(M\right)=80$. This means Mathew is paired with Amala.

Step 2: Analyzing the Test Scores
The test scores are all multiples of 10. The minimum is 40, the maximum is 80, and the average is 60.
Four of the test scores are distinct, and the remaining two are equal to the average (60).
Thus, the set of test scores is: $\{40,50,60,60,70,80\}$.
From the facts:
1. Shyamal scored the second highest in the test, so $T\left(S\right)=70$.
2. Biman scored the second lowest in the test, so $T\left(B\right)=50$.
3. Koli scored 20 more than Amala in the test: $T\left(K\right)=T\left(A\right)+20$.
Since 50 and 70 are already taken, the only possible pairs for $\left(T\right(A),T(K\left)\right)$ from the remaining scores $\{40,60,60,80\}$ are:
- Case 1: $T\left(A\right)=40,T\left(K\right)=60$
- Case 2: $T\left(A\right)=60,T\left(K\right)=80$

Step 3: Calculating weights and aggregate scores
Let $w$ be the weight of the project component, and $1-w$ be the weight of the test component.
From fact 4, Shyamal scored 2 more than Koli but 2 less than Amala in the aggregate:
$S\left(S\right)=S\left(K\right)+2$ and $S\left(S\right)=S\left(A\right)-2$
This implies that:
$S\left(A\right)-S\left(K\right)=4$

Let's check Case 2: $T\left(A\right)=60,T\left(K\right)=80$
The aggregate score equations are:
$S\left(A\right)=80w+60(1-w)=20w+60$
$S\left(K\right)=40w+80(1-w)=80-40w$
Substituting these into the difference equation:
$(20w+60)-(80-40w)=4$
$60w-20=4\Rightarrow 60w=24\Rightarrow w=0.4$
So, the weights are 0.4 for the project and 0.6 for the test.

Using these weights, we calculate the aggregate scores:
- Amala's aggregate score (which is the maximum):
$S\left(A\right)=20\left(0.4\right)+60=68$
- Koli's aggregate score:
$S\left(K\right)=80-40\left(0.4\right)=64$
- Shyamal's aggregate score:
$S\left(S\right)=64+2=66$

Thus, the maximum aggregate score obtained by any student is 68.