31) A

Let marks awarded for research = R

Let marks awarded for design = D

Let marks awarded for construction = C

Let marks awarded for evaluation = E

According to the information provided,

1) D – C = 2 (Rearrange to get C = D – 2)

2) D – E = 11 (Rearrange to get E = D – 11)

We know he got a total of 48 marks.

R + D + C + E = 48

R + D + D – 2 + D – 11 = 48 (substitution of C and E)

R + 3D = 61

3D = 61 – R

D = ^{61 – R}/_{3}

Now substitute the values of R given in the options.

** Option A**R = 10

D =

^{61 – R}/

_{3}D =

^{61 – 10}/

_{3}D = 17

When D = 17,

C = 15, E = 6

Total = 17 + 15 + 10 + 6 = 48

So option A is correct.

If you try all other options, you will either get decimals (which is incorrect as one of the conditions is that all marks are expressed in whole numbers) or incorrect totals.

Exception: Option D

When R = 13,

D = 16, C = 14 and E = 5

This also gives us a total of 48.

However, one of the conditions is that the smallest difference between the marks awarded to any two of Bruno’s sections was 2. If R = 13, C = 14. Therefore, the smallest difference would be 1 and not 2. So option D is also incorrect.