Q 6.

Let $s$ be the sum of all scores so far. Let $N$ be the total exams taken so far. Then, from the problem statement and definition of an average score:

$\begin{cases} \frac{s + 71}{N + 1} = 83 \\ \frac{s + 99}{N + 1} = 87 \end{cases}$

Multiplying each side by $N + 1$ :

$\begin{cases} s + 71 = 83N + 83 \\ s + 99 = 87N + 87 \end{cases}$

Subtracting the top equation from the bottom equation yields:

$28 = 4N + 4$

So, the number of exams taken is $N = 6$

Let the marks in different subjects be denoted by english alphabets.(a, b, c...) and the total number of subjects be denoted by x.

Then,

(71 + a + b + ....) / x = 83

or, 71 + a + b + .... = 83x -----------------------(i)

(99 + a + b + ....) / x = 87

or, 99 + a + b + .... = 87x ------------------------(ii)

Now,

Subtracting eauation (i) from (ii) we get,

99 + a + b ...... = 87x

-(71 + a + b ..... = 83x)

99 + a + b ...... = 87x

-71 - a - b ..... = -83x

28 = 4x

x = 7

Wait !!! our answer is not 7. Read the question once again. It asks how many exams has been already given not how many subjects are there.

Since, the marks of one subject is not confirmed, it has not been given.

Thus, our answer is 7 - 1 = 6