Let original strength = y
Then , 40y + 12 x 32 = ( y + 12) x 36
? 40y + 384 = 36y + 432
? 4y = 48
? y = 12

Let the number of students in the class = N
According to the question,
(35N + 6 x 33) / (N + 6) = 35 - (1/2)
? 35N + 198 = 34.5 (N + 6)
? 35N +198 = 34.5N + 207
? 0.5N = 207 -198N = 9/0.5 = 18
? N = 18 days

M + Tu + W + Th = 4 x 48 = 192
Tu + W + Th + F = 4 x 46 = 184
M = 42
Tu + W + Th = 192 - 42 = 150
F = 184 ? 150 = 34

Given total number of passengers in the bus = 45

First average weight of 45 passengers = 52 kgs

Average weight of 5 passengers who leave bus = 48

Average weight of 5 passengers who joined the bus = 54

Therefore, the net average weight of the bus is given by

$$\begin{gathered} = 45 x 52 - 5 x 48 + 5 x 54 45 \\ = 2370 45 \\ = 158 3 \\ = 52 . 66 kgs . \end{gathered}$$

Let average expenditure was Rs. R

13 x 79 + 6x(R + 4) = 19R

=> R = Rs. 80.84

Total money = 19 x 80.84 = Rs. 1536.07.

Using Trial and error method,

we get the number of girls in the hall = 78

We have : ( a + b + c) / 3 = M   or (a + b + c) = 3M.

Now.  a + b + c 2 = 3 M 2 = 9 M 2 .

< = > a 2 + b 2 + c 2 + 2 a b + b c + c a = 9 M 2

a 2 + b 2 + c 2 = 9 M 2 , ( a b + b c + c a = 0 )     

Required mean = a 2 + b 2 + c 2 3 = 9 M 2 3 = 3 M 2 .

Total money decided to contribute = 750 x 12 = 9000

Let 'b' boys dropped

The rest paid 150/- more

=> (12 - b) x 900 = 9000

=> b = 2

Hence, the number of boys who dropped out is 2.

Let the number of boys = b

Let the number of girls = g

From the given data,

81b + 83g = 81.8(b + g)

81.8b - 81b = 83g - 81.8g

0.8b = 1.2g

b/g = 1.2/0.8 = 12/8 = 3/2

=> g : b = 2 : 3

Hence, ratio between the number of girls to the number of boys = 2 : 3.

The total marks of all 5 subjects = 82 x 5 = 410

Total marks of 1st two subjects = 86.5 x 2 = 173

Total marks of last two subjects = 84 x 2 = 168

Now, marks in 3rd subject = 410 - (173 + 168) = 410 - 341 = 69.