Let the daily wage of a man be Rs. x.

Then, daily wage of a woman = Rs. (x - 5).

Now, 600x + 400 (x - 5) = 25.50 * (600 + 400) <=>  1000x = 27500 <=>  x = 27.50.

Man's daily wages = Rs. 27.50; Woman's daily wages = (x - 5) = Rs. 22.50.

 Let the ratio be k : 1. Then,

k * 16.4 + 1 * 15.4 = (k + 1) * 15.8

<=> (16.4 - 15.8) k = (15.8 - 15.4)  <=> k =  0.4/0.6 = 2/3.

Required ratio  = 2/3 : 1 = 2 : 3.

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

Clearly, we have (3 + 11 + 7 + 9 + 15 + 13 + 8 + 19 + 17 + 21 + 14 + x) / 12 = 12.

or  137 + x = 144 or x = 144 - 137 = 7.

 Since the month begins with a Sunday, so there will be five Sundays in the month,

Required average = (510 * 5 + 240 * 25) / 30 = 8550/30 = 285.

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 .