Consider the following frequency distribution:

x     f

8     6

5     4

6     5

10    8

9     9

4     6

7     4

What is the median for the distribution?

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.

Clearly, we  have : x = (3y + 3z ) / 6 or 2x = y + z.

Average = total runs / no.of innings = 32 

So, total = Average x no.of innings = 32 x 10 = 320.

Now increase in avg = 4runs. So, new avg = 32+4 = 36runs 

Total runs = new avg x new no. of innings = 36 x 11 = 396 

Runs made in the 11th inning = 396 - 320 = 76

Manager's monthly salary Rs. (1600 * 21 - 1500 * 20) = Rs. 3600.

i) Let the ages of the five members at present be a, b, c, d & e years. 

And the age of the new member be f years. 

ii) So the new average of five members' age = (a + b + c + d + f)/5 ------- (1) 

iii) Their corresponding ages 3 years ago = (a-3), (b-3), (c-3), (d-3) & (e-3) years 

So their average age 3 years ago = (a + b + c + d + e - 15)/5 = x ----- (2) 

==> a + b + c + d + e = 5x + 15 

==> a + b + c + d = 5x + 15 - e ------ (3) 

iv) Substituting this value of a + b + c + d = 5x + 15 - e in (1) above, 

The new average is: (5x + 15 - e + f)/5 

Equating this to the average age of x years, 3 yrs, ago as in (2) above, 

(5x + 15 - e + f)/5 = x 

==> (5x + 15 - e + f) = 5x 

Solving e - f = 15 years. 

Thus the difference of ages between replaced and new member = 15 years.

Weight of the teacher = (35.4 x 25 - 35 x 24) kg = 45 kg.

Age of the 15th student = [15 * 15 - (14 * 5 + 16 * 9)] = (225-214) = 11 years.

Let the numbers be x, x + 2, x + 4, x + 6 and x + 8.

Then  [x + (x + 2) + (x + 4) + (x + 6) + (x + 8) ] / 5 = 61.

or 5x + 20 = 305 or x = 57.

So, required difference = (57 + 8) - 57  = 8