The average weight of 60 students in a class was calculated to be 40 kg. Later it was found out that the weight of one of the students was calculated as 36 kg, whereas his actual weight was 33 kg. What is the actual average weight of the students in the class?

Here initial average = 7
As we know that, if all the numbers are multiple by a certain number, then their average must be a multiple of that number.
? New average = 7 x 12 = 84

Total of 6 observation = 6 x 12 = 72
Total of 7 observation (including the 7th) = 7 x (12 - 1) = 77
? 7th observation = 77 - 72 = 5

Required number = (Sum of all numbers) - [(Sum of first 5 numbers) + (Sum of last 3 numbers)
= 50 x 9 - ( 54 x 5 + 3 x 52)
= 450 - (270 + 156) = 450 - 426 ? 24

Average of all prime numbers between 60 and 90
= (61 + 67 + 71 + 73 + 79 + 83 + 89)/7
= 523/7 = 74.7

Required average = ( Sum of scores) / (No of scores)
= (124 + 856 + 331 + 227 + 963 + 338 + 259 + 662)/8
= 3760/8 = 470

Let the average of 14th innings = x
? Average of 15th innings = x+2
Then, 14x+72 = 15(x+2)
? 14x+72 = 15x +30
? x = 42
? Average for 15th innings = x+2 = 42 + 2 = 44

Total of 5 observations = 5 x 15 = 75

New sum = 75 + 16.5 + 18 + 14.5 = 124

New Average = 124 ? 8 = 15.5

x₁ + x₂ + x₃ = 3 x 14 = 42 ....(i)
and (x₁ + x₂) x 2 = 30
? x₁ + x₂ =15 ....(ii)
From Eqs. (i) and (ii), we get
x₁ + 15 = 42
? x₁ = 42 - 15 = 27

Sum of the age of 45 students = 45 x 13 = 585 yr

Sum of the age of 15 students = 15 x18 = 270 yr

Sum of the age of all students = sum of the age of 45 students + sum of the age of 15 students
= 585 yr + 270 yr
=855 yr

? Average age of all students = 855 ? 60
= 14.25 yr

Let the number of non-officers be y.

Then number of entire staff = 16 + y

Sum of total salary = 200(16 + y)
=16 x 550 + 120y
? 120y + 550 x 16 = 200x16 + 200y
? 3y +220 = 80 + 5y
? 2y = 140
? y = 140/2 = 70