A batsman has a certain average of runs for 14th innings. He scores 72 runs in the 15th innings, thus his average increased by 2. Find the average after 15th innings.

Total correct weight = (40 x 60) - 36 + 33 = 2400 - 3 = 2397 kg
? Required average weight = 2397/60 = 39.95 kg

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

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

( A + B + C ) / 3 = 11111 ....(i)

Also, ( A+C ) / 2 = 11111 ....(ii)

From Eqs. (i) & (ii), we get
B = ? 11111 crore