The average age of 45 students in a class in 13 yr, 15 new students with an average age of 18 yr join the class. What is the average age of all the students in the class?

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

Total of 5 observations = 5 x 15 = 75

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

New Average = 124 ? 8 = 15.5

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 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

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

Total decrease in the age = 25 x 1/2 =12.5 yr

? Age of new girl = 20 - 12.5
= 7.5 yr

Let number of males = y
Then, total number of family members = 15 + y

Sum of age of family members = sum of age of males + sum of age of females
25(15 + y) = 30y + 15x20
? 375 + 25y = 30y + 300
? 5y = 75
? y = 15

Total required marks in the three examinations = (3 x 600 x 70)/100 =1260

Marks secured in first examination = 55% of 600 = ( 600 x 55)/100 =330

Marks secured in second examination = 60% of 600 = ( 600 x 60)/100 =360

Thus, the required marks in third examination = 1260 - (330 + 360)
= 1260 - 690
= 570