The average salary of the entire staff in an office is ? 200 per day. The average salary of officers is ? 550 and that of non officers is ? 120. If the number of officers is 16, then find the number of non officers in the office.

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

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

( 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

Correct mean = ( sum of total marks - 57 + 75 ) / 30
= ( 30 x 58.5 - 57 + 75 ) / 30
= (1755 + 18 ) / 30
= 59.1