Average- Aptitude Questions and Answers

Total number of all students of the class = 30 x 45 = 1350

After adjustment effective increment in marks=45–15=30

New total marks=1350+30=1380

Thus, New average after adjustment=1380/30=46

Let the required number of non officers = x
Then, 110*x + 460 * 15 = 120 * ( x + 15 )
or, 120x - 110x = 460 *15 - 120 * 15
10x = 15 *( 460 - 120 )
10x = 15 *340
x = 15*34
So the number of non ­officers in the office is, x = 510

Average = Sum of quantities/ Number of quantities
Let the remaining two numbers are 2x and 3x
Then ,Sum of remaining two numbers = Sum of 11 numbers - Sum of 9 numbers
2x+3x = 11*36 - 9*34
5x = 396 - 306
5x= 90
x=18

Therefore,the numbers are 2x=>36 and 3x=>54
the value of the smallest number=>36

Present average age of family = 25 year
3 year ago average of family = 25 - 3 = 22 years

Let the Number of executives be 2X and Number of non-executives be X
As is given, 2X+X=60
X=20
Therefore, Number of executives, 2X = 40 , and number of non-executives employees, X = 20
Now,
Total salary = 40 * salary of executives + 20 * salary of non-executives
60 * 12000 = 40 * k + 20 * l, here k, l are unknowns
so we con't determine the required average salary

Age of friend left = Previous average of the age - Number of persons present x Increase in average age
= 15 - ( 10-1 ) x ( 16-15 )
= 15- 9 x 1
= 15 - 9 = 6yr

Age of the teacher = Sum of age of 24 students and teacher - Sum of age of 24 students.
= 25 x 11 - 24 x 10
= 35 years.
Thus, Age of teacher = 35 years

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

Required average =7*(1+2+3+4+5+6) / 6
=7*(6*(6+1)/2) /6
=7*42 / 6*2
=24.5

Let x1, x2, x3, ......., x8, x9 be the numbers

x1+ x2+ x3+ x4+ x5+ x6+ x7+ x8+ x9= 18*9
x1+ x2+ x3+ x4+ x5+ x6+ x7+ x8+ x9= 162 .... (1)

x1+ x2+ x3+ x4+ x5= 19*5
x1+ x2+ x3+ x4+ x5= 95 ....(2)

x5+ x6+ x7+ x8+ x9= 17*5
x5+ x6+ x7+ x8+ x9= 85 ....(3)

Adding equations (2) and (3) and then subtracting equation (1) from it, we get
x5= 18

Sum of the ages of the four men five year ago= (48 *4) year = 192 year
Present sum of the ages of the four men= 192 + (5* 4) year = 212 year

Now, a new man joins and the average age increases by 2 years = 48+2 =50

So, 50 =(212+ Age of the new man) / 5
Age of the new man = (50* 5)- 212 year = 38 year

Let the number of males be x and the number of females be y.
Sum of the salaries of the men = 20800x
Sum of the salaries of the females = 16800y
Now, 20000 = 20800x + 16800y / (x + y)
20000 (x + y) = 20800x + 16800y
800x = 3200y
x/y =4/1
x : y = 4 : 1

Let the number be a and b. Then Average is, (a +b) / 2 = 14.5
a + b = 29

as given square root of their product is √ab = 10
ab = 100

(a - b)² = (a + b)² - 4ab
(29)² -(4 *100)= 441
(a - b) = 21
On solving a + b = 29 and a - b = 21, we get a =25 and b = 4.

Let the ages of the four men be 2x, 3x, 4x and 5x year respectively.
As given, Sum of ages four men (2x + 3x + 4x + 5x) = 42 * 4
14x = 42 *4
x =168/14 = 12
Hence, the age of the eldest person is, 5x = (5 *12) years = 60 years

Sum of (1 + 2 + 3 +…. + n) =n(n+1)/2

Mean = [n(n+1)/2] / n => (n+1)/2

As given, (n+1)/2 = 13
n+1 =26
n=25

The median of a series of numbers is the number that appears in the middle .
arranged series from smallest to largest => 28,34,38,40,42,48,50

Median =40

Note:If the number of members of the series was even, the average of the two middle numbers would be the median

Ordering the data from least to greatest, we get:
15, 20, 35, 45, 65, 70, 80, 95

highest - lowest = 95 - 15 = 80
The range is 80