AVERAGES

1. Next Class: AVERAGE

3. Short-Cuts
Average of first n natural no. = n+1/2
Average of square of first n natural no. = (n+1)(2n+1)/6
Average of first n odd no. = n
Average of first n even no. = n+1
Average of consecutive no. = (first term + last term)/2

4. I Formula
Numbers Sum Average
(Sum/n)
First n numbers n(n+1)/2 (n+1)/2
First n odd numbers n^2 n
First n even numbers n(n+1) (n+1)
First n square
numbers
n(n+1)(2n+1)/6 (n+1)(2n+1)/6
Consecutive numbers n(first term + last term)/2 (first term + last term)/2

5. Short-Cuts
The average of consecutive numbers is always
the middle number.

6. Find the average of following number:
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21
A. 9
B. 11
C. 12
D. 13
Question

7. Average of Consecutive Numbers = Middle Number
Given numbers
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21
are 11 consecutive odd numbers.
So Average is middle no
i.e. 6th number
And 6th Number is 11.
Solution

8. The average of 20 numbers is zero. Of them, at the
most, how many may be greater than zero?
A. 0
B. 1
C. 10
D. 19
Question

9. Answer: Option D
Explanation:
Average of 20 numbers = 0.
Sum of 20 numbers (0 x 20) = 0.
It is quite possible that 19 of these numbers may be
positive and if their sum is a then 20th number is (-a).
Solution

10. The average weight of 8 person's increases by 2.5 kg
when a new person comes in place of one of them
weighing 65 kg. What might be the weight of the new
person?
A. 76 kg
B. 76.5 kg
C. 85 kg
D. Data inadequate
E. None of these
Question

11. Answer: Option C
Explanation:
Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.
Or
Factor = Difference / Total
2.5 = Difference / 8
Difference = 20 Kg since word increase was mentioned in
statement so weight of new person= 65+20 =85
Solution

12. A library has an average of 510 visitors on Sundays and
240 on other days. The average number of visitors per
day in a month of 30 days beginning with a Sunday is:
A. 250
B. 276
C. 280
D. 285
Question

13. Answer: Option D
Explanation:
Since the month begins with a Sunday, so there
will be five Sundays in the month.(i.e on
1,8,15,22,29)
Required average =
(510 x 5 + 240 x 25) / 30
= 8550 / 30 = 285
Solution

14. A pupil's marks were wrongly entered as 83
instead of 63. Due to that the average marks for
the class got increased by half (1/2). The
number of pupils in the class is:
A. 10
B. 20
C. 40
D. 73
Question

15. Answer: Option C
Explanation:
Let there be x pupils in the class.
I Total increase in marks = 83 – 63 = 20
II Total increase in marks are x * ½
So, x * ½ = 20 and x = 40.
Or
Factor=Difference / Total
½=20/ T
Solution

16. The captain of a cricket team of 11 members is 26 years
old and the wicket keeper is 3 years older. If the ages of
these two are excluded, the average age of the
remaining players is one year less than the average age
of the whole team. What is the average age of the
team?
A. 23 years
B. 24 years
C. 25 years
D. None of these
Practice

17. Answer: Option A
Explanation:
Let the average age of the whole team by x years.
11x - (26 + 29) = 9(x -1)
11x - 9x = 46
2x = 46
x = 23.
So, average age of the team is 23 years.
Solution

18. In the first 10 overs of a cricket game, the run
rate was only 3.2. What should be the run rate
in the remaining 40 overs to reach the target of
282 runs?
A. 6.25
B. 6.5
C. 6.75
D. 7
Practice

19. Answer: Option A
Explanation:
Required run rate = (282 - (3.2 x 10))/40
= 250 / 40
= 6.25
Solution

20. The average of five consecutive odd numbers is
61. What is difference between the highest and
lowest number?
A. 6
B. 7
C. 8
D. 9
Practice

21. Answer: Option C
Solution

22. Example: There are 36 students in class A whose average is 30kg and 24 students
in class B whose average is 40kg. What will be the average if the classes are
combined?
General equation
Avg36= 30, Avg24 = 40
Total weight of class A = 30 x 36
Total weight of class B = 40 x 24
Overall average = (Total weight of class A) + (Total weight of class B) / Number of
students in class A + Number of students in class B
= (30 x 36) + (40 x 24) / 36 + 24
= 34

23. BALANCE METHOD:
The principle of this concept is that the weight in
the balance is inversely proportional to the distance of the pivot.
Where,
Wa – Class A weight Wb – Class B weight
Avga – Class A average Avgb – Class B average
Avgc – combined average
Wa Wb
Avga Avgb
Avgc

24. There are 36 students in class A whose average is 30kg and 24 students in
class B whose average is 40kg. What will be the average if the classes are
combined?
Step 1: Find the ratio of the weights of A and B
= 36:24
= 3:2
Step 2: Inverse the weights to get the distance ratio
= 2:3
36
36
24
24
3:2
3:2
2:3
30
40
30
40

25. Step 3: Split the distance between the averages in the ratio 2:3
Here the distance from 30 to 40 is 10. So 10 should be split in the ratio 2:3
as 4 and 6.
The combined average is (30+4) or (40-6) = 34
30
36
40
24
3:2
2:3
34
4 6

26. . In class A there are 63 students whose average is 32,
and in class B there are 21 students whose average is 44
, then find the overall average?
A)33
B)35
C)36
D)38

27. Example: Average of 5 students marks is 30. If one student having 90 mark is added to the
team then what will be the new average ?
3.i General equation
Avg5 = 30
No of students = 5
Sum5 = 30x5
= 150
Sum6 = 150+90
= 240
Avg6 = 240/6
= 40
Or F=D/T
F=60/6=10 so new average=10+ 30=40

28. . Average of 4 students marks is 50 and one student
having marks has 200, is added to the team , what is new
average ?
A)70
B)80
C)90
D)None

29. Example 8. Average of 6 students marks is 60 , what is
the new average if a student of marks 110 is taken out ?
A50
B)55
C)60
D)None

30. .A batsman having average 40 makes 90 runs in his last
inning thereby his average increases by 2. Find the
number of matches he has played.
A)10
B)50
C)25
D)24

31. The average weight of 8 men having average weight 40
kg is increased by 2 kg when a new man is included.
The weight of the new man is
A) 56
B) 58
C) 96
D) 98