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Next Class: AVERAGE
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
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
Short-Cuts
The average of consecutive numbers is always
the middle number.
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
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
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
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
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
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
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
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
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
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
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
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
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
Answer: Option A
Explanation:
Required run rate = (282 - (3.2 x 10))/40
= 250 / 40
= 6.25
Solution
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
Answer: Option C
Solution
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
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
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
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
. 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
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
. 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
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
.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
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
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Average.pdf

  • 2.
  • 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
  • 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