AVERAGE
The average or mean or arithmetic mean of a Therefore, x = 2y = 3z.
number of quantities of the same kind is equal to 𝑦 = 𝑥̅ and 𝑧 = 𝑥̅
their sum divided by the number of those 2 3
𝑥̅ 𝑥̅
𝑥̅+ +
quantities Now, 2 3 = 44
3
11𝑥̅
Arithmetic average is used for all or = 44 or 𝑥̅ = 72
18
averages like: average income, average
profit, average age, average marks etc
Example 3:
It is defined as the sum total of all
The average of five consecutive odd numbers is
volumes of items divided by the total
61. What is the difference between the highest
number of items
and lowest numbers?
In individual series.
𝑠𝑢𝑚𝑜𝑓𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛 Solution:
Average = Let the numbers be x, x + 2, x+ 4, x + 6 and x +
𝑁𝑢𝑚𝑏𝑒𝑟𝑜𝑓𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
𝑥̅1 +𝑥̅2+𝑥̅3+⋯+𝑥̅𝑛 8
or 𝑥̅ =
𝑛 𝑥̅+(𝑥̅+2)+(𝑥̅+4)+(𝑥̅+6)+(𝑥̅+8)
To calculate the sum of observations, Then, = 61
5
they should be in the same unit or 5x + 20 = 305 or x = 57.
So, required difference = (57 + 8) – 57 = 8.
Average of a group consisting two
Example 1: different groups when their averages are
known:
A man purchased 5 toys at the rate of Rs 200 (a) Let Group A contains m quantities
each, 6 toys at the rate of Rs 250 each and 9 toys and their average is A and group B
at the rate of Rs 300 each. contains n quantities and their average is
b, then average of group C containing a
Calculate the average cost of one toy.
+ b quantities
Solutions: 𝑚𝑎 + 𝑚𝑏
Price of 5 toys = 200 x 5 = 1000
=
𝑚+𝑛
Price of 6 toys = 250 x 6 = 1500 Example 4:
Price of 9 toys = 300 x 9 = 2700 There are 30 student in a class. The average age
Aveage cost of one toy of the first 10 students is 12.5 years. The average
1000 +1500 +2700 5200
= = = 𝑅𝑠260\− age of the next 20 students is 13.1 years. The
20 20
average age of the whole class is:
Example 2: Solution:
In three numbers, the first is twice the second Total age of 10 students = 12.5x10 = 125 years
and thrice the third, if the average of these three Total age of 20 students = 13.1x20 = 262 years
numbers is 44, then the first number is: Average age of 30 students = 125+262 = 12.9
30
Solution: years
Let the three numbers be x, y and z
1
, Example 5: Solution:
The average age of students of a class is 15.8 Here, m = 50, n =5, a = 850, b = 2500
𝑚𝑎 −𝑛𝑏
years. The average age of boys in the class is ∴ Average salary of remaining staff =
𝑚−𝑛
16.4 years and that of the girl is 15.4 years. The 50 × 850 − 5 × 2500
ration of the number of boys to the number of =
50 − 5
girls in the class is 42500 − 12500
Solution: =
45
Let the number of boys in a class be x. = 667 (approx)
Let the number of girls in a class be y.
WEIGHTED AVERAGE
∴ Sum of the ages of the boys = 16.4 x
If we have two or more groups of members
Sum of the ages of the girls = 15.4 y
whose individual averages are know, then
∴ 15.8(𝑥̅ + 𝑦) = 16.4𝑥̅ + 15.4𝑦
𝑥̅ 2 combined average of all the members of all the
⇒ 0.6𝑥̅ = 0.4𝑦 ⇒ = groups is known as weighted average. Thus if
𝑦 3
there are k groups having member of number n1,
∴ Required ratio = 2: 3
n2, n3… nk with averages A1, A2, A3, ……. Ak
Shortcut method:
respectively then weighted average.
𝑛1𝐴1 + 𝑛2𝐴2 + 𝑛3𝐴3 + ⋯ + 𝑛𝑘 𝐴𝑘
𝐴𝑤 =
15.4 𝑛1 + 𝑛 2 + 𝑛3 + ⋯ + 𝑛𝑘
16.4
15.8 Example 7:
The average monthly expenditure of a family
16.4-
15.8- 15.8= was Rs 2200 during the first 3 months; Rs 2250
15.4= 0.6 during the next 4 months and Rs 3120 during the
0.4
last 5 months of a year. If the total saving during
the year were Rs 1260, then the average monthly
income was
∵ required ratio = 0.4 = 2 Solution:
0.6 3
(b) It average of m quantities is a and the Total annual income
average of n quantities out of them is b then the = 3 × 2200 + 4 × 2250 + 5 × 3120 + 1260
average of remaining group rest of quantities is = 6600 + 9000 + 15600 + 1260 = 32460
𝑚𝑎 −𝑛𝑏 ∴ Average monthly income
𝑚−𝑛 32460
= = 𝑅𝑠 2705
12
Example 6: If X is the average 𝑥̅1 + 𝑥̅2 + 𝑥̅3 + ⋯ +
Average salary of all the 50 employees 𝑥̅𝑛 then
increasing 5 officers of a company is Rs 850. If (a) The average of 𝑥̅1 + 𝑎, 𝑥̅2 + 𝑎, 𝑥̅3 +
the average salary of the officers is Rs 2500. 𝑎, … 𝑥̅𝑛 + 𝑎 is 𝑋 + 𝑎
Find the average salary of the remaining staff of (b) The average of 𝑥̅1 − 𝑎, 𝑥̅2 − 𝑎, 𝑥̅3 −
the company. 𝑎, … 𝑥̅𝑛 − 𝑎 is 𝑋 − 𝑎
(a) 560 (b) 660 (c) The average of 𝑎𝑥̅1, 𝑎𝑥̅2, … 𝑎𝑥̅𝑛 is
(c) 667 (d) 670 aX, provide 𝑎 ≠ 0
2
The average or mean or arithmetic mean of a Therefore, x = 2y = 3z.
number of quantities of the same kind is equal to 𝑦 = 𝑥̅ and 𝑧 = 𝑥̅
their sum divided by the number of those 2 3
𝑥̅ 𝑥̅
𝑥̅+ +
quantities Now, 2 3 = 44
3
11𝑥̅
Arithmetic average is used for all or = 44 or 𝑥̅ = 72
18
averages like: average income, average
profit, average age, average marks etc
Example 3:
It is defined as the sum total of all
The average of five consecutive odd numbers is
volumes of items divided by the total
61. What is the difference between the highest
number of items
and lowest numbers?
In individual series.
𝑠𝑢𝑚𝑜𝑓𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛 Solution:
Average = Let the numbers be x, x + 2, x+ 4, x + 6 and x +
𝑁𝑢𝑚𝑏𝑒𝑟𝑜𝑓𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
𝑥̅1 +𝑥̅2+𝑥̅3+⋯+𝑥̅𝑛 8
or 𝑥̅ =
𝑛 𝑥̅+(𝑥̅+2)+(𝑥̅+4)+(𝑥̅+6)+(𝑥̅+8)
To calculate the sum of observations, Then, = 61
5
they should be in the same unit or 5x + 20 = 305 or x = 57.
So, required difference = (57 + 8) – 57 = 8.
Average of a group consisting two
Example 1: different groups when their averages are
known:
A man purchased 5 toys at the rate of Rs 200 (a) Let Group A contains m quantities
each, 6 toys at the rate of Rs 250 each and 9 toys and their average is A and group B
at the rate of Rs 300 each. contains n quantities and their average is
b, then average of group C containing a
Calculate the average cost of one toy.
+ b quantities
Solutions: 𝑚𝑎 + 𝑚𝑏
Price of 5 toys = 200 x 5 = 1000
=
𝑚+𝑛
Price of 6 toys = 250 x 6 = 1500 Example 4:
Price of 9 toys = 300 x 9 = 2700 There are 30 student in a class. The average age
Aveage cost of one toy of the first 10 students is 12.5 years. The average
1000 +1500 +2700 5200
= = = 𝑅𝑠260\− age of the next 20 students is 13.1 years. The
20 20
average age of the whole class is:
Example 2: Solution:
In three numbers, the first is twice the second Total age of 10 students = 12.5x10 = 125 years
and thrice the third, if the average of these three Total age of 20 students = 13.1x20 = 262 years
numbers is 44, then the first number is: Average age of 30 students = 125+262 = 12.9
30
Solution: years
Let the three numbers be x, y and z
1
, Example 5: Solution:
The average age of students of a class is 15.8 Here, m = 50, n =5, a = 850, b = 2500
𝑚𝑎 −𝑛𝑏
years. The average age of boys in the class is ∴ Average salary of remaining staff =
𝑚−𝑛
16.4 years and that of the girl is 15.4 years. The 50 × 850 − 5 × 2500
ration of the number of boys to the number of =
50 − 5
girls in the class is 42500 − 12500
Solution: =
45
Let the number of boys in a class be x. = 667 (approx)
Let the number of girls in a class be y.
WEIGHTED AVERAGE
∴ Sum of the ages of the boys = 16.4 x
If we have two or more groups of members
Sum of the ages of the girls = 15.4 y
whose individual averages are know, then
∴ 15.8(𝑥̅ + 𝑦) = 16.4𝑥̅ + 15.4𝑦
𝑥̅ 2 combined average of all the members of all the
⇒ 0.6𝑥̅ = 0.4𝑦 ⇒ = groups is known as weighted average. Thus if
𝑦 3
there are k groups having member of number n1,
∴ Required ratio = 2: 3
n2, n3… nk with averages A1, A2, A3, ……. Ak
Shortcut method:
respectively then weighted average.
𝑛1𝐴1 + 𝑛2𝐴2 + 𝑛3𝐴3 + ⋯ + 𝑛𝑘 𝐴𝑘
𝐴𝑤 =
15.4 𝑛1 + 𝑛 2 + 𝑛3 + ⋯ + 𝑛𝑘
16.4
15.8 Example 7:
The average monthly expenditure of a family
16.4-
15.8- 15.8= was Rs 2200 during the first 3 months; Rs 2250
15.4= 0.6 during the next 4 months and Rs 3120 during the
0.4
last 5 months of a year. If the total saving during
the year were Rs 1260, then the average monthly
income was
∵ required ratio = 0.4 = 2 Solution:
0.6 3
(b) It average of m quantities is a and the Total annual income
average of n quantities out of them is b then the = 3 × 2200 + 4 × 2250 + 5 × 3120 + 1260
average of remaining group rest of quantities is = 6600 + 9000 + 15600 + 1260 = 32460
𝑚𝑎 −𝑛𝑏 ∴ Average monthly income
𝑚−𝑛 32460
= = 𝑅𝑠 2705
12
Example 6: If X is the average 𝑥̅1 + 𝑥̅2 + 𝑥̅3 + ⋯ +
Average salary of all the 50 employees 𝑥̅𝑛 then
increasing 5 officers of a company is Rs 850. If (a) The average of 𝑥̅1 + 𝑎, 𝑥̅2 + 𝑎, 𝑥̅3 +
the average salary of the officers is Rs 2500. 𝑎, … 𝑥̅𝑛 + 𝑎 is 𝑋 + 𝑎
Find the average salary of the remaining staff of (b) The average of 𝑥̅1 − 𝑎, 𝑥̅2 − 𝑎, 𝑥̅3 −
the company. 𝑎, … 𝑥̅𝑛 − 𝑎 is 𝑋 − 𝑎
(a) 560 (b) 660 (c) The average of 𝑎𝑥̅1, 𝑎𝑥̅2, … 𝑎𝑥̅𝑛 is
(c) 667 (d) 670 aX, provide 𝑎 ≠ 0
2
