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In this section, we will take up the Average (Basics and Based on Number System) topic and deal with all the concepts and formulae in it.

• Average: An average of the given data is the sum of all the observations in it divided by number of observations. Average is also called the Arithmetic Mean.

Average = Sum of observations/ No. of observations

Suppose we have a data {a, b, c, d} and we have to find the average of it, then-

Average = Sum of observations/ No. of observations

= (a + b + c + d)/4

Let’s take few examples to understand the concept.

• E.g. 1: If Rajesh gets 50, 55, 51, 57 and 62 marks in Physics, Chemistry, Mathematics, History and Geography respectively. Find his average marks.

Average = Sum of observations/ No. of observations

= Sum of marks/No. of subjects

= (50 + 55 + 51 + 57 + 62)/5

= 275/5

= 55 (Ans).

• E.g. 2: Raj scored 67, 69, 78 and 88 in 4 of his subjects. What should be his score in the 5th subject so that his average equals to 80?

Let the score in 5th subject be x.

∵ Average = Sum of observations/ No. of observations

⇒ 80 = (67 + 69 + 78 + 88 + x)/5

⇒ 80 × 5 = 302 + x

⇒ 400 = 302 + x

⇒ x = 400 – 302

x = 98 (Ans).

• Properties of Average:
1. If we have a given data {a, b, c, d} and a and d are the smallest and largest observation in it respectively. Then, average (A) lies between the smallest and largest observation, i.e.

a < A < d

1. If the observations of a given data are same i.e. {a, a, a, a}, then the average (A) of the data will be a.

A = a

1. If all the number gets increased or decreased by x, then the average is also increased or decreased by x.
2. If all the numbers are multiplied or divided by x, then the average is also multiplied or divided by x.
3.

These properties can be used either to verify the answers/ to quickly get to the answers in certain cases. We don’t get to see problems directly on these properties but they help us to clear our concepts.

Now we will see some specific type of problems based on Average and try to solve them with direct formula.

• If average weight of m observations is a and the average weight of n observations is b. Then –

Average weight of all the observations = {(m × a) + (n × b)} (m + n)

Let’s understand this by solving a problem.

E.g.: If the average age of 20 students in class I is 10 years and average age of 25 students in Class II is 12 years, find the average age (in years) of all the students.

Sum of age of students in Class I = 20 × 10 = 200 years

Sum of age of students in class II = 25 × 12 = 300 years.

Average = Sum of age of students in both the class/ No. of students in both the class

= (200 + 300)/(20 + 25)

= 500/45

= 100/9

= 11.111 (Ans).

This process can be summed up in a formula,

Average of all the students = {(m × a) + (n × b)} (m + n)

= {(20 × 10) + (25 × 12)} (20 + 25)

= (200 + 300)/45

= 500/45

= 100/9

= 11.111 (Ans).

• If average of n observation is a, but the average becomes b when a new observation is added to it. Then the Value of Added Observation = n(b – a) + b

Let’s take an example to see the application of this formula.

• E.g.: The average weight of 30 students in a class is 25 kgs. If the weight of the class teacher is added, the average weight increases by 2 kgs. The weight of the class teacher (in kgs) is:

Let the weight of the teachers be x kgs.

Weight of students = (No. of students × Average weight)

= 30 × 25

= 750 kgs

A/q,

(25 + 2) = (Sum of weight of students + x)/31

27 = (750 + x)/31

27 × 31 = 750 + x

837 = 750 + x

837 – 750 = x

x = 87 kgs (Ans).

By formula,

Value of added observation = n (b – a) + b

Here, n = 30, a = 25 and b = 27

Weight of teacher = 30 (27 – 25) + 27

= 30 × 2 + 27

= 60 + 27

= 87 kgs (Ans).

• Error Problems:

We get to see certain problems in which any observation is misread and that affects the average value. Error problems can be solved by the following formula-

Correct Average = {(Average × No. of observations) ± Error}/No. of observation

Where, Error = Value counted – Actual value

Note: If the error is a positive (+) value, then use negative sign in the formula and vice versa.

Let’s see a problem of this type and try to solve it.

• E.g.: The average weight of 25 bags is 55 kg. The weight of one bag was misread as 65 instead of 56. Find the correct mean value.

Total weight of the bags = 25 × 55

= 1375 kgs.

The error occurred while finding the average = 65 – 56

= +9 kgs

So the total weight was 9 kg more than the correct weight.

Hence,

Correct weight = (1375 – 9) kgs

= 1366 kgs

∴ Correct mean value = Correct weight/No. of bags

= 1366/25

= 54.64 (Ans).

By formula,

Error = Value counted – Actual value

= 65 – 56

= +9 kgs

∴ Correct Average = {(Average × No. of observations) ± Error}/No. of observation

= {(55 × 25) – 9}/25

= (1375 – 9)/25

= 1366/25

= 54.64 (Ans).

• Average Speed:

If a man/car covers a certain distance by A km/h and again covers the same distance by B km/h, then-

Average speed = 2AB/(A + B)

• E.g.: A car covers a certain distance with 60 kmph and comes back on the same path with 80 kmph. Find the average speed during the journey.

Average speed = 2AB/(A + B)

= 2 × 60 × 80/(60 + 80)

= 120 × 80/140

= 120 × 4/7

= 480/7

= 68.57 (Ans).

[BASED ON NUMBER SYSTEM]

Average topic deals with the number system as well. Number system based average problems are also frequent in some exams.

We have some important formulae related to average of numbers.

Let’s solve few examples to understand the use of these formulae.

E.g. 1: Find the average of all the odd umbers from 1 to 33.

Odd numbers from 1 to 33 are: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, and 33

Average of odd numbers from 1 to 33 = Sum of odd numbers from 1 to 33/No. of odd numbers from 1 to 33

∴ Average = (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33)/17

= 289/17

= 17 (Ans).

By formula,

Average of 1 to n odd numbers = (Last odd number + 1)/2

= (33 + 1)/2

= 34/2

= 17 (Ans).

E.g. 2: The average of first 20 multiples of 12 is?

By formula,

Average of first n multiples of any number = Number × (n + 1)/2

= 12 × (20 + 1)/2

= 6 × 21

= 126 (Ans).

• Miscellaneous:

Let’s see one miscellaneous problem from Average topic asked in the exams.

E.g.: The average of first three out of four numbers is 18. The average of last three numbers is 14. The sum of the first and last number is 16. The last number is?

Let the numbers be a, b, c, and d.

Total sum of first three numbers, a + b + c = 18 × 3 = 54            – (i)

Total sum of last three numbers, b + c + d = 14 × 3 = 42            – (ii)

Sum of first and last number, a + d = 16                                       – (iii)

Adding (i) and (ii),

a + b + c + b + c + d = 54 + 42

(a + d) + 2b + 2c = 96        [Putting value of (a + d) from (iii)]

16 + 2(b + c) = 96

2(b + c) = 96 – 16

2(b + c) = 80

(b + c) = 80/2

b + c= 40

Putting value of (b + c) in (ii),

b + c + d = 42

40 + d = 42

d = 2 (Ans).

∴ The last number = 2 (Ans).