Average: Definition, Formula, Examples, Properties, Tips and Tricks to solve Average Questions
Posted on : 03-01-2019 Posted by : Admin

Average meaning and formula

Geerally speaking in layman terms, an average is a single value that represents the whole set of unequal values. An average is the sum of given observations divided by the number of observations "n". It is denoted by the letter A. An average is also known as arithmetic mean. Arithmetic mean is the central tendency of the given set of data observations. Below is the formula for average,

$Average (A) = \frac{Sum of given observations (S)}{Number of given observations (N)}$

Let us consider (x1, x2, X3, X4…. Xn) as given set of observations, now we can represent the Average as,

$Average (A) = \frac{x 1 + x 2 + x 3 + x 4 \dots xn}{n}$

Examples of Average

Calculate the average of 3, 6, 8, 9

$Average (A) = \frac{3 + 4 + 8 + 9}{4} = \frac{24}{4} = 6$

Calculate the average of 50, 40, 30

$Average (A) = \frac{50 + 40 + 30}{3} = \frac{120}{3} = 40$

Properties of Average

The following are the properties of Average

Average of the given data is less than the greatest observation and greater than the smallest observation of the given data

For example: Find the average of 4, 8, 10, 14

$Average (A) = \frac{4 + 8 + 10 + 14}{4} = \frac{36}{4} = 9$

Here, the average is 9 which is less than the greatest observation (14) and greater than the smallest observation (4)

If all the observations of the given data are equal, then the average will also be the same as the observations.

For example: Find the average of 2, 2, 2, 2, 2

$Average (A) = \frac{2 + 2 + 2 + 2 + 2}{5} = \frac{10}{5} = 2$

Here, the average is same as the observations.

Is zero is one of the observations of the given set of data, then that zero should also be included while calculating the average.

For example: Find the average of 2, 7, 0

$Average (A) = \frac{2 + 7 + 0}{3} = \frac{9}{3} = 3$

TYPE OF AVERAGE QUESTIONS

Type 1: Calculating average of given set of data

This is one of the simplest types of questions asked in the competitive exams. Here, we can directly apply the formula. In this type of questions, sum of all given values is divided by the total number of values.

$Average (A) = \frac{Sum of given values (S)}{Total number of values (N)}$

For example:

Question: Find the average of 102, 340, 98, 10, 50

Solution:

Sum of the given values (S)= 102+340+98+10+50= 600

Total number of values (N)= 5

Applying the formula we get,

$Average (A) = \frac{102 + 340 + 98 + 10 + 50}{5} = \frac{600}{5} = 120$

Type 2: Finding average of specific numbers

These types of questions are similar to type 1 questions but these questions can be solved by direct formulae. It is good to know and remember the below given direct formulae rather than adding the given set of values one by one. Applying the formula will rather take much less time and we know time management is very crucial in any competitive exam.

Average of first "n" natural numbers	$(\frac{n + 1}{2})$
Average of first "n" even numbers	$n + 1$
Average of first "n" odd numbers	$n$
Average of consecutive numbers	$\frac{First no . + Last no .}{2}$
Average of 1 to "n" odd numbers	$\frac{Last odd number + 1}{2}$
Average of 1 to "n" even numbers	$\frac{Last e v e n number + 2}{2}$
Average of squares of first "n" natural numbers	$\frac{(n + 1) (2 n + 1)}{6}$
Average of cubes of first "n" natural numbers	$\frac{n (n + 1)^{2}}{4}$
Average of "n" multiples of any numbers	$\frac{Number \times (n + 1)}{2}$

Practice test on this topic

Type 3: Addition or elimination of items

In these questions, certain items are to be ADDED and then the impact of these additions on final average value is calculated. If average of "n" observations is "a" but the average becomes "b" when one observation is added, then the value of added observations is n(b-a)+b.

In these questions, certain items are to be ELIMINATED and then the impact of these eliminations on final average value is calculated. If average of "n" observations is "a" but the average becomes "b" when one observation is eliminated, then the value of eliminated observations is n(a-b)+b.

For example:

Question 1: The average weight of 21 boys was recorded as 64 kg. If the weight of the teacher was added, the average increased by 1 kg. What was the teachers weight?

Solution:

Here we must use the formula n(b-a)+b

here, n= 21, a = 64, b= (64+1)=65 (original average plus new obseration)

upon substitution of values we get,

⇒21(65-64)+65

⇒21(1)+65

⇒21+65=86

Therefore, the weight of teacher is 86 kg.

Question 2: In a garden, the average height of 100 big trees and a Banyan tree is 3 meters. If the heigth of Banyan tree is eliminated, then the average decreases by 2 meters. Find out the heigth of Banyan tree?

Solution:

Here we must use the formula n(a-b)+b

here, n= 100, a = 3, b= (3-2)=1 (original average plus new obseration)

upon substitutio of values we get,

⇒101(3-1)+1

⇒101(2)+1

⇒203

Therefore, the height of banyan tree is 203 meters.

Type 4: Calculating average after replacement of data

In these questions, some items of the data are to be replaced. Sometimes there is a difference in the actual value and replaced value. Consider we have "n" observations out of which some observations (a1, a2, a3....) are replaced by new observations. And if the average increases or decreases by "b", then

Value of new observations= a ± nb

Question 1: The average height of 3 girls is increased by 4 cm, when one of them whose height is 126 cm is replaced by another girl. What is the height of the new girl?

Solution:

Here we must use the formula,

Value of new observations= a + nb

(we have taken + sign as the average increased)

n= 3, a = 100cm, b= 4 cm

⇒height of new girl=100+3×4=112 cm

Type 5: Calculating weighted average

In these questions, two or more different groups of data are joined to form a new group. This type of average is known as weighted average. Here it is important to note that the average of the individual group is known.

$Weighted average = \frac{Sum of data of all groups}{number of data of all the groups}$

Let us consider, Z number of groups with averages A1, A2, A3, A4, Ax and having n1, n2, n3, n4….nx elements, then the weighted average is given by,

$Weighted Average (Aw) = \frac{n 1 A 1 + n 2 A 2 + n 3 A 3 + n 4 A 4 \dots nxAx}{n 1 + n 2 + n 3 + . . . nx}$

Question: A man drives to his office at 60 km/hr and returns home along the same route at 30 km/hr. Find the average speed.

Solution:

Here we must use the formula,

$Weighted Average (Aw) = \frac{n 1 A 1 + n 2 A 2}{n 1 + n 2}$

$n_{1} = 60, n_{2} = 40, a_{1} = 55, a_{2} = 45$

⇒ $\frac{60 \times 55 + 40 \times 45}{60 + 40} = 51$

Type 6: Calculating Average Speed

Type 1: If a person covers a certain distance at a speed of A km/h and again covers the same distance at a speed of B km/h, then the average spped during the whole journey will be given by $\frac{2 AB}{A + B}$

Question: In an examination, a batch of 60 students made an average score of 55 and another batch of 40 made it only 45. What is the overall average score?

Solution:

According to the question,

A=60, B=30

we know the formula,

Average speed= $\frac{2 AB}{A+B}$

The average speed

$= \frac{2 \times 60 \times 30}{60 + 30} = \frac{3600}{90} = 40 km / hr$

Type 2: If a person covers three equal distances at a speed of A km/h, B km/h and C km/h, then the average speed during the whole journey will be given by $\frac{3 AB C}{AB + BC + CA}$

Question: A man covers $\frac{1}{3}$ rd of his journey by train at 60 km/hr, next $\frac{1}{3}$ by bus at 30 km/hr and the rest by cycle at 10 km/hr. Find his average speed during whole journey?

Solution:

According to the question,

A=60, B=30, C=10

we know the formula,

Average speed during the whole journey = $\frac{3 ABC}{AB+BC+CD}$

Average speed the whole journey

$= \frac{3 \times 60 \times 30 \times 10}{60 \times 30 + 60 \times 10 + 30 \times 10} = \frac{54000}{2700} = 20 km / hr$

Type 3: If dsitance P is covered with speed x, distance Q is covered with speed y and distance R is covered with speed z, then for the whole journey, then the average speed is given by,

\frac{P + Q + R + . . .}{\frac{P}{x} + \frac{Q}{y} + \frac{R}{z} + . . .}

Question: If andrew travels in his car and covers $\frac{1}{4}$ part of his journey with 8 km/hr, $\frac{3}{5}$ part with 6 km/hr and remianing $\frac{3}{20}$ part with a speed of 10 km/hr. Find out his average speed during the whole journey??

Solution:

According to the average speed formula,

P= $\frac{1}{4}$ , Q= $\frac{3}{5}$ , R= $\frac{3}{20}$

X= 8 km/hr, Y= 6km/hr, Z=10 km/hr

∴ Required average speed

$= \frac{1}{\frac{P}{X} + \frac{Q}{Y} + \frac{R}{Z}} = \frac{1}{\frac{1}{4 \times 8} + \frac{3}{5 \times 6} + \frac{3}{20 \times 10}} = \frac{1}{\frac{1}{32} + \frac{1}{10} + \frac{3}{200}} = \frac{1}{\frac{25 + 80 + 12}{800}} = \frac{800}{117} = 6.83 km / hr$

Practice test on this topic

Average properties, definition of average, Average in mathematics, Average Formula, Average examples, Average questions, tips and tricks to solve aver

Average: Definition, Formula, Examples, Properties, Tips and Tricks to solve Average Questions Posted on : 03-01-2019 Posted by : Admin

Average meaning and formula

Examples of Average

Properties of Average

TYPE OF AVERAGE QUESTIONS

Type 1: Calculating average of given set of data

Type 2: Finding average of specific numbers

Practice test on this topic

Type 3: Addition or elimination of items

Type 4: Calculating average after replacement of data

Type 5: Calculating weighted average

Type 6: Calculating Average Speed

Login to post your comment here...

- or with social Account -

Average: Definition, Formula, Examples, Properties, Tips and Tricks to solve Average Questions
Posted on : 03-01-2019 Posted by : Admin