How to solve Averages Questions Quickly

Today we will discuss first basic Arithmetic Mathematics topic which is very important according to our Govt. exams perspective.

Basic Definition

• Average means a number expressing the central value in a set of data which is calculated by the sum of values in the set by their numbers. 
• The main term of average is equal sharing of values among all, where it may share persons or things. 

Formula

We obtain the average of a number using the formula that is:
Average = (Sum of observations / Number of observations)

Examples:

Question 1

Find the average of numbers? 23, 25, 65, 12, 50
Solution:
According to the formula:
Average = (23+25+65+12+50)/5
= 175/5 = 35
Hence 35 is the average of these numbers
But this is the basic Average formula which helps us in our School days a lot to clear the exams but now we are in Competitive exams so we want to quick and precise. For these we need our basic and also we have to learn something new.

TYPE 1:

AVERAGE OF NUMBERS HAVING DIFF IN ARITHMETIC:
There are two cases:

• When a number of terms is odd then the average will be middle term. 
• When a number of terms is even then the average will be the average of two middle terms. 

Examples:

Question 2
Find the average of  12, 13, 14, 15, 16

Solution:

• Here the difference is in AP and number of terms is odd. So the average will be the middle number. 
• There is another way to calculate the average for numbers having diff in Arithmetic 
• Average can be calculated as ½ (first term + last term) 

Examples:

Question 3
Find the average of : 12, 13, 14, 15, 16, 17

Solution:

Here the number of terms is even and diff is in arithmetic So the average will be the average of middle two terms.
That is, 14 and 15
Average = (14 + 15)/2 = 14.5
ALTERNATIVE:
Average = ½ (first term + last term)
= ½ (14 + 15) = 14.5

TYPE 2:

If the average of some numbers is M and another number x is added, subtracted, multiplied or divided from each number then then average will be added, subtracted, multiplied or divided respectively.

• If M is average and x is added to each number, then new average is “M + x” 
• If M is average and x is multiplied to each number, then new average is “M * x” 
• If M is average and x is subtracted to each number, then new average is “M - x” 
• If M is average and x is divided to each number, then new average is “M / x” 

TYPE 3

If the average of n numbers is “M” and one number is included, then two cases arise:

If the included number ”X” is greater than average then average is subtracted from the included number and results will be divided by “n+1” 

Extra value = X – MDistributed value = (X – M) / (n+1)
New average = M + Distributed value

If the included number ”X” is smaller than average then the included number is subtracted from the average and resultant is divided by the “n+1”

Extra value = M – XAdjusted value = (M – X) / (n+1)
New average = M - Adjusted value

TYPE 4

If the average of n numbers is “M” and one number is excluded, then two cases arise: 

If the excluded number ”X” is greater than average then average is subtracted from the excluded number and resultant will be divided by “n-1” 

Extra value = X – MAdjusted value = (X – M) / (n-1)
New average = M - Adjusted value

If the excluded number ”X” is smaller than average then the excluded number is subtracted from the average and resultant is divided by the “n-1”

Extra value = M – XAdjusted value = (M – X) / (n-1)
New average = M + Adjusted value

TYPE 5:

If the average of n numbers is “M” and one number is excluded and another number is included then two cases arise: 

If the excluded number ”X” is greater than included number “Y” then included number is subtracted from the excluded number and resultant will be divided by “n”

Extra value = X – YAdjusted value = (X – Y) / (n)
New average = M - Adjusted value

If the excluded number ”X” is smaller than included number then the excluded number is subtracted from the included number and resultant is divided by the “n” 

Extra value = Y – XAdjusted value = (Y – X) / (n)
New average = M + Adjusted value

EXAMPLES:

Question 4
The average age of 15 students in class is 42 and the age of teacher which is 58 is also included with them. Find the new average age of teacher and students.

Solution:

So the included age of teacher is greater than the average age of students.
Extra value = 58 – 42 = 16
Adjusted value = 16/16 = 1
New average = 42 + 1 = 43
Hence required new average is 43.

Question 5

An average marks o a student in 7 subjects is 94 and the mark of English which is 70 is excluded. Find the new average marks of a student.

Solutions:

So the excluded mark is smaller than the average marks of a student.
Extra value = 94 – 70 = 24
Adjusted value = 24/6 = 4
New average = 94 – 4 = 90
Hence the new average marks of 6 subjects of student are 90.

Question 6

The average weight of 10 persons is 65 kg. But one person weighing 75 kg left and another person weighing 65 kg joined the group. Find the new average weight.

Solutions:

So the weight of leaving person is greater than the weight of joining person. So the average will also decrease.

Extra value = 75 – 65 = 10
Adjusted value = 10/10 = 1
New average = 65 – 1 = 64
Hence the new required average weight of 10 persons is 64.

Question 7

The average age of 10 students is 25. But the age of teacher is also included and the new average age is increased to 29. Find the age of teacher.

Solutions:

We can also find any missing value through this process.
Acc. to the ques.
New average age is greater than the previous average age. This implies that the age of teacher must be greater than the average. We have to find how much.
Extra value = 29 – 25 = 4
Adjusted value = 4 * 11 = 44
Age of teacher = 25 + 44 = 69
Hence the required age of teacher is 69.

Some More Examples

Example 1

The average age of the family of 6 members is 20 years. If the age of the youngest member is 6 years. Find the average age of the family at the birth of youngest member?

Solution:

During the birth of the youngest child, there will be only 5 members of the family. According to Type 2 if a particular number is subtracted from every number then the number is subtracted from the average.
So the average of 6 members = 20 – 6 = 14
But there will be only 5 members during birth.
So, Required average = Total age/ no. of members
=`\frac{14\times 6}{5} = {84}/{5}` = 16.8
Hence the average age is 16.8 years

Example 2. 

The average score of batsmen in 10 innings is 21.5 runs.
How many runs a batsmen must score in his next innings So that his average score will jump to 25 runs?
Solution:
This  ques. Is based on TYPE 3 where one innings is to be included and with that average will increase.

Increased average= 25 – 21.5 = 3.5
This increase will be due to more runs scored then the average score.
Hence the 11th inning score = 21.5 + 3.5`\times `11 = 60

Try out some Examples

1. A batsman in his 16th inning scores 80 runs and thereby increases the average by 4 runs. Find the average after the 16th inning. (HINT: Firstly find average of 16 innings)
2. The average age of 30 boys in the class is equal to 15 years. If the age of teacher is also included the average becomes 16 years. Find the age of teacher.
3. The average of 45 observations is 35. Later it was found that one observation which is 56 is wrongly taken as 24. Find the actual average of 45 observations.

• Till now we have considered only one person either included or excluded or both. But there can be a possibility of more than one person.

Example 3.

The average age of 8 persons is increased by 3 years when 2 men whose age is 40 years and 50 years are substituted by two other men. Find the average age of two new men.
Solution:
In this case, there are two persons including and excluding. So we have to consider both simultaneously.
Total age of excluding men= 40 + 50 = 90 years
Increased age = 3 `\times ` 8 = 24 years
This increased age is because the of including men is more than the exclexcluded.
Average age of new men =` \frac{(90 + 24)}{2}` = 57
In this case we cannot calculate individual value of each included men.

Some of the problems of average can directly be solved through formulas without using any trick

• 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 + LAST  NUMBER)}{2}`
• Average of the square of first n natural no. = `\frac{[(n+1)(2n+1)]}{6}`
• Average of a cube of first n natural no. = `\frac{[n(n+1)^{  2}]}{4}`
• If the average of m numbers is a and the average of n numbers (m>n) is b then the average of remaining numbers will be: `\frac{(ma-nb)}{(m-n)}`

Try out some Examples:

• Find the average of first 100 natural numbers.
• Find the average of first 100 even numbers.
• Find the average of first 100 odd numbers
• Find the average of all natural numbers between 101 and 200.
• Find the average of square of first 20 natural numbers
• Find the average of cube of first 20 natural numbers 
• The average of 25 numbers is 45 and the average of first 15 numbers is 35. Find the average of remaining numbers.

Some More Tricky Questions:

Example 1.

The average score of 45 exams is 470. Out of which average score of first 23 exams is 490 and the average score of last 23 exams is 445. Find the score of the 23rd exam?
Solution:
Here 23rd exam is included in both first and last averages. So firstly we will find the change in average of both from overall average and add both the differences
Then we will multiply the added difference with 23 as we have to find the 23rd exam score.
And finally, we will add the result to the overall average. That will provide us with the required value.
ExamAverage
\begin{matrix} & Exam & Average \\ Total & 45 & 470 \\ First & 23 & 490 & +20 \\  Last & 23 & 445 & -25  \end{matrix}
-5 `\times `23 = 115
23rd exam score = 470-115 = 355

Example 2. 

In a class average age of some students is 40 years. After some time 12 students with an average age of 32 years joins the class due to which average age of class will decrease by 4 years. Then find the number of students in the starting.
Solution:
Students average
\begin{matrix} X & 40 \\ X + 12 & 36 \end{matrix}
Average decreases due to the inclusion of these 12 students. And this is because there average age is less then the average age of class.
Adjusted value = `\frac{4x}{2}` = `\frac{x}{3}`
New average = average of included students + adjusted value
Hence, 36 = 32 + `\frac{x}{3}`
`\frac{x}{3}` = 4
x = 12
Hence the students in the starting is 12.