Important formula and tricks about Averages

Average type of questions is most easy to solve and score. This types of questions can be found out in all the question papers.

$Average = \frac{Total}{No. of items}$

$Total = Average * No. of items$

This above formula is very much enough to crack the questions which arrives while answering the questions. The best way to crack any question is to understand the question in an easy manner and derive at an easy equation which makes the solution less complicated and helps you to arrive at the solution in a quick manner.

Important Formulae:
1. The average of first ‘N’ natural numbers = (N+1)/2
2. The average of first ‘N” natural odd numbers = N
3. The average of first ‘N’ natural even numbers = N+1

Note – Natural numbers are those which start from 1 to infinity ….

Let’s dive into some Average question examples :
Question 1 ) What is the average of the first five natural numbers?

Solution: The first five natural numbers are 1, 2, 3, 4, and 5.

$Average = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3.$

OR
By formula for the first ‘N’ natural numbers.
Here the ‘N’ is 5.

$Average = \frac{5+1}{2} = \frac{6}{2} =3$

Answer: 3

Question 2 ) What is the average of the first 65 natural numbers?

Solution: We can’t go through the traditional method. So, let’s use the formula

$Average = \frac{65+1}{2} = \frac{66}{2} = 33$

Answer: 33

Question 3 ) What is the average of the first 3 odd natural numbers?

Solution: The first three odd numbers are 1, 3 & 5.

$Average = \frac{1+3+5}{3} = \frac{9}{3} = 3$

OR
Let’s go through the formula
The formula for the average of the first ‘N’ natural odd numbers = N
In the above question ‘N’ = 3.
So, the answer = 3

Answer: 3

Question 4 ) What is the average of first 79 odd numbers?

Solution: The average of first ‘N’ natural odd numbers = N
In the above question ‘N’ is 79.

Answer: 79

Question 5 ) What is the average of first 3 natural even numbers?

Solution: The first three even natural numbers are 2, 4 and 6.

$Average = \frac{2+4+6}{3} = \frac{12}{3} = 4$

OR
The formula: The average of the first ‘N’ natural even numbers = N+1
In the above question ‘N’ = 3.
So, the answer = 3+1 = 4

Answer: 4

Question 6 ) What is the average of first 57 even natural numbers?

Solution: The formula for the first ‘N’ natural even numbers = N+1
In the above question ‘N’ is 57
The answer: 57 + 1 = 58

Answer: 58

The average of the consecutive numbers = (First number + Last number) / 2
Consecutive numbers are those numbers that possess each other in the same order from small to large or having the same difference between adjacent numbers
Example of the consecutive numbers are 1,2,3,4 and 5. Here, the numbers are in same order and they are from small to large.
Example of non consecutive numbers are 1,3,4 and 6. Here, there is no order. So, this is called as non consecutive numbers

Example:
Question 7 ) What is the average of 1,2,3,4,5?

Solution: 1,2,3,4 and 5 is a consecutive numbers.

$Average = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3$

$Formula = \frac{First number+Last number}{2}$

$Average = \frac{1+5}{2} = \frac{6}{2} = 3$

Answer: 3

Question 8 ) What is the average of 10,20,30?

Solution: 10,20,30 is a consecutive numbers.

$Average = \frac{10+20+30}{3} = \frac{60}{3} = 20$

$Formula = \frac{First number + Last number}{2}$

$Average = \frac{10+30}{2} = \frac{40}{2} = 20$

Answer: 20

Question 9 ) What is average of 5,10,15,20,….. 100?

Solution: It is an example of consecutive numbers.

$Average = \frac{5+100}{2} = 52.5$

Answer: 52.5

So far covered all basic formulas of Average and very basic questions of average. Lets dive into some intermediate questions.

Question 10 ) Find the average of 7 consecutive odd numbers and if the smallest of those numbers is denoted by k is:

Solution: If the smallest odd number is k, then the next odd number will be k+2.
For example: If the smallest number is 1 then the next odd number is 3 = 1+2.

So, the equation will be

$= \frac{k+k+2+k+4+k+6+k+8+k+10+k+12}{7}$

There are 7 numbers that is why we divide it by 7.

$= \frac{7k+42}{7}$

Let’s take the 7 outside which is common from the numerator. So, the equation will be

$= \frac{7(k+6)}{7}$

7 gets cancelled.
Remaining part is k+6.
Answer: k+6

Let go with the formula – 7 consecutive odd numbers – First number is k and 7th number is k+12
Average of consecutive number is = First number + last number /2 = k + k + 12 /2 = 2k + 12/2 = k + 6

Question 11) The average of A, B & C is Rs.12,000 per month and the average of B, C & D is Rs.15,000 per month. If the average salary of D be twice that of A, then the average salary of B&C is (in Rs.)

Solution: We already know

$Total = Average * No. of items$

The average salary of A, B and C = 12,000
The average salary of B, C and D = 15,000
Then,

$A + B + C = 12,000 * 3 = 36,000$

We are dividing it with 3 because there are 3 people namely A, B and C.

$B + C + D = 15,000 * 3 = 45,000$

We are dividing it with 3 because there are 3 people namely B, C and D.
So, the equation becomes

$A + B + C = 36,000 ----- (1)$

$B + C + D = 45,000 ----- (2)$

It is said that the average salary of D is twice that of A.

$D = 2A ----- (3)$

Now, let’s do (2) – (1)

$(B + C + D = 45000) - (A + B + C = 36,000$

The remaining left will be

$D - A = 9,000 ----- (4)$

Substitute (3) in (4)

$2A - A = 9,000$

Because D is said to be 2A. Verify equation number (3)

$A = 9000$

We know that

$A + B + C = 36,000$

$B + C = 36,000 - A$

$B + C = 36,000 - 9,000$

$B + C = 27,000$

We are supposed to find the average of B & C.

$Average = \frac{27,000}{2}$

There are only two people namely B&C. So only we are dividing it by 2.

$Average = 13,500$

Answer: 13,5000

Question 12 ) The average weight of 15 students in a group is 26 kg. When the teacher’s weight is added it becomes 30 kg. What is the weight of the teacher?

Solution: There are 15 students with an average weight of 26 kg. Then the teacher weight is added and it becomes 30 kg and the number of people increases to 16 persons. So, the equation becomes.

$\frac{15 * 26 + T}{16} = 30$

There are 15 students with an average weight of 26 and then there is T which is added. The denominator is 16 because there are 16 people (15 students + 1 teacher). The average weight becomes 30 when teacher is added.

Now, the equation becomes

$15 * 26 + T = 16 * 30$

$T = (16 * 30) - (15 * 26)$

$T = 90$

The weight of the Teacher is 90.

Answer: 90

Question 13 ) If the average weight of 4 men increases by 3 kg when one of them weighing is 90 kg is replaced by another man, then the weight of new man is?

Solution: There are 4 men and one of them weighing 90 kg is replaced by a new man (N). The average weight is increased by 3. The equation becomes

$\frac{4A - 90 + N}{4} = A +3$

90 is subtracted and N is added because one of them weighing 90 is replaced by N. There are still 4 members. So only there is 4 in the denominator. The average is increased by 3 and it is denoted by A + 3.

$4A - 90 + N = 4 (A + 3)$

4 is multiplied in the RHS

$4A - 90 + N = 4A + 12$

$N = 4 A + 12 - 4A + 90$

4A gets cancelled

$N = 90 + 12$

$N = 102$

The weight of the new man is 102

Answer: 102

Question 14) If the average weight of 4 men increases by 3 kg when one of them is replaced by another man 90 kg, then the weight of the replaced man is?

Solution: One of the 4 men’s is replaced with a 90 kg weight man. So, the equation becomes

$\frac{ 4A - O + 90}{4} = A + 3$

The replaced man is denoted by O since his weight is unknown. 90 is added because the replaced man is replaced with the man whose weight is 90.

$4A - O + 90 = 4 ( A + 3 )$

$4A - O + 90 = 4A + 12$

$-O = 4A + 12 - 4A - 90$

4A gets cancelled

$-O = 12 - 90$

$-O = -78$

‘Minus get cancelled’

$O = 78$

The replaced man weight is 78.

Answer: 78

Question 15 ) The average weight of 3 men A, B & C is 84 kg. Another man D joins the group & the average now becomes 80 kg. If another man E, whose weight is 3 kg more than that of D, replaces A, then the average weight of B, C, D & E becomes 79kg. Then the weight of A is:

Solution: When D join A, B & C the average weight is reduced by 4 kg.
Previously before D joining the weight was 84 kg.
After D joining the weight was 80 kg.
So,

$D = 84 - (4 * 4) = 84 - 16 = 68$

There are now 4 people A, B, C & D. That is why 4 has been multiplied by 4. The other 4 is the average weight reduced.
It is said that E is 3 kg more than D.

$E = 68 + 3 = 71$