## Averages Part 1

Important Formulae:

• Average = $\frac{Sum&space;Of&space;Observations&space;}{No.of&space;Observations}$
• The average of first ‘n’ natural numbers = $\frac{n+1}{2}$
• The average of first ‘n’ even numbers = n+1
• The average of first ‘n’ odd numbers = n
• The average of square of first ‘n’ natural numbers = $\frac{(n+1)(2n+1)}{6}$
• The average of consecutive
• a , a+1 , a+2 , a+3 ,………….,a+(n-1) = $a+\frac{(n-1)}{2}$
• a is smallest number   a+(n-1) is largest number
• Consecutive even or odd numbers
• a,a+2,a+4,a+6,………….a+(n-1) = a+2(n-1)

## Averages Part 2

Problem 1: Find the average of 1,2,3,….91?

Solution:

$\frac{n+1}{2}=\frac{91+1}{2}=46$

Problem 2: Find the average of first 39 even numbers?

Solution:

2,4,6,8,…….

n+1 = 39+1 =40

Problem 3: Find the average 4+16+36+……..+400?

Solution:

4+16+36+…….+400

$2^{2}+4^{2}+6^{2}+.......+20^{2}$

$2^{2}(1+2^{2}+3^{2}+......+10^{2})$

$\frac{(n+1)(2n+1)}{6}$

$\frac{4\times&space;11\times&space;21}{6}=154$

Problem 4: The average of 6 consecutive even number is 63. Find the largest number?

Solution:

63=a+6-1

a = 63-5 = 58

## Averages Part 3

Problem 1: The average of 5 numbers is 52. If one number is excluded, the average becomes 50. Find the excluded number?

Solution:

Sum = 52*5 = 260

Sum = 50*4 = 200

260-200 = 60

Shortcut

52+4*2=60

Problem 2: The average age of 30 students is 12 years. When the teacher’s age is included in it, the average increased by 1. What is the teacher’s age?

Solution:

12+31*1=43

Problem 3: Average marks 0f 22 students is 75. The average is reduced by 2 if we exclude the highest and lowest marks. Find the sum of the highest and lowest marks?

Solution:

75*2+2*20 = 150+40 =190

## Averages Part 4

Problem 1: The average age of class A and Class B is 11 years, the average age of class A is 8 years and the average age of class B is 15 years. If the number of students in class A is 52. Then find the number of students in class B?

Solution:

8*52 + 15*B =(52+B)*11

8*52 +15B = 52*11 +11B

4B=52(11-8)

B=39

Problem 2: The average of 20 numbers is 25. If one of the numbers is considered as 35 instead of 75, what will be the new average?

Solution:

$25+\frac{35-75}{20}$

$25-\frac{40}{20}=23$

Problem 3: The Average age of A and B is 50 years and that of B and C is 38 years. What is the difference between the age of A and C?

Solution:

A+B = 50*2 =100

B+C = 38*2 = 76

A-C = 24

Problem 4: The average of 30 numbers is 20. The average of the first 12 numbers is 26. Find the average of remaining numbers?

Solution:

$20-\frac{6\times&space;12}{18}=16$

## Averages Part 5

Problem 1: Average of n numbers is‘a’. The first number is increased by 2, the second one is increased by 4, the third one is increased by 8 and so on. The average of the new numbers is?

Solution:

Average of n numbers is a

Total =an+2+4+8+16

a=2, r=2, n=n

$S_{n}=\frac{a\times&space;(r^{n}-1)}{r-1}$ , r>1

$\frac{a^{n}+\frac{2\times&space;(2^{n}-1)}{2-1}}{n}$

$a+2\times&space;\frac{2^{n-1}}{n}$

Problem 2: In a class of 50 boys, the average height of 30 boys is 160 cm. If the average height of remaining boys is 100 cm, then what will be the average height of boys(in cm)?

Solution:

$\frac{30\times&space;160+20\times&space;100}{50}$

$\frac{30\times&space;160}{50}+\frac{20\times&space;100}{50}=136$

Problem 3: The average of the first three numbers is thrice the fourth number. If the average of all the four numbers is 10, then find the fourth?

Solution:

$\frac{a+b+c}{3}=3d$

$a+b+c=9d$

$\frac{a+b+c+d}{4}=10$

$9d+d=4\times&space;10$

$10d=40$

$d=4$

Problem 4: The average of some natural numbers is 15. If 30 is added to the first number and 5 is subtracted from the last number, the average becomes 17.5. Then the numbers of natural numbers are?

Solution:

15n+30-5 = 17.5n

25 = 2.5n

n = 10

Shortcut:

$\frac{30-5}{2.5}=\frac{25}{2.5}=\frac{250}{25}=10$