Average is a straight- forward concept and it can be solved easily by equal distribution method.

**Equal distribution method:** It is the easiest way to solve problems

- If each element is increased by same value, then the average also gets increased by same value.

**Example**:- 32, 46, 25, 57

Suppose each value is increased by 5, the average also gets increased by 5.

Average of given numbers 32, 46, 25, 57 is 40.

When each number is increased by 5 i.e. 37, 51, 30, 62, the average is also increased by 5,Â and so the new average is 45. - If the new value added is same as the average value, then the average is same.

**Example**:- 32, 46, 25, 57,

**40. Here 40 is the new number that is added which is same as the original average.**

So, new Average is 40 - If new value added is different as the average value, then the difference is equalized on all the values.

**Example**:- 32, 46, 25, 57, 50

Average is 40, newly added value is 50

Now their difference is 10

Equally distributed value is \(\frac{10}{(no .of values)} = \frac{10}{5}\) = 2

Hence, the average is 40 + 2 = 42

- Where ‘A’ â†’ Average
- \(‘x’\) â†’ Sum of given elements
- \( ‘y’ \) â†’ Total number of elements

**Example 1**:

Find the average of all prime numbers between 30 and 50.

**Solution**:

- There are five prime numbers between 30 and 50.

They are 31, 37, 41, 43 and 47.

Therefore, Required average = (\(\frac{31 + 37 + 41 + 43 + 47}{5}\)) = \(\frac{199}{5}\) = 39.8.

**Example 2**:

Find the average of first 20 multiples of 7.

**Solution**:

- Required average = \(\frac{7(1 + 2 + 3 + ….. + 20)}{20}\) = (\(\frac{7 \times 20 \times 21}{20 \times 2}\)) = (\(\frac{147}{2}\)) = 73.5.

**Example 3**:

Find the average of first 40 natural numbers.

**Solution**:

- Sum of first n natural numbers = \(\frac{n(n + 1)}{2}\)

So, sum of first 40 natural numbers = \(\frac{40 \times 41}{2}\) = 820.

Therefore, Required average = \(\frac{820}{40}\) = 20.5

- Where ‘x and y’ â†’ Distance

**Example 1**:

Distance between two stations A and B is 778 km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56 km per hour. Find the average speed of the train during the whole journey.

**Solution**:

- Required average speed = \(\frac{2 x y}{x + y} km/hr \) = \(\frac{2 \times 84 \times 56}{(84 + 56)} km/hr\)

= \(\frac{2 \times 84 \times 56}{140} km/hr\) = 67.2 km/hr.

**Example 2**:

A motorist travels to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/hr. His average speed for the whole journey in km / hr is.

**Solution**:

- Average Speed = \(\frac{2 x y}{x + y} km/hr \) = \(\frac{2 \times 50 \times 30}{50 + 30} km/hr\) = 37.5 km/hr.

- Average(A) = \(\frac{Sum \, of \, given \, elements \, (x)}{(Total \, number \, of \, elements \,(y))}\)

Where ‘A’ â†’ Average

\(‘x’\) â†’ Sum of given elements

\( ‘y’ \) â†’ Total number of elements - If we need to find sum of “n” elements then

Sum = n x \( x \) - Average Speed = \(\frac{2 x y}{x + y} km/hr \)
- When two sections are given then average is

Average = \(\frac{n_{1}a_{1} + n_{2}a_{2}}{n_{1} + n_{2}}\)

- Given that

Rahul made 1000 runs in 20 matches

Pradeep made 500 runs in 10 matches

Now, Average = \(\frac{(sum of values)}{(total number of values)}\)

by substituting the values,

â‡’Average = \(\frac{(1000 + 500)}{(10 + 20)}\)

â‡’Average = \(\frac{(1500)}{(30)}\)

â‡’Average = 50

Therefore, Average made by them = 50

**2. The sum of five numbers is 555.The average of first two numbers is 75 and third number is 115. Find the average of last two numbers?**

**Solution**:

- Let the five numbers be a, b, c, d, e

Given that

a + b + c + d + e = 555

Average of first two numbers is = 75

Third number is 115

By substituting the given values,

75 + 115 + d + e = 555

â‡’d + e = 555 â€“ 265

â‡’d + e = 290

Therefore, average of last two numbers (d, e) = \(\frac{290}{2}\) =145

**3. The average age of 20 students of a section is 12 years.The average of 25 students of another section is 12 years. Find average age of both the sections?**

**Solution**:

- Given that

Average age of 20 students of a section = 12 years

Average age of another section of 25 students = 12 years

Average = \(\frac{n1a1 + n2a2}{n1 + n2}\)

Here n1 is number of students in first section

a1 is average age of first section

n2 is number of students in second section

a2 is average age of second section

by substituting the given values in above formula,

Average = \(\frac{20 * 12 + 25 * 12}{20 + 25}\)

â‡’Average = \(\frac{540}{45}\)

â‡’Average = 12

Therefore, average age of both the sections = 12.

**4. The total of ages of class of 75 girls is 1050. Average age of 25 of them is 12 years and another 25 of them is 16 years. Find the average age of remaining girls?**

**Solution**:

- Given that

Total ages of class =1050

Average age of 25 girls = 12 years

Average age of another 25 girls = 16 years

Now, average age of remaining girls = \(\frac{(1050 – (25 * 12 + 25 * 16))}{(75 â€“ 50)}\)

â‡’Average = \(\frac{(1050 â€“ (25*28))}{(25)}\)

â‡’Average = \(\frac{1050}{25}\) – \(\frac{(25 * 28)}{25}\)

â‡’Average = 42 – 28

â‡’Average = 12

Therefore, average age of remaining girls = 12 years

**5. Average age of 50 girls 58. If the weight of one of the girl is taken as 45 instead of 65. What is the actual average?**

**Solution**:

- Given that

Average age of 50 girls = 58

One of them is taken as 45 instead of 65

So, 65 – 45 = 20

\(\frac{20}{50}\) = 0.4

Hence, actual average is 58 + 0.4 = 58.4