# Weighted Average Impotant Formulas, Definitions & Exmaples:

#### Overview:

 Topic Included: Formulas, Definitions & Exmaples. Main Topic: Quantitative Aptitude. Quantitative Aptitude Sub-topic: Average Aptitude Notes & Questions. Questions for practice: 10 Questions & Answers with Solutions.

#### What is Weighted Average:

If there are N number of groups and average of each group is known $$x_1, x_2, x_3,........x_n$$ and number of elements of each group are also known $$y_1, y_2, y_3,........y_n$$, then Weighted average

$$\frac{x_1 y_1 + x_2 y_2 + x_3 y_3 +........x_n y_n}{y_1 + y_2 + y_3 +........y_n}$$

Example (1): A student named Bhim scored average, $$65$$ marks in $$3$$ tests of History, $$50$$ marks in $$4$$ tests of geography, and $$70$$ marks in $$6$$ tests of sports, then find weighted average marks of Bhim scored?

Solution: Given values, average marks $$x_1 = 65, x_2 = 50, x_3 = 70$$ and elements in each group $$y_1 = 3, y_2 = 4, y_3 = 6$$, then Weighted average

$$\frac{x_1 y_1 + x_2 y_2 + x_3 y_3}{y_1 + y_2 + y_3}$$ $$= \frac{65 \times 3 + 50 \times 4 + 70 \times 6}{3 + 4 + 6}$$ $$= \frac{195 + 200 + 420}{13}$$ $$\frac{815}{13} = 62.692 \ marks$$

Example (2): A cricketer named Mr.John scored average runs, $$120$$ runs in $$2$$ test cricket matches, $$180$$ runs in $$4$$ one-day cricket matches, and $$100$$ runs in $$4$$ T-twenty cricket matches, then find weighted average runs Mr. John scored?

Solution: Given values, average runs $$x_1 = 120, x_2 = 180, x_3 = 100$$ and elements in each group $$y_1 = 2, y_2 = 4, y_3 = 4$$, then Weighted average

$$\frac{x_1 y_1 + x_2 y_2 + x_3 y_3}{y_1 + y_2 + y_3}$$ $$= \frac{120 \times 2 + 180 \times 4 + 100 \times 4}{2 + 4 + 4}$$ $$= \frac{240 + 720 + 400}{10}$$ $$\frac{1360}{10} = 136 \ runs$$

Example (3): A student scored average, $$45$$ marks in $$6$$ tests of Mathematics, $$40$$ marks in $$5$$ tests of Science, and $$50$$ marks in $$4$$ tests of English, then find average marks the student scored?

Solution: Given values, average marks $$x_1 = 45, x_2 = 40, x_3 = 50$$ and elements in each group $$y_1 = 6, y_2 = 5, y_3 = 4$$, then Weighted average

$$\frac{x_1 y_1 + x_2 y_2 + x_3 y_3}{y_1 + y_2 + y_3}$$ $$= \frac{45 \times 6 + 40 \times 5 + 50 \times 4}{6 + 5 + 4}$$ $$= \frac{270 + 200 + 200}{15}$$ $$\frac{670}{15} = 44.67 \ marks$$

### Important Formulae:

The Sum of the first $$n$$ natural numbers: If we have the first $$n$$ natural numbers $$1, 2, 3, 4,........,n$$, then the sum of first $$n$$ natural numbers $$(1 + 2 + 3 + 4 +........+ n)$$ will be $$S_n = \frac{n \ (n + 1)}{2}$$

Example: Calculate the sum of first $$6$$ natural numbers?

Solution: First $$6$$ natural numbers are $$1, 2, 3, 4, 5, 6$$, and the sum of the first $$6$$ natural numbers will be $$S_n = \frac{n \ (n + 1)}{2}$$ $$S_6 = \frac{6 \ (6 + 1)}{2} = \frac{42}{2} = 21$$

The Sum of the first $$n$$ odd numbers: If we have the first $$n$$ odd numbers $$1, 3, 5, 7,........,(2n - 1)$$, then the sum of first $$n$$ odd numbers $$[1 + 3 + 5 + 7 +........+ (2n - 1)]$$ will be $$S_n = n^2$$

Example: Calculate the sum of first $$4$$ odd numbers?

Solution: First $$4$$ odd numbers are $$1, 3, 5, 7$$, and the sum of the first $$4$$ odd numbers will be $$S_n = n^2$$ $$S_4 = 4^2 = 16$$

The Sum of the first $$n$$ even numbers: If we have the first $$n$$ even numbers $$2, 4, 6, 8,........,2n$$, then the sum of first $$n$$ even numbers $$[2 + 4 + 6 + 8 +........+ 2n]$$ will be $$S_n = n \ (n + 1)$$

Example: Calculate the sum of first $$5$$ even numbers?

Solution: First $$5$$ even numbers are $$2, 4, 6, 8, 10$$, and the sum of the first $$5$$ even numbers will be $$S_n = n \ (n + 1)$$ $$S_5 = 5 \ (5 + 1) = 30$$

The Sum of the squares of first $$n$$ natural numbers: If we have the squares of first $$n$$ natural numbers $$1^2, 2^2, 3^2, 4^2,........,n^2$$, then the sum of the squares of first $$n$$ natural numbers $$[1^2 + 2^2 + 3^2 + 4^2 +........+ n^2]$$ will be $$S_n = \frac{n \ (n + 1) \ (2n + 1)}{6}$$

Example: Calculate the sum of squares of the first $$3$$ natural numbers?

Solution: squares of the First $$3$$ natural numbers are $$1^2, 2^2, 3^2$$, and the sum of squares of first $$3$$ natural numbers will be $$S_n = \frac{n \ (n + 1) \ (2n + 1)}{6}$$ $$S_3 = \frac{3 \ (3 + 1) \ (2 \times 3 + 1)}{6} = 14$$

The Sum of the cubes of first $$n$$ natural numbers: If we have the cubes of first $$n$$ natural numbers $$1^3, 2^3, 3^3, 4^3,........,n^3$$, then the sum of the cubes of first $$n$$ natural numbers $$[1^3 + 2^3 + 3^3 + 4^3 +........+ n^3]$$ will be $$S_n = \left[\frac{n \ (n + 1)}{2}\right]^2$$

Example: Calculate the sum of cubes of the first $$4$$ natural numbers?

Solution: cubes of the First $$4$$ natural numbers are $$1^3, 2^3, 3^3, 4^3$$, and the sum of cubes of the first $$4$$ natural numbers will be $$S_n = \left[\frac{n \ (n + 1)}{2}\right]^2$$ $$S_4 = \left[\frac{4 \ (4 + 1)}{2}\right]^2 = 100$$