| 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. |
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
$$\frac{x_1 y_1 + x_2 y_2 + x_3 y_3}{y_1 + y_2 + y_3} $$ $$= \frac{45 \times 6 + 40 \times 5 + 50 \times 4}{6 + 5 + 4} $$ $$ = \frac{270 + 200 + 200}{15} $$ $$ \frac{670}{15} = 44.67 \ marks $$
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 $$S_n = \left[\frac{n \ (n + 1)}{2}\right]^2$$ $$S_4 = \left[\frac{4 \ (4 + 1)}{2}\right]^2 = 100$$