| 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. |
In mathematics the average is the arithmetic mean value of N numbers. We can understand this as, suppose we have N numbers \(K_1, K_2, K_3,.........K_n\), then the average
$$ = \frac{K_1 + K_2 + K_3 +.........K_n}{N} $$
$$ Average = \frac{Sum \ of \ observations}{Number \ of \ observations} $$ $$ = \frac{K_1 + K_2 + K_3 +.........K_n}{N} $$
Example (1): Find the average value of the first five natural numbers?
Solution: The first five natural numbers are 1, 2, 3, 4, 5, then$$ Average = \frac{1 + 2 + 3 + 4 + 5}{5} \\ = \frac{15}{5} = 3 $$
Example (2): Find the average value of the first four even numbers?
Solution: The first four even numbers are 2, 4, 6, 8, then$$ Average = \frac{2 + 4 + 6 + 8}{4} \\ = \frac{20}{4} = 5 $$
Case (1):
If a man covers a certain distance \(d_1\) with the speed of \(x \ m/sec\) and the same man also covers equal distance with a speed \(y \ m/sec\), then
$$ Average \ Speed = \frac{2 \ x \ y}{(x + y)} \ m/sec $$
Case (2):
If we have, total distance travelled by a man \(D = d_1 + d_2 + d_3 +......d_n\) and total time taken by the man by travelling the total distance \(T = t_1 + t_2 + t_3 +......t_n\), then $$ Average \ Speed = \frac{Total \ distance \ (D)}{Total \ time \ (T)}$$
Example (1): A person named Ram, covers \(500\) \(m\) distance at the speed of \(5\) \(m/minute\) and covers another equal distance at the speed of \(10\) \(m/minute\), find the average speed of Ram?
Solution: Given values, \(x = 5 \ m/minute\) and \(y = 10 \ m/minute\) by using average speed formula from case (1),$$ Average \ Speed = \frac{2 x y}{(x + y)} $$ $$ Average \ Speed = \frac{2 \times 5 \times 10}{(5 + 10)} \\ = \frac{100}{15} = 6.667 \ m/minute $$
Example (2): A man travelles \(50 \ km\) distance, one half distance travelles at the speed of \(100 \ m/sec\) and another half distance at the speed of \(150 \ m/sec\) find the average speed of the man?
Solution: Given values, \(x = 100 \ m/sec\) and \(y = 150 \ m/sec\) by using average speed formula from case (1),$$ Average \ Speed = \frac{2 x y}{(x + y)} $$ $$ Average \ Speed = \frac{2 \times 100 \times 150}{(100 + 150)} \\ = \frac{30000}{250} = 120 \ m/sec $$
Example (3): Mr. George travelled \(10 \ km\) distance in \(20 \ minutes\), another \(20 \ km\) distance in \(30 \ minutes\) and another \(20 \ km\) distance in \(20 \ minutes\), then find the average speed of Mr. George?
Solution: Given values, distances travelled by Mr. George, \(d_1 = 10 \ km\), \(d_2 = 20 \ km\), and \(d_3 = 20 \ km\), time taken by Mr. George to cover the distance, \(t_1 = 20 \ minutes\), \(t_2 = 30 \ minutes\), and \(t_3 = 20 \ minutes\), by using average speed formula from case (2),$$ Average \ Speed = \frac{Total \ distance \ (D)}{Total \ time \ (T)}$$ $$ = \frac{(10 + 20 + 20) \ km}{(20 + 30 + 20) \ minutes} $$ $$ = \frac{50 \ km}{70 \ minutes} $$ $$ = 0.714 \ km/minute $$
Example (4): A woman travelled \(20 \ km\), \(25 \ km\), and another \(25 \ km\) and the time taken by the woman \(50 \ minutes\), \(60 \ minutes\), and \(70 \ minutes\), successively, find the average speed of the woman?
Solution: Given values, distances travelled by woman, \(d_1 = 20 \ km\), \(d_2 = 25 \ km\), and \(d_3 = 25 \ km\), time taken by the woman to cover the distance, \(t_1 = 50 \ minutes\), \(t_2 = 60 \ minutes\), and \(t_3 = 70 \ minutes\), by using average speed formula from case (2),$$ Average \ Speed = \frac{Total \ distance \ (D)}{Total \ time \ (T)}$$ $$ = \frac{(20 + 25 + 25) \ km}{(50 + 60 + 70) \ minutes} $$ $$ = \frac{70 \ km}{180 \ minutes} $$ $$ = 0.389 \ km/minute $$
The average of the first ten natural numbers is $$ = \frac{1 + 2 + 3 + 4 +....+ 10}{10} $$ $$ = \frac{55}{10} = 5.5 $$
The average of the first ten whole numbers is $$ = \frac{0 + 1 + 2 + 3 +....+ 9}{10} $$ $$ = \frac{45}{10} = 4.5 $$
The average of the first ten even numbers is $$ = \frac{2 + 4 + 6 + 8 +....+ 20}{10} $$ $$ = \frac{110}{10} = 11 $$
The average of the first ten odd numbers is $$ = \frac{1 + 3 + 5 + 7 +....+ 19}{10} $$ $$ = \frac{100}{10} = 10 $$
The average of the first ten prime numbers is $$ = \frac{2 + 3 + 5 + 7 +....+ 29}{10} $$ $$ = \frac{129}{10} = 12.9 $$
The average of the first n natural numbers is $$ = \frac{n + 1}{2} $$
The average of the squares of the first n natural numbers is $$ = \frac{(n + 1) (2n + 1)}{6} $$
The average of the cubes of the first n natural numbers is $$ = \frac{n \ (n + 1)^2}{4} $$
The average of first n odd natural numbers is \(= n\)
The average of first n even natural numbers is \(= n + 1\)