A bus is travelling the first one third distance at a speed of $10\;kmh^{-1}$, the next one third at $20\;kmh^{-1}$ and the last one third at $60\;kmh^{-1}$. The average speed of the bus is:
A. $9\;kmh^{-1}$
B. $16\;kmh^{-1}$
C. $18\;kmh^{-1}$
D. $48\;kmh^{-1}$

Hint: Assume that the total distance covered by the bus is $x\;km$. Then, given the speed with which it covers each $\dfrac{x}{3}$ of distance, calculate the time the bus takes to travel each third of the distance. Sum all of these times together to get the total time. Now, that you have the total distance and total time, calculating the average velocity should be pretty straightforward.

Formula Used: Distance travelled $= velocity \times time$
Average velocity $= \dfrac{Total\;distance}{Total\;time}$

Complete answer:
Let the total distance travelled by the bus be $x\;km$
Then, average velocity will be $ = \dfrac{Total\;distance}{Total\;time}$
Let us calculate the total time by calculating the time the bus takes to travel each of the thirds.
The distance it travels in each of the thirds will be $\dfrac{1}{3}x = \dfrac{x}{3}$
We know that time taken to travel a distance can be given as $time = \dfrac{distance}{speed}$
Let us now calculate the time the bus takes to travel each of the thirds.
For the first third: $t_1 = \dfrac{\dfrac{x}{3}}{10} = \dfrac{x}{30}\; hrs$
For the second third: $t_1 = \dfrac{\dfrac{x}{3}}{20} = \dfrac{x}{60}\; hrs$
For the final third: $t_1 = \dfrac{\dfrac{x}{3}}{60} = \dfrac{x}{180}\; hrs$
Therefore, total time taken
$= t_1 +t_2 +t_3 = \dfrac{x}{30} +\dfrac{x}{60} +\dfrac{x}{180} = \dfrac{6x + 3x+x}{180} = \dfrac{10x}{180} = \dfrac{x}{18}\;hrs$
Therefore, the average velocity with which the bus travels $ = \dfrac{Total\;distance}{Total\;time} = \dfrac{x}{\dfrac{x}{18}} = 18\;kmh^{-1}$
Therefore, the correct choice will be C. $18\;kmh^{-1}$

Note: Do not assume that the average speed is the arithmetic mean of individual speed. This is because the time taken to travel the same distance will be different under different speeds. Therefore, in order to account for this variation in time, we take average speed as the ratio of the total distance covered to the total time taken. Thus, average speed is a weighted average over time.