A bus travelling the first one third distance at a speed of $10 km/h$, the next one third at$20 km/h$ and the last one third at $60 km/h$. The average speed of the bus is

For a journey where equal distances are traveled at different speeds, the average speed is calculated using the formula 3abc divided by (ab + bc + ca). Substituting the speeds 10, 20, and 60 gives a numerator of 3 * 10 * 20 * 60 equaling 36000. The denominator is (10 * 20) + (20 * 60) + (60 * 10) equaling 2000, making the average speed 36000/2000, which is 18 km/h.

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