A man travels from place A to place B at a speed of 10 m/s in X minutes. He than moves from place B to C with a speed of 45 km/hr in Y minutes. If the ratio of the times X and Y is 2:3, then the average speed of the whole journey is:
41.4 km/hr
Average speed = total distance ÷ total time. First convert both speeds to the same unit: \(10 \text{ m/s} = 10 \times \frac{18}{5} = 36\) km/hr.
Since X : Y = 2 : 3, let the times be 2t and 3t (in the same unit). Distance A to B = 36 × 2t = 72t and distance B to C = 45 × 3t = 135t.
Total distance = 72t + 135t = 207t and total time = 2t + 3t = 5t, so average speed = \(\frac{207t}{5t} = 41.4\) km/hr.
Check: it is the time-weighted mean of the speeds, \(\frac{2 \times 36 + 3 \times 45}{5} = \frac{72 + 135}{5} = 41.4\), and it lies between 36 and 45, closer to 45 because more time is spent at that speed.
Hence, 41.4 km/hr.
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Important Questions from Average Speed
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