A chartered bus carrying office employees travels everyday in two shifts- morning and evening. In the evening, the bus travels at an average speed which is 50% greater than the morning average speed; but takes 50% more time than the amount of time it takes in the morning. The average speed of the chartered bus for the entire journey is greater/less than its average speed in the morning byÂ
Distance = Speed x Time
In morning slot, Let speed be S and time taken be T, then distance covered = ST
Then, in evening slot, speed will be 1.5S and time taken will be 1.5T, then distance covered = 2.25ST
Average Speed = $$\frac{Total \space distance \space covered}{Total \space time \space taken}$$
For entire journey, average speed is:Â
Average Speed =$$\frac{ST + 2.25ST}{T + 1.5T}$$ = $$\frac{3.25ST}{2.5T}$$ = 1.3S
For morning, Average Speed is S.
Hence, the average speed of the bus for the entire journey is greater than its average speed in the morning by 30%
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