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 by 

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: 

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%