Question

A bus started from bus stand at 8am and after 30 min staying at destination, it returned back to the bus stand. The destination is 27 miles from the bus stand. The speed of the bus is 18mph. In return journey bus travels with 50% fast speed. At what time it return to the bus stand?

Answer 1

As the bus is travelling 18mph it takes 1 & 1/2 hr to reach destination.  

Bus stays there for 1/2 hr.  

Distence = 27m

Bus travels 50% of fast speed

Time = distance/speed = 27/18 = 3/2h = 1h 30m

As in the return journy bus travelled with the 50% fast speed meance with the speed of 

18mph.

time=(27/(9+18)) = 1h

total  time 1h + 1 h 30m + 30m = 3h

Bus returns as 11 a.m.

An alternate approach:

11am
bus covers 27 miles with 18mph in= 27/18 = 1 hr. 30 min.
it waits at stand = 30min.
after this, speed of return increases by 50 % so 50 % of 18 mph = 9 mph
total speed of returning = 18+9 = 27 mph
then in return it takes 27/27 = 1hr.
thus total time in journey = 1+1.30+0.30 = 3hr
so it will come at 8+3hr=11am.

Or Simply:

Here,D= 27 miles.

let speed = v = 18 mph(Given)
Time to go from P to Q = 27/18 = 3/2 hours = 1.5 hours
Stay there for 30 minutes = 0.5 hours
Time to go from Q to P = 27/(27) = 1 hour.
Total time taken to round a circle = 3 hours.
The time it return to bus stand = 8:00+3 hrs = 11:00 am.

Hope This Helps :)

Answer 2

It's 11 a.m.The bus will reach the destination at 9:30 a.m.The bus started returning at 10:00 a.m.As the bud was travelling at a speed of 18+(15/100*18)mph.The time taken for returning will be 1 hour.And, after 1 hour it will be 11 a.m.