Sat Feb 16, 2013 9:10 am by tartle 


If a man travels at speed of 20km/hr, he reaches the office 20 minutes late. If he travels at 30 km/hr, he reaches the office 15 minutes early. If he travels at a speed of 25 km/hr, then, when does he arrive at the office? 




Fri Mar 08, 2013 5:47 pm by ste 


Journey 1:
Speed = 20km/hr = 1/3km/min
Time = 20 mins late
Journey 2:
Speed = 30km/hr = 1/2km/min
Time = 15 mins early
Journey 3:
Speed = 25km/hr = 5/12km/min
y = the distance to work
speed (s) = distance (d) / time (t)
If the distance to work was (1/3 x 20)km shorter for journey 1 and (1/2 x 15)km longer for journey 2 both journeys would take the same time. Therefore, using s=d/t:
t = (y  1/3 x 20)/(1/3) = (y + 1/2 x 15)/(1/2)
=> The distance to work (y) = 35km.
=> Journey 1 took 35/(1/3) = 105 mins.
105 mins is 20 mins late therefore the journey should take 85 mins.
Journey 3 takes 35/(5/12) = 84 mins. Therefore he's 1 minute early for work. 






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