Monday, 2 July 2012

Journey to home

A student leaves the coaching everyday at 11. His brother leaves the house some time before 11 to reach coaching at 11 to pick the student. One day, the coaching got over at 10 and the student started walking towards home at 8km/hr. In the way, he found his brother coming and then they went together reaching home 20 mins earlier than usual time. Find the speed of his brother?


  1. v*T=v*(T-2/6)+8*5/6 gives v=20km/hr
    the student walks for 50 min and the above equation holds because the distance covered on that day and a normal day is same.

    1. The answer is 40km/hr
      Since they reached 20 mins earlier than the usual time. The brother would have traveled 10 min lesser on either side of the trip. The brother was supposed to reach the coaching by 11, but since he travel 10 mins lesser on either side of the trip, he would find the student at 10:50. Now the student took 50 mins to reach a position where the brother would have took 10 mins. Thus, the speed of the brother would be 5 times the speed of the student (i.e) 40 km/hr