gmatkkvinu wrote:Sam leaves home everyday at 4 p.m to pick his son from school and returns home at 6 p.m. One day, the school was over at 4 p.m and the son started walking home from school. Sam, unaware of this, started from home as usual and met his son on the way and returns home with him 15 minutes early. If the speed of Sam is 30 km\hr, find the speed of his son.
It means Sam by and large reaches the school at 5 PM, and this is the same time his son is there waiting, today his son was freed at 4 PM and started walking.
The time saved by the son walking was 15 minutes, which is the time that Sam required to reach the school and come back from the point of meeting, it means that the time required from the point he reached to the school is 7½ minutes, if Sam usually reaches the school on 5 PM sharp, and now they met 7½ minutes before reaching the school, this means they met on 04:52:30 PM.
The son started walking on 4 PM, hence he walked for 52 minutes and 30 seconds (i.e. 7/8 hours) from 4 PM to 04:52:30 PM. In this time, the son covered a distance that Sam could have covered in 7½ minutes at 30 km\hr (calculations give this distance equal to 15/4 km). Hence, the speed of his son is
= (15/4) Km ÷ (7/8) hours
= [spoiler]
30/7 kmph
[/spoiler]
