Question 52051
Question 
Steven left Town A and walked towards Town B at a speed of 100 m/min. At the same time, Jason and Melvin started from Town B and walked towards Town A at a speed of 80 m/min and 75 m/min respectively. 
RELATIVE SPEED OF STEVEN AND JASON = 100+80 = 180 M/MT.
RELATIVE SPEED OF STEVEN AND MELVIN = 100+75=175 M/MT.
If Steven met Melvin six minutes after passing Jason, find the distance between Town A and Town B. 
LET DISTANCE BETWEEN A AND B = D
STEVEN MEETS JASON IN = D/180 MT
STEVEN MEETS MELVIN IN =D/175 MTS
DIFFERENCE = D/175 - D/180 =5D/(180*175) MTS.= 6 MTS.
D = 6*180*175/5=6*180*35=37800 M. =37.8 KM

Thank you very much.