Question 756264
let after t min & s m both meet
then A travels for t min & the distance=50*50*t/3 m 
[as 1km/hr=50/3m/min]
& B travels for t-9 min & moves a distance= 55*50(t-9)/3 m
now 50*50*t/3=55*50(t-9)/3
or 50t=55(t-9)=55t-55*9
or 5t=55*9
or t=11*9=99 min
so the distance=50*50*t/3=50*50*99/3=82500m=82.5 km