You can
put this solution on YOUR website!

He'll get there in 1/3 hr
Now convert 1/3 hr to minutes
(60 min/hr)(1/3 hr) = 20 min
If he has 10 min to get there, he won't make it on time
You can
put this solution on YOUR website!Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
r=15
t=10 min or 10/60 hr=1/6 hr
d=15*(1/6)=2 1/2 hr----------------------No! He will not be on time
Hope this helps---ptaylor