SOLUTION: Bill and Roger want to race. Bill runs an average of 12 mph and Roger runs an average of 10 mph. To 'even things up' Bill lets Roger have a head start of .15 hour. How long before
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: Bill and Roger want to race. Bill runs an average of 12 mph and Roger runs an average of 10 mph. To 'even things up' Bill lets Roger have a head start of .15 hour. How long before
Log On
Question 381977: Bill and Roger want to race. Bill runs an average of 12 mph and Roger runs an average of 10 mph. To 'even things up' Bill lets Roger have a head start of .15 hour. How long before Bill catches up?
this is what I did:
12(x)=10(.15)
x=.125
or 7.5 minutes
Is that correct?
I also came up with 45 minutes... Answer by mananth(16949) (Show Source):
You can put this solution on YOUR website! Bill runs an average of 12 mph
Roger runs an average of 10 mph.
head start of .15 hour. for roger
..
make up speed for Bill is 12-10 = 2mph.
..
Roger has head start of 0.15 hour
so distance = 0.15 *10= 1.5 miles
..
t = d/r
1.5/2
=0.75 hours OR 45 minutes to catch up
...
m.ananth@hotmail.ca
