SOLUTION: a long-distance runner left his house running at an average rate of 8 miles per hour. fifteen minutes (one-fourth of an hour) later, his son left his house on his bike traveling on
Algebra ->
Customizable Word Problem Solvers
-> Numbers
-> SOLUTION: a long-distance runner left his house running at an average rate of 8 miles per hour. fifteen minutes (one-fourth of an hour) later, his son left his house on his bike traveling on
Log On
Question 451035: a long-distance runner left his house running at an average rate of 8 miles per hour. fifteen minutes (one-fourth of an hour) later, his son left his house on his bike traveling on the same route at an average rate of 12 miles per hour. how long will it take the son to catch up with his father? Found 2 solutions by rwm, edjones:Answer by rwm(914) (Show Source):
You can put this solution on YOUR website! 8*(t+.25)=12t
8t+2=12t
2=4t
.5=t half an hour for the son to catch up.
8*.75=12*.5
6=6
ok
In the 1/4 hour the runner went 2 miles
In the next half hour the runner went 4 miles for a total of 6 miles.
The biker went 6 miles in half an hour
You can put this solution on YOUR website! speed * time = distance
Let t=son's time
8(t+.25)=father's dist.
12t=son's dist.
12t=8t+2 distances are equal when the son catches up.
4t=2
t= 1/2 hour.
.
Ed
(
