SOLUTION: At 7 am Joe starts jogging at 6mi/hr. At 7:10 am Ken starts off after him. How fast must Ken run in order to overtake him at 7:30 am?
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Question 362407: At 7 am Joe starts jogging at 6mi/hr. At 7:10 am Ken starts off after him. How fast must Ken run in order to overtake him at 7:30 am?
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
At 7 am Joe starts jogging at 6mi/hr. At 7:10 am Ken starts off after him. How fast must Ken run in order to overtake him at 7:30 am?
...
Joe --- 7.00 am 6mph
..
Ken ----2.10am---x mph
..
difference in speed = x-6 mph
..
in 10 minutes joe has jogged 1/6*6 = 1mile.
..
time = distance /speed
time = 20 minutes ( 7.10 to 7.30)=2/3 hours
speed = x-6. ( Ken catches up x-6 mph.
he has to catch up 1 mile
1/(x-6)=2/3
2(x-6)=3
2x-12=3
2x=12+3
2x=15
x= 7.5 mph.
ken should run at 7.5 mph
...
m.ananth@hotmail.ca
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