SOLUTION: Mike can run the mile in 6 minutes, and Dan can run the mile in 9 minutes. If Mike gives Dan a head start of 1 minute, how far from the start will Mike pass Dan? How long does it t

Question 1207770: Mike can run the mile in 6 minutes, and Dan can run the mile in 9 minutes. If Mike gives Dan a head start of 1 minute, how far from the start will Mike pass Dan? How long does it take?
Found 3 solutions by mananth, ikleyn, greenestamps:
Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
Mike can run the mile in 6 minutes, and Dan can run the mile in 9 minutes. If Mike gives Dan a head start of 1 minute, how far from the start will Mike pass Dan? How long does it take?
Mike can run the mile in 6 minutes, speed = 1/6 mi /minute
and Dan can run the mile in 9 minutes. = 1/9 mi /minute
Mike gives Dan a head start of 1 minute
It means the Dan is nine miles ahead of mike when Mike starts to run
So Mile has to catchup 9 miles . catchup distance = 1/9 miles
Catch up speed = 1/6-1/9= 1/18 miles / minute
d/r = t
catchup time = catchup distance / catchup speed
(1/9)/(1/18)= 2 minutes
Distance covered by Mike in 2 minutes: 1/6×2=2/6=1/3
Mike will pass Dan 1/3 miles from the start, and it will take him 2 minutes to do so.

Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.

This problem was solved at this forum several years ago (perhaps, 5-6 years ago).

For solution (which is different from that by @mananth) see the link

https://www.algebra.com/algebra/homework/word/travel/Travel_Word_Problems.faq.question.199978.html#google_vignette

Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!

Mike can run the mile in 6 minutes, so his speed is 1/6 of a mile each minute.

Dan can run the mile in 9 minutes, so his speed is 1/9 of a mile each minute.

Mike gives Dan a head start of 1 minute; in that time Dan runs 1/9 of a mile.

When Mike starts running, the difference in their speeds is the rate at which Mike will catch up to Dan. The difference in their speeds is (1/6)-(1/9) = (3/18)-(2/18) = (1/18) mile per minute.

So Mike needs to make up a difference of 1/9 mile at a rate of 1/18 mile each minute. The time needed to do that (time = distance divided by rate) is (1/9)/(1/18) = (1/9)*(18/1) = 2 minutes.

ANSWER (second part): It will take Mike 2 minutes to catch up to Dan.

When Mike catches up to Dan, Dan will have run for 3 minutes at (1/9) mile per minute, covering a distance of 3/9 = 1/3 of a mile; Mike will have run for 2 minutes at a rate of (1/6) mile per minute, covering a distance of 2/6 = 1/3 of a mile.

ANSWER (first part): Mike will catch up to Dan 1/3 of a mile from the start.

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