Question 1207770
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Mike can run the mile in 6 minutes, so his speed is 1/6 of a mile each minute.<br>
Dan can run the mile in 9 minutes, so his speed is 1/9 of a mile each minute.<br>
Mike gives Dan a head start of 1 minute; in that time Dan runs 1/9 of a mile.<br>
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.<br>
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.<br>
ANSWER (second part): It will take Mike 2 minutes to catch up to Dan.<br>
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.<br>
ANSWER (first part): Mike will catch up to Dan 1/3 of a mile from the start.<br>