SOLUTION: Janet leaves home at 10:15a jogging 8mph. Sue leaves the same home at 10:35am biking 18mph. What time do they meet and at what mile? solved for their ratio of speed S4:J9

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Question 1194658: Janet leaves home at 10:15a jogging 8mph.
Sue leaves the same home at 10:35am biking 18mph.
What time do they meet and at what mile?
solved for
their ratio of speed S4:J9 and distance of J9:S4
and at 10:25 Janet would have gone 2 2/3 miles.
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Sue leaves 20 minutes or one-third hour after Janet. In that time, the distance between the two is 8%28miles%2Fhour%29%2A%281%2F3%29%2Ahour=highlight_green%28%288%2F3%29%2Amiles%29.

If Janet catches up in x hours, then %2818-8%29x=8%2F3;
10x=8%2F3
x=8%2F30hours;

For number of minutes, %288%2F30%29%2Ahour%2A60%28minutes%2Fhour%29=highlight_green%2816%2Aminutes%29
and you can compute the time on the clock.
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What time met?
35%2B16
51
10:51AM

What distance when met?
18 miles per hour for 16 minutes
18%28miles%2Fhour%29%2A%2816%2F60%29hour
%2818%2A16%29%2F60miles
%283%2A2%2A3%2A4%2A4%29%2F%282%2A3%2A2%2A5%29
%282%2A3%2A4%29%2F5
24%2F5
highlight%284%264%2F5%29highlight%28miles%29