Question 1208606
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Jane driving:
distance = rate*time
d = r*t
22 = 50t
t = 22/50 = 0.44 of an hour
0.44 hr = 0.44*60 = 26.4 minutes
Each decimal value is exact and hasn't been rounded.


Jane on the subway:
d = r*t
14 = 30t
t = 14/30 of an hour 
That converts to (14/30)*60 = 28 minutes


Jane drives for 26.4 minutes, there's the 10 minute gap from car to subway, and then she's on the subway for 28 minutes.
Total = 26.4+10+28 = 64.4 minutes.
Round this up to 65 minutes. 
If we went with 64 minutes, then we'd miss out on that extra 0.4 of a minute.


8:30 AM - 65 min
= (8 hrs + 30 min) - (60 min + 5 min)
= (8 hrs + 30 min) - (1 hr + 5 min)
= 8 hrs + 30 min - 1 hr - 5 min
= (8 hrs - 1 hr) + (30 min - 5 min)
= 7 hr + 25 min
= <font color=red>7:25 AM</font> is when Jane should leave home.
In other words, subtract off 1 hour from 8 AM to drop to 7 AM. Then we subtract the extra 5 minutes from the 30 to get 25.
Verification with <a href="https://www.wolframalpha.com/input?i=8%3A30+AM+-+65+min">WolframAlpha</a>


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Bill driving:
d = r*t
30 = 60t
t = 30/60 = 0.5 hr
0.5 hr = 0.5*60 = 30 min


Bill on the subway:
d = r*t
6 = 20t
t = 6/20 = 0.3 hr
0.3 hr = 0.3*60 = 18 min


Bill's total duration is 30+10+18 = 58 minutes


Tutor ikleyn has made an error in her calculation
{{{30/((red(30)/60)) + 6/((20/60)) + 10 = 88}}}
should be
{{{30/((red(60)/60)) + 6/((20/60)) + 10 = 58}}}
The item in red is Bill's driving speed in km/hr.



8:30 AM - 58 min
= (8 hrs + 30 min) - (60 min - 2 min)
= (8 hrs + 30 min) - (1 hr - 2 min)
= 8 hrs + 30 min - 1 hr + 2 min
= (8 hrs - 1 hr) + (30 min + 2 min)
= 7 hrs + 32 min
= 7:32 AM is when Bill should leave home
Verification with <a href="https://www.wolframalpha.com/input?i=8%3A30+AM+-+58+min">WolframAlpha</a>



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To recap, 
Jane should leave home at 7:25 AM
Bill should leave home at 7:32 AM
Jane should leave 7 minutes earlier compared to bill (because 32-25 = 7).



Answer: 
<font color=red>Jane</font> leaves the earliest, and does so at <font color=red>7:25 AM</font>
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