Question 1195978
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Part A


The gap from 5:03 AM to 8:11 PM is 15 hours, 8 minutes
This is because...<ul><li>The jump from 5:03 AM to 5:03 PM is 12 hours</li><li>The jump from 5:03 PM to 8:03 PM is another 3 hours (12+3 = 15 hours so far)</li><li>The jump from 8:03 PM to 8:11 PM is another 8 minutes (since 11-3 = 8)</li></ul>That's how I got a total duration of 15 hrs, 8 min. This accounts for actual driving, and the break when the person isn't driving.


The driver took a 36 minute break.
Rewrite 36 as 8+28
I'm doing it like this so we can pull out the 8.


We'll subtract off those first 8 minutes to get
(15hr,8min) - (8min) = 15 hrs


Then subtract off the remaining 28 minutes of that break
15 hrs - 28 min = (14 hrs + 1 hr) - 28 min
15 hrs - 28 min = (14 hrs + 60 min) - 28 min
15 hrs - 28 min = 14 hrs + (60 min - 28 min)
15 hrs - 28 min = 14 hrs + 32 min


After ignoring the break, s/he actually spent 14 hrs & 32 min driving.


Let's convert that figure to hours only
14 hrs + 32 min = 14 hrs + (32/60 hr)
14 hrs + 32 min = 14 hrs + 0.533333 hr
14 hrs + 32 min = 14.533333 hrs
The decimal value is approximate.


The total distance traveled was:
163 mi + 328 mi = 491 mi


Now we can then say:
distance = rate*time
rate = distance/time
rate = (491 mi)/(14.533333 hrs)
rate = 33.784404 mph
This value is approximate.


Round to the nearest whole number to get 34 mph


Answer: <font color=red>34 mph</font>
This answer is approximate.


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Part B


In part A, we found his/her speed to be roughly 34 mph.
If they go 5 mph slower, then they now have a speed of 34-5 = 29 mph.


The total distance stays the same.
Let's compute the time duration spent driving (ignore the 36 min break for now)


distance = rate*time
time = distance/rate
time = (491 mi)/(29 mph)
time = 16.931034 approximately
This is the number of hours spent driving.


Let's convert to hours,minutes format
16.931034 hr = 16 hr + 0.931034 hr
16.931034 hr = 16 hr + (0.931034 hr)*(60 min/1 hr)
16.931034 hr = 16 hr + (0.931034*60) min
16.931034 hr = 16 hr + 55.86204 min
16.931034 hr = 16 hr + 56 min
I rounded to the nearest minute.


We've yet to include the 36 minute break. 
Let's do so now.
(16 hr + 56 min) + 36 min
16 hr + (56 min + 36 min)
16 hr + (92 min)
16 hr + (60 min + 32 min)
16 hr + (1 hr + 32 min)
(16 hr + 1 hr) + 32 min
17 hr + 32 min
This is the total duration from when the motorist left Caribou City to when s/he arrived at Shellberg. 


Rewind the clock back to 5:03 AM
<ul><li>Add on 12 hours to get to 5:03 PM</li><li>Add on 5 more hours (12+5 = 17 hrs so far) to get to 10:03 PM</li><li>Then add on the final 32 minutes to get to 10:35 PM</li></ul>


Answer: <font color=red>10:35 PM</font>
This answer is approximate.


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Part C


The steps shown here are very similar to part B.
In fact, because of this similarity, I effectively copy/pasted the main outline of the steps. The only main changes are the numbers.


In part A, we found his/her speed to be roughly 34 mph.
If they go 1 mph faster, then they now have a speed of 34+1 = 35 mph.


The total distance stays the same.
Let's compute the time duration spent driving (ignore the 36 min break for now)


distance = rate*time
time = distance/rate
time = (491 mi)/(35 mph)
time = 14.028571 approximately
This is the number of hours spent driving.


Let's convert to hours,minutes format
14.028571 hr = 14 hr + 0.028571 hr
14.028571 hr = 14 hr + (0.028571 hr)*(60 min/1 hr)
14.028571 hr = 14 hr + (0.028571*60) min
14.028571 hr = 14 hr + 1.71426 min
14.028571 hr = 14 hr + 2 min
I rounded to the nearest minute.


We've yet to include the 36 minute break. 
Let's do so now.
(14 hr + 2 min) + 36 min
14 hr + (2 min + 36 min)
14 hr + 38 min
This is the total duration from when the motorist left Caribou City to when s/he arrived at Shellberg.


Rewind the clock back to 5:03 AM
<ul><li>Add on 12 hours to get to 5:03 PM</li><li>Add on 2 more hours (12+2 = 14 hrs so far) to get to 7:03 PM</li><li>Then add on the final 38 minutes to get to 7:41 PM</li></ul>


Answer: <font color=red>7:41 PM</font>
This answer is approximate.
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