Question 1206222
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The wording "2/3 of the trip" seems a bit vague in my opinion. 
It could mean "2/3 of the total distance" or it could mean "2/3 of the total time".


I'll consider both scenarios.


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Scenario A: The wording "2/3 of the trip" means "2/3 of the total distance" 


20 miles total
20*(2/3) = 40/3 = 13.3333333 miles on the freeway
20*(1/3) = 20/3 = 6.6666667 miles on the residential streets
The decimal values are approximate.


15 min = 15/60 = 0.25 hour


I'll assume you drive the max speed allowed on the residential section.
distance = speed*time
6.6666667 = 35*time
time = 6.6666667/35
time = 0.1904762 of an hour is the amount spent on the residential section.


That leaves you with 0.25 - 0.1904762 = 0.0595238 of an hour spent on the freeway.
distance = speed*time
speed = distance/time
speed = 13.3333333/0.0595238
speed = 224.00003528 mph
There is some rounding error since the result should be 224 mph exactly (as the tutor ikleyn has shown).
You'll have to break the speed limit, and perhaps the laws of physics, to be able to make it on time.


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Scenario B: The wording "2/3 of the trip" means "2/3 of the total time".


2/3 of 15 min = (2/3)*15 = 10 min
10 min = 10/60 = 1/6 of an hour spent on the freeway


The other 15-10 = 5 min (aka 5/60 = 1/12 of an hour) is spent on the residential section.
I'll assume you drive the max speed allowed here.
distance = speed*time
distance = (35 mph)*(1/12 of an hour)
distance = 35/12 miles
distance = 2.9166667 miles approximately is the amount traveled along the residential section.


There are 20 - 2.9166667 = 17.0833333 miles remaining on the commute.
For the freeway section:
distance = speed*time
speed = distance/time
speed = (17.0833333 miles)/( 1/6 of an hour )
speed = 102.4999998 mph approximately
The result should be 102.5 mph exactly. 
Like before there's some rounding error going on.


It's not the same result, but it's still a very large speed value. 
I don't know if your teacher intended to have such a result in either scenario.
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