Draw segment AB.
Let C be the midpoint of segment AB
This means AC = CB
C is at the halfway point between the two endpoints A and B.
Let's say that the distance from A to C is x miles. This would mean from C to B is also the same distance x.
Furthermore, AB = 2*AC = 2x
x is some positive real number.
The bus travels from A to C at a speed of 40 mph going x miles. This means it travels for x/40 hours. I'm using the idea that
distance = rate*time
to solve for time to get
time = distance/rate
The bus goes from C to B going 50 mph
time = distance/rate = x/50
When going from A to B, the total time taken is:
x/40+x/50 = 5x/200+4x/200 = 9x/200 hours
We're told that "The trip from B to A is an hour longer than the trip from A to B"
So whatever 9x/200 is equal to, we add on 1 hour to get the expression
9x/200 + 1 = 9x/200 + 200/200 = (9x+200)/200
The time it takes to go from B to A is (9x+200)/200 hours.
We'll use this expression later.
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As the bus is traveling on the return trip, it travels for 2 hours going 35 mph.
This means the bus travels a distance of
distance = rate*time
distance = 35*2
distance = 70
After these 70 miles are up, let's say the bus arrives at point D.
This point D may or may not be at the same location as point C.
We have defined point D such that segment BD = 70 miles.
This makes AD = 2x-70
You start with the total distance 2x, and subtract off what you've done so far.
Because the bus travels from D to A at 40 mph, we can say,
distance = rate*time
2x-70 = 40*time
time = (2x-70)/40
This represents the amount spent on the remaining part of the trip
It represents the number of hours that elapse when going from D to A.
Add this to the 2 hours spent, when we traveled BD = 70 miles, and we get
(2x-70)/40 + 2 = (2x-70)/40 + 80/40 = (2x-70+80)/40 = (2x+10)/40 = 2(x+5)/40 = (x+5)/20
The bus spends (x+5)/20 hours going from B to A.
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Recall earlier, at the end of the first section, we found that the time spent going from B to A was (9x+200)/200 hours.
At the end of the second section we also found the time spent going from B to A was (x+5)/20 hours.
Equate the two time values and solve for x
(x+5)/20 = (9x+200)/200
200(x+5) = 20(9x+200)
200x+1000 = 180x+4000
200x-180x = 4000-1000
20x = 3000
x = 3000/20
x = 150
The distance from A to C is 150 miles. So is the distance from C to B.
So AB = AC+CB = 150+150 =
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Let's check our answer:
If you travel AC = 150 miles at 40 mph, then you spend 150/40 = 3.75 hours doing so.
If you travel CB = 150 miles at 50 mph, then you spend 150/50 = 3 hours doing so.
The total time going from A to B is 3.75+3 = 6.75 hours.
If you take 1 hour longer going from B to A, then you travel for 6.75+1 = 7.75 hours
2 of those hours is spent traveling that 70 miles mentioned earlier (from point B to point D), so we have 7.75-2 = 5.75 hours left to travel the remaining distance of 300-70 = 230 miles.
The last leg of the trip, from D to A, has us going 40 mph for 5.75 hours
distance = rate*time
distance = 40*5.75
distance = 230
This matches with the 230 we found in the previous paragraph, so this confirms our answer.
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Answer: Distance from A to B =
300 miles