Kevin drove from A to B at a constant speed of 60 mph. Once he reache

Let the distance between A and B = d. Therefore, the time from A to B is d/60 and the time from B to A is d/80, and thus the total time for the round trip is d/60 + d/80 = 4d/240 + 3d/240 = 7d/240. Since we are given that exactly 4 hours before the end of his trip, he was still approaching B (and was thus still traveling at 60 mph), only 15 miles away from it, we can create the equation:

60(7d/240 - 4) = d - 15

7d/4 - 240 = d - 15

7d - 960 = 4d - 60

3d = 900

d = 300

Alternate Solution:

Let’s concentrate only on the 4-hour time period. During this time, he was still going from A to B at a rate of 60 mph (or 1 mile per minute), and he traveled 15 miles to actually get to B. Thus, he traveled 15 miles in 15 minutes. This means that the remaining 3 hours and 45 minutes of the 4-hour travel time was all used to get from B back to A. During these 3.75 hours, he traveled at a rate of 80 mph; thus, the distance from B to A = 3.75 x 80 = 300 miles.

Answer: B

He returned with 80 mph so 80*3.75 hours that is 300

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\(? = d\,\,\,\left[ {{\text{miles}}} \right]\)

Excellent opportunity to use UNITS CONTROL, one of the most powerful tools of our course!

\(\left[ {\text{h}} \right] = \frac{{\left[ {{\text{miles}}} \right]}}{{\left[ {{\text{mph}}} \right]}}\)

From "Exactly 4 hours before the end of his trip, he was still approaching B, only 15 miles away from it. " we have (see image attached):

\(4 = {T_{{\text{CB}}}} + {T_{{\text{BC}}}} + {T_{{\text{CA}}}} = \frac{{15}}{{60}} + \frac{{15}}{{80}} + \frac{{d - 15}}{{80}} = \frac{{1 \cdot \boxed{20}}}{{4 \cdot \boxed{20}}} + \frac{d}{{4 \cdot 20}} = \frac{{d + 20}}{{80}}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\text{h}} \right]\)

\(\frac{{d + 20}}{{80}} = 4\,\,\,\, \Rightarrow \,\,\,\,\,? = d = 4 \cdot 80 - 20 = 300\,\,\,\,\,\left[ {{\text{miles}}} \right]\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Use the answers and avoid calculations. 60 miles per hrs is the constant speed on the way to b and 80 on the way back.
Lets use a multiple of both numbers such as 320 the total distance.

If this was the case, at 60 miles per hour it would take him 8 hrs to do the trip and 4 hrs to do on the way back.

The question mentioned that 4 hrs before the end of the trip he was still on the way to be, which means that total distance must be a bit less that 320.

Correct answer is 300

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