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

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Kevin drove from A to B at a constant speed of 60 mph. Once he reached B, he turned right around without pause, and returned to A at a constant speed of 80 mph. Exactly 4 hours before the end of his trip, he was still approaching B, only 15 miles away from it. What is the distance between A and B?

A. 275 miles
B. 300 miles
C. 320 miles
D. 350 miles
E. 390 miles

The OA is B.

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by Scott@TargetTestPrep » Tue Sep 11, 2018 4:25 pm
swerve wrote:Kevin drove from A to B at a constant speed of 60 mph. Once he reached B, he turned right around without pause, and returned to A at a constant speed of 80 mph. Exactly 4 hours before the end of his trip, he was still approaching B, only 15 miles away from it. What is the distance between A and B?

A. 275 miles
B. 300 miles
C. 320 miles
D. 350 miles
E. 390 miles
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

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by [email protected] » Tue Sep 11, 2018 7:36 pm
Hi All,

We're told that Kevin drove from A to B at a constant speed of 60 mph, then he turned right around without pause, and returned to A at a constant speed of 80 mph. In addition, exactly 4 hours before the end of his trip, he was STILL approaching B, only 15 miles away from it. We're asked for the distance between A and B. This question can be solved in a number of different ways, including by TESTing THE ANSWERS.

To start, we know that with 4 hours remaining in the total travel time, Kevin was still approaching B. This means that the time he spent driving back from B to A was LESS than 4 hours. IF he had spent the FULL 4 hours driving from B to A, then he would have traveled 4(80mph) = 320 miles. That did not happen though - since he drove LESS time, the distance between B and A is LESS than 320 miles. We can eliminate Answer C, D and E. Let's TEST Answer B...

Answer B: 300 miles
IF... the distance from A to B is 300 miles....
then it takes 300 miles/60 mph = 5 hours to get from A to B...
and it takes 300 miles/80 mph = 30/8 = 15/4 = 3 3/4 hours to get from B to A
Since Kevin drove another 15 MILES to get to B (on the way from A to B), then that would have taken 15 MINUTES (since he was going 60 miles/hour). Those 15 minutes, plus the 3 hours 45 minutes he took to go from B to A, totals exactly 4 hours of drive time. This is an exact match for what we were told, so this MUST be the answer.

Final Answer: B

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by fskilnik@GMATH » Sun Sep 16, 2018 7:14 pm
swerve wrote:Kevin drove from A to B at a constant speed of 60 mph. Once he reached B, he turned right around without pause, and returned to A at a constant speed of 80 mph. Exactly 4 hours before the end of his trip, he was still approaching B, only 15 miles away from it. What is the distance between A and B?

A. 275 miles
B. 300 miles
C. 320 miles
D. 350 miles
E. 390 miles

Source: Magoosh
\[? = d\,\,\,\left[ {{\text{miles}}} \right]\]
Excellent opportunity to use UNITS CONTROL, one of the most powerful tools of our course!
\[\left[ {\text{h}} \right] = \frac{{\left[ {{\text{miles}}} \right]}}{{\left[ {{\text{mph}}} \right]}}\]
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From "Exactly 4 hours before the end of his trip, he was still approaching B, only 15 miles away from it. " we have:
\[4 = {T_{{\text{CB}}}} + {T_{{\text{BC}}}} + {T_{{\text{CA}}}} = \frac{{15}}{{60}} + \frac{{15}}{{80}} + \frac{{d - 15}}{{80}} = \frac{{1 \cdot \boxed{20}}}{{4 \cdot \boxed{20}}} + \frac{d}{{4 \cdot 20}} = \frac{{d + 20}}{{80}}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\text{h}} \right]\]
\[\frac{{d + 20}}{{80}} = 4\,\,\,\, \Rightarrow \,\,\,\,\,? = d = 4 \cdot 80 - 20 = 300\,\,\,\,\,\left[ {{\text{miles}}} \right]\]

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

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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