Bunuel 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 mi

B. 300 mi

C. 320 mi

D. 350 mi

E. 390 mi

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

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.

Attachments

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Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or GMATH.com.br (Portuguese version)

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