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

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Re: Kevin drove from A to B at a constant speed of 60 mph. Once he reache  [#permalink]

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New post 12 Dec 2017, 02:00
Bunuel wrote:
Bunuel wrote:
Kevin drove from A to B at a constant speed of 60 mph. Once he reached B, he turned right around with 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



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MAGOOSH OFFICIAL SOLUTION:

In the last 15 miles of his approach to B, Kevin was traveling at 60 mph, so he traveled that distance in ¼ hr, or 15 minutes. That means, when he arrived at B, 15 minutes had elapsed, and he took (4 hr) – (15 min) = 3.75 hr to drive the distance D at 80 mph. It will be easier to leave that time in the form (4 hr) – (15 min).

D = RT = (80 mph)[ (4 hr) – (15 min)] = 320 mi – 20 mi = 300 mi

Answer = (B)


mikemcgarry
shouldn't the question say "without pause"? Because if there is a pause, then without knowing for how much time he paused, we can't find the distance
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Re: Kevin drove from A to B at a constant speed of 60 mph. Once he reache  [#permalink]

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New post 12 Dec 2017, 03:35
Bunuel wrote:
Kevin drove from A to B at a constant speed of 60 mph. Once he reached B, he turned right around with 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



Kudos for a correct solution.


Let the distance between A and B be x.
So, Time taken from A to B = x/60
Time taken from B to A = x/80
Total time taken = x/60 + x/80

So, time taken 4 hours before completing the trip = x/60 + x/80 - 4

distance traveled 4 hours before completing the trip = (x/60 + x/80 - 4)*60 as Kevin was approaching B.
Also acc. to question, distance traveled 4 hours before completing the trip = x - 15

So, (x/60 + x/80 - 4)*60 = x - 15
x = 300 mi

Answer B
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Re: Kevin drove from A to B at a constant speed of 60 mph. Once he reache  [#permalink]

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New post 12 Dec 2017, 17:22
ShashankDave wrote:
mikemcgarry
shouldn't the question say "without pause"? Because if there is a pause, then without knowing for how much time he paused, we can't find the distance

Dear ShashankDave,

Yes, that's a typo. Apparently it wasn't copied correctly from the source. I made the correction to the problem at the top of the thread. Thank you for pointing this out.

Mike :-)
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Re: Kevin drove from A to B at a constant speed of 60 mph. Once he reache  [#permalink]

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New post 11 Sep 2018, 17:28
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


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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Re: Kevin drove from A to B at a constant speed of 60 mph. Once he reache  [#permalink]

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New post 11 Sep 2018, 17:33
He returned with 80 mph so 80*3.75 hours that is 300

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Re: Kevin drove from A to B at a constant speed of 60 mph. Once he reache  [#permalink]

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New post 16 Sep 2018, 20:20
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.
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Re: Kevin drove from A to B at a constant speed of 60 mph. Once he reache   [#permalink] 16 Sep 2018, 20:20

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