Rate and Distance

Roland2rule wrote:Kevin drove from X to Y at a constant speed of 60 mph. Once he reached Y, he turned right around with pause, and returned to X at a constant speed of 80 mph. Exactly 4 hours before the end of his trip, he was still approaching Y, only 15 miles away from it. What is the distance between X and Y?
A. 275 mi
B. 300 mi
C. 320 mi
D. 350 mi
E. 390 mi
qa is b.

I'm confused how to set up the formulas here. I am getting E as the answer. Can any experts help?

Say the distance between X and Y is D miles

Thus,

Time taken to reach Y from X = D/60 hours
Time taken to reach X from Y = D/80 hours

Total travel time = D/60 + D/80 = 7D/240 hours

Since 4 hours before the end of his trip, he was still approaching Y, only 15 miles away from it, we can deduce that in (7D/240 – 4) hours, he traveled (D – 15) miles at a speed of 60 mph.

Thus, 7D/240 – 4 = (D – 15)/60

=> D = 300 miles.

The correct answer: B

Hope this helps!

-Jay

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Roland2rule wrote:Kevin drove from X to Y at a constant speed of 60 mph. Once he reached Y, he turned right around with pause, and returned to X at a constant speed of 80 mph. Exactly 4 hours before the end of his trip, he was still approaching Y, only 15 miles away from it. What is the distance between X and Y?

A. 275 mi
B. 300 mi
C. 320 mi
D. 350 mi
E. 390 mi
qa is b.

I'm confused how to set up the formulas here. I am getting E as the answer. Can any experts help?

Another approach can be...

Say exactly 4 hours before the end of his trip, Kevin was at point A. Thus, From point A to Y and from Y to X, he takes 4 hours. See the below diagram.

X -------60mph-------A-------15 miles@60mph-------Y----------80mph------------X

-------------------------<-------------------------4 hours---------------------------------->

Time taken to travel from A to Y = 15/60 = 1/4 hours

Thus, time taken to travel from Y to X = 4 – 1/4 = 15/4 hours

X -------60mph-------A-------15 miles@60mph-------Y----------80mph------------X

--------------------------------------------------------------<--15/4 hours@80mph---->

Thus, the distance between Y and X = 80*(15/4) = 300 miles

The correct answer: D

Hope this helps!

-Jay

Download free ebook: Manhattan Review GMAT Quantitative Question Bank Guide
_________________
Manhattan Review GMAT Prep

Locations: New York | Barcelona | Manila | Melbourne | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.

BTGmoderatorRO wrote: ↑

Sat Sep 30, 2017 7:55 pm

Kevin drove from X to Y at a constant speed of 60 mph. Once he reached Y, he turned right around with pause, and returned to X at a constant speed of 80 mph. Exactly 4 hours before the end of his trip, he was still approaching Y, only 15 miles away from it. What is the distance between X and Y?
A. 275 mi
B. 300 mi
C. 320 mi
D. 350 mi
E. 390 mi
qa is b.

I'm confused how to set up the formulas here. I am getting E as the answer. Can any experts help?

Solution:
Note: The question should read “he turned right around without pause”.

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