It takes the high-speed train x hours to travel the z miles from Town A to Town B at a constant rate, while it takes the regular train y hours to travel the same distance at a constant rate. If the high-speed train leaves Town A for Town B at the same time that the regular train leaves Town B for Town A, how many more miles will the high-speed train have traveled than the regular train when the two trains pass each other?
A)z(y - x) / (x + y)
B)z(x - y) / (x + y)
C)z(x + y) / (y - x)
D)xy(x - y) / (x + y)
E)xy(y - x) / (x + y)
Plug in easy numbers for the two rates.
Then plug in a distance that is a multiple of both the two individual rates and the COMBINED rate.
Let the high speed rate = 3 miles per hour.
Let the regular rate = 2 miles per hour.
Combined rate for the trains = 2+3 = 5 miles per hour.
Let z = 30 miles.
Time for the high speed train to travel 30 miles = x = 30/3 = 10 hours.
Time for the regular train to travel 30 miles = y = 30/2 = 15 hours.
Time for the trains to meet = 30/5 = 6 hours.
Distance traveled by the high speed train in 6 hours = r*t = 3*6 = 18 miles.
Distance traveled by the regular train in 6 hours = r*t = 2*6 = 12 miles.
Distance for the high speed train - distance for the regular train = 18-12 = 6. This is our target.
Now plug x=10, y=15, and z=30 into the answers to see which yields our target of 6.
Only answer choice
A works:
z(y - x)/(x+y) = 30*(15-10)/(10+15) = 6.
Algebra:
Rule:
If A takes x hours to do a job, and B takes y hours to a job, the combined rate for A and B working together = (x+y)/(xy), and the total time for A and B working together = (xy)/(x+y).
In the problem above:
When the two trains travel toward each other, they WORK TOGETHER to cover the z miles between them.
Since the time for the high-speed train = x, and the time for the regular train = y, the total time for the two trains working together = (xy)/(x+y).
Thus:
Since the rate for the high-speed train = d/t = z/x, the distance traveled by the high-speed train = rt = (z/x) * (xy)/(x+y) = (zy)/(x+y).
Since the rate for the regular train = d/t = z/y, the distance traveled by the regular train = rt = (z/y) * (xy)/(x+y) = (zx)/(x+y).
Difference between the distances = (zy)/(x+y) - (zx)/(x+y) = [(z)(y-x)]/(x+y).
The correct answer is
APrivate tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at
[email protected].
Student Review #1
Student Review #2
Student Review #3
