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

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?

(1) Pick numbers and plug in
(2) Make sure you pick numbers that make the prompt true. IE: The rate of the faster train must be faster than that of the regular train.
(3) Set up an rate*time=distance chart
(4) They travel for the same time, so T is the time for each one
(5) They travel distance differences. Set these columns up in terms of "T" (RATE * T)
(6) However, we know the total distance they traveled combined is equal to D
(7) Pick a value to be D. I recommended you the value you chose for Z
(8) Set the value of the distances equal to Z, to solve for T
(9) Plug in the value for T into the items you set up in step 5
(10) Subtract what you get in step 9 from each other to find the difference
(11) Now plug in your variables into the answer choices and look for one that matches

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(Step 1) X=4, Y=6, Z=12
(Step 2) "Ok this holds true to the prompt, check!"
(Steps 3,4,5,6, and 7)
Train...............R.......*.......T.......=......D
Fast...............Z/X..............T...............D
Regular..........Z/Y..............T...............D
Total.............Combine.......T...............D

Train...............R.......*.......T.......=......D
Fast.................3...............T...............3T
Regular............2...............T...............2T
Total................5..............T...............12

(Step 8)
5T = 12
T = 12/5

(Step 9)
Train...............R.......*.......T.......=......D
Fast.................3.............12/5............7.2
Regular............2.............12/5............4.8
Total................5..............T...............12

(Step 10)
7.2 - 4.8 = 2.4

(Step 11)
Only answer A = 2.4 when you plug in our values for Z,X, and Y.

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A key to being able to solve this problem on GMAT Day is to understand that in a situation where trains or people are meeting, the total distance is going to be D (unless one explicitly traveled more) and the total time is going to be T (unless one left before the other).
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Benjiboo

I solved it this way:

Speed of high speed train =z/x
Speed of normal train = z/y

Let d be the distance covered by high speed train when they meet.
So distance covered by normal train is z-d

When they meet, the time taken by both is same. So,

d/x = (z-d)/y
d*y = z*x - d*x
d(y+x) = z*x
d = z*x / x+y

the difference between the distance would be d - (z-d) which is equal to 2d - z
= 2z*x / (x+y) - z
= 2z*x - z*x - z*y / (x+y)
= z(x-y)/x+y

which is option b

please let me know where am I going wrong.

ohh..thanks

Though I had calculated this during my CAT, the best way to solve is as stated above.
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You can also use ratios here.
Ratio of time taken by high speed:regular = x:y
Ratio of distance covered in same time by high speed:regular = y:x (inverse of ratio of speed)
So distance covered by high speed train will be y/(x+y) * z
and distance covered by regular train will be x(x+y) * z
High speed train will travel yz/(x+y) - xz/(x+y) = z(y-x)/(x+y) more than regular train.
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z represents distance, x and y represent time.
The answer requires an expression with units of distance.
We can immediately eliminate answers D and E, both have units of time squared and not distance.

The regular train is slower than the high-speed train, so necessarily y > x. We can eliminate choice B, being negative.

Since (x + y)/(y - x) > 1, we can eliminate choice C, as neither of the two trains could have traveled a distance greater than z until they passed each other.

We are left with the only choice A.
The above posts confirm that it is the correct answer.

Answer A.
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Rate of high speed train: \frac{z}{x}
Rate of regular train: \frac{z}{y}

t: time it takes for two trains to pass each other from opposite direction

\frac{z}{x}(t)+\frac{z}{y}(t)=z

Calculate t: t = \frac{x+y}{xy}

Distance travelled by high-speed train minus regular train:

(\frac{z}{x})(\frac{xy}{x+y})-(\frac{z}{y})(\frac{xy}{x+y})

Answer: z(\frac{y-x}{x+y})

Two trains are traveling to meet each other.

Distance = 100 miles;
Rate of train A = 20 miles per hour;
Rate of train B = 30 miles per hour.

In how many hours will they meet?

(Time) = (Distance)/(Combined rate) = 100/(20+30) = 2 hours.

Does this make sense?
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\frac{z}{\frac{z}{x}+\frac{z}{y}}=\frac{z}{\frac{zy+zx}{xy}}=\frac{xyz}{zy+zx}=\frac{xy}{x+y}.
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