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# It takes the high-speed train x hours to travel the z miles

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Director
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It takes the high-speed train x hours to travel the z miles [#permalink]

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04 Oct 2008, 07:08
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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) z(y – x) / x + y
(2) z(x – y) / x + y
(3) z(x + y) / y – x
(4) xy(x – y) / x + y
(5) xy(y – x) / x + y
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Re: MGMAT - TownA to TownB [#permalink]

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04 Oct 2008, 12:59
Speed of the High Speed Train = Z/X
Speed of the regular train = Z/Y

The combined speed of the two trains multiplied by the time T that it takes for them to pass each other is equal to the total distance, Z.

So

$$T(Z/X + Z/Y) = Z$$
$$T((ZY + ZX)/XY) = Z$$
$$T((X+Y)/XY)=1$$
$$(X+Y)/XY=1/T$$
$$XY/(X+Y) = T$$

To find the distance each train has traveled at time T, simply multiply T by the speed of each train. To answer the question, subtract the distances.

$$(Z/X)(XY/(X+Y)) - (Z/Y)(XY/(X+Y))$$
$$XYZ / X(X+Y) - XYZ / Y(X+Y)$$
$$(XY^2Z(X+Y) - X^2YZ(X+Y))/XY(X+Y)^2$$

Cancel out a Y, an X and an (X+Y) and you get

$$(YZ-XZ)/(X+Y)$$

Factor out the Z in the numerator and you get

$$Z(Y-X)/(X+Y)$$ which is the first answer. However, none of the denominators in any of the answers have parentheses around the denominator, so I don't see how any of the given answer choices could actually be right, I guess I might be missing something.

But I got $$Z(Y-X)/(X+Y)$$
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Re: MGMAT - TownA to TownB [#permalink]

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04 Oct 2008, 14:45

Speed(High Speed) = z/x
Speed(Regular) = z/y

Lets say 'a' is the distance high speed train travelled
Since they both started at the same time, t1 = t2

a/z/x = (z-a)/z/y
ax = zy -ay
a = z/(x+y)

Difference is a-(z-a) = 2a-z = 2zy/x+y -z = z(y-x)/(x+y)
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Re: MGMAT - TownA to TownB [#permalink]

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06 Oct 2008, 06:53
How do you solve this in 2 mins? Does plugin work?
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Re: MGMAT - TownA to TownB [#permalink]

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06 Oct 2008, 12:59
pawan203 wrote:
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) z(y – x) / x + y
(2) z(x – y) / x + y
(3) z(x + y) / y – x
(4) xy(x – y) / x + y
(5) xy(y – x) / x + y

Both train spend the same time traveling = t
Velocity(X) = Z/X
Velocity(Y) = Z/Y

Distance(X) = rate X * time = (Z/X)t
Distance(Y) = rate Y * time = (Z/Y)t

Distance(X) + Distance(Y) = Z
(Z/X)t + (Z/Y)t = Z
(t/X) + (t/Y) = 1 ---------- divides the above equation with Z
t(1/X + 1/Y) = 1
t((X+Y) / XY) = 1
t = XY / (X+Y) ---------- (1)

The question want to know Distance(X) - Distance(Y)
= (Z/X)t - (Z/Y)t ---------- (2)
= (Z/X)(XY / (X+Y) - (Z/Y)(XY / (X+Y)) ---------- (1) & (2) combined
= ZY/(X+Y) - ZX/(X+Y)
= Z(Y-X)/(X+Y)

The answer is A ( or (1) in this case).
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Re: MGMAT - TownA to TownB [#permalink]

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06 Oct 2008, 22:13
I tried to solve this by plug ins...
z= 100 (distance)
x=2 (Time taken High speed train)
speed of high speed train =50
y=5 (Time taken Regular train speed)
speed of regular train =20
Let d be the distance high speed train covers before they meet.
Since both the trains will meet at time t, we can put the equation as
d/50=(100-d)/20
solving this we get d=500/7
so (100-d)=200/7
Difference=300/7
subsituting the values of x , y, z only z(y-x) / x+y gives the value 300/7...n so is the ans
Re: MGMAT - TownA to TownB   [#permalink] 06 Oct 2008, 22:13
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