Colliding Trains

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[email protected]: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Mon Apr 26, 2010 9:30 pm; Location: Johannesburg, South Africa

by [email protected] » Sun Nov 14, 2010 12:08 am

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)

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Source: Beat The GMAT — Problem Solving | Sun Nov 14, 2010 12:08 am

beat_gmat_09: Master | Next Rank: 500 Posts; Posts: 437; Joined: Sat Nov 22, 2008 5:06 amFollowed by: 1 member; Location: India; Thanked: 50 times; GMAT Score:580

by beat_gmat_09 » Sun Nov 14, 2010 12:19 am

[email protected] 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?

(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)

S(f) = z/x
S(r) = z/y
Relv speed = z/x+z/y
Time taken for dis z = z/ [z(1/x+1/y )] = xy/(x+y)
dis by Faster = S(f) * t = z/x * (xy/(x+y)) = zy/(x+y)
dis by Regular = S(r) * t = z/y * (xy/(x+y)) = zx/(x+y)
Distance travelled by faster is more than regular by = zy/(x+y) - zx/(x+y) = z(y-x)/(x+y)

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by Rahul@gurome » Sun Nov 14, 2010 12:34 am

[email protected] 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?
(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)

This question is a good example of option elimination method.
We can proceed for option elimination as follows,

(1) Option D and E are not possible because of dimension mismatch.
(2) Option C is not possible because [(x + y)/(y - x)] is always greater than 1, thus option C is greater than z, which is not possible.
(3) Option B results in negative distance (as x is smaller than y), which is also not possible.

Only possible option is option A.
[spoiler]
The correct answer is A.[/spoiler]

Last edited by Rahul@gurome on Sun Nov 14, 2010 1:52 am, edited 1 time in total.

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[email protected]: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Mon Apr 26, 2010 9:30 pm; Location: Johannesburg, South Africa

by [email protected] » Sun Nov 14, 2010 12:54 am

beat_gmat_09 wrote:
[email protected] 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?

(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)
S(f) = z/x
S(r) = z/y
Relv speed = z/x+z/y
Time taken for dis z = z/ [z(1/x+1/y )] = xy/(x+y)
dis by Faster = S(f) * t = z/x * (xy/(x+y)) = zy/(x+y)
dis by Regular = S(r) * t = z/y * (xy/(x+y)) = zx/(x+y)
Distance travelled by faster is more than regular by = zy/(x+y) - zx/(x+y) = z(y-x)/(x+y)

what is 'Relv speed = z/x+z/y'? you lost me.

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by GMATGuruNY » Sun Nov 14, 2010 3:38 am

[email protected] 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?

(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)

Since there are variables in the answer choices, we can plug in values.

Let x = 2/hr
Let y = 3/hr
Combined rate for x+y = 2+3 = 5/hr
Let z =10 miles
Time for trains to meet = d/r = 10/5 = 2hrs.
Distance for x = r*t = 2*2 = 4 miles.
Distance for y = r*t = 2*3 = 6 miles.
Distance for y - distance for x = 6-4 = 2. This is our target answer.

Only answer choice A works:
z(y - x)/(x+y) = 10*(3-2)/(2+3) = 10/5 = 2.

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beat_gmat_09: Master | Next Rank: 500 Posts; Posts: 437; Joined: Sat Nov 22, 2008 5:06 amFollowed by: 1 member; Location: India; Thanked: 50 times; GMAT Score:580

by beat_gmat_09 » Sun Nov 14, 2010 4:48 am

[email protected] wrote: what is 'Relv speed = z/x+z/y'? you lost me.

I just got the proof, thought i'd share, z/x is speed for faster train and z/y is for regular and their directions are opposite.

Relative speed -
1) If A with speed x km/hr goes in one direction and B with speed y km/hr goes in the same direction as A then
relative speed - x-y km/hr (x>y)

2) If A with speed x km/hr goes in one direction and B with speed y km/hr goes in the opposite direction i.e. towards A then
relative speed - x+y km/hr

Example -
For 2) A and B in opposite direction. Distance between A and B = D. Refer fig.

Assume distance covered by A = d km then distance covered by B when A and B meet = (D-d) km
S(A) = d/t; S(B)=D-d/t
S(A) = x = d/t; S(B) = y = (D-d)/t
time taken by both is same, therefore equate for t.
d/x = (D-d)/y
yd = x(D-d)
yd = xD-xd
yd+xd = xD
d(y+x) = xD
d = xD/(y+x)

time t = d/S(A) = xD/(y+x)/x = D(y+x)
Therefore Relative speed = (x+y) = Distance/Time = D/t

For 1) A and B in same direction. Refer fig.

Suppose train A starts from the point show and B starts from its point at the same time and B is D km (this distance will be know in the problem) ahead of A. Assume that A and B meet at certain point d km from starting point of B. Then total distance to be covered by A is D + d
Distance to be covered by A in time 't' , in this time frame 't' B has moved d km from its point. Hence time taken by A and B is same in order to meet at some point.
x = D+d/t ;
y = d/t
As time 't' taken by both A & B is same equate t.
(D+d)/x = d/y
y(D+d) = xd
yD = d(x-y) (obviously x has to be > y in order to catch up with B)
d = yD/(x-y)
Time t = d/y = yD/y(x-y) = D/(x-y).

Therefore relative speed = (x-y) = Distance/time = D/t

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