BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Colliding Trains

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 27
Joined: Mon Apr 26, 2010 9:30 pm
Location: Johannesburg, South Africa

Colliding Trains

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)
Top
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 437
Joined: Sat Nov 22, 2008 5:06 am
Location: India
Thanked: 50 times
Followed by:1 members
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)
Hope is the dream of a man awake
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pm
Location: Milpitas, CA
Thanked: 447 times
Followed by:88 members

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.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Top
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.
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

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.
Private 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
Top
Master | Next Rank: 500 Posts
Posts: 437
Joined: Sat Nov 22, 2008 5:06 am
Location: India
Thanked: 50 times
Followed by:1 members
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.

Image


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.

Image


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
Hope is the dream of a man awake
Top
Post new topic Post Reply
• Page 1 of 1