Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 04 Dec 2008
Posts: 66

It takes the highspeed train x hours to travel the z miles
[#permalink]
Show Tags
Updated on: 07 Feb 2012, 06:34
Question Stats:
55% (03:00) correct 45% (02:59) wrong based on 856 sessions
HideShow timer Statistics
It takes the highspeed 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 highspeed 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 highspeed 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
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by joyseychow on 19 May 2009, 22:36.
Last edited by Bunuel on 07 Feb 2012, 06:34, edited 1 time in total.
Edited the question and added the OA




Math Expert
Joined: 02 Sep 2009
Posts: 64951

Re: Manhattan CAT math question
[#permalink]
Show Tags
21 May 2010, 03:10
Prax wrote: Hi,
I have another doubt:
It takes the highspeed 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 highspeed 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 highspeed train have traveled than the regular train when the two trains pass each other?
z(y – x)/x + y
z(x – y)/x + y z(x + y)/y – x
xy(x – y)/x + y
xy(y – x)/x + y
Please help me with this question. It takes the highspeed train x hours to travel the z miles > rate of highspeed train is \(rate_{highspeed}=\frac{distance}{time}=\frac{z}{x}\); It takes the regular train y hours to travel the same distance > rate of regular train is \(rate_{regular}=\frac{distance}{time}=\frac{z}{y}\); Time in which they meet is \(time=\frac{distance}{combinedrate}=\frac{z}{\frac{z}{x}+\frac{z}{y}}=\frac{xy}{x+y}\). Difference in distances covered: {Time}*{Rate of highspeed train}  {Time}{Rate of regular train} > \(\frac{xy}{x+y}*\frac{z}{x}\frac{xy}{x+y}*\frac{z}{y}=\frac{z(yx)}{x+y}\). Answer: A.
_________________




Manager
Joined: 08 Jan 2009
Posts: 200

Re: Man Cat 4 #12High speed train v. Regular Train
[#permalink]
Show Tags
19 May 2009, 23:04
Let Sh be the speed of the faster train Let Sl be the speed of the slower train.
Sh = z/x and Sl = z/y
Since the move towards each other relative speed = Sl + Sh = z/x + z/y
The time the both trains meet t = z / z/x + z/y = xy / x + y
The distance travelled by Sh = Sh * t = z/x * xy / x + y = zy/ x +y The distance travelled by Sl = Sl * t = z/y * xy / x + y = zx/ x +y
Now the Number of mile more = zy/ x +y  zx/ x +y = z( yx)/ x +y
Ans A




Intern
Joined: 24 Sep 2009
Posts: 22

Re: Man Cat 4 #12High speed train v. Regular Train
[#permalink]
Show Tags
29 Nov 2009, 13:26
It takes the highspeed 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 highspeed 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 highspeed 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
 (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.
 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).  Benjiboo



Intern
Joined: 14 Feb 2010
Posts: 47

Re: Manhattan CAT math question
[#permalink]
Show Tags
21 May 2010, 03:30
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 zd
When they meet, the time taken by both is same. So,
d/x = (zd)/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  (zd) which is equal to 2d  z = 2z*x / (x+y)  z = 2z*x  z*x  z*y / (x+y) = z(xy)/x+y
which is option b
please let me know where am I going wrong.



Math Expert
Joined: 02 Sep 2009
Posts: 64951

It takes the highspeed train x hours to travel the z miles
[#permalink]
Show Tags
21 May 2010, 03:56
Prax wrote: 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 zd
When they meet, the time taken by both is same. So,
d/x = (zd)/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  (zd) which is equal to 2d  z = 2z*x / (x+y)  z = 2z*x  z*x  z*y / (x+y) = z(xy)/x+y
which is option b
please let me know where am I going wrong. You can solve this way too, but you made a mistake in calculation: \(time=\frac{distance}{rate}=\frac{d}{\frac{z}{x}}=\frac{zd}{\frac{z}{y}}\) > \(dx=zydy\) (not d/x = (zd)/y) > \(d=\frac{zy}{x+y}\). Difference \(2dz=\frac{2zy}{x+y}z=\frac{z(yx)}{x+y}\). Answer: A. You could spot that answer B cannot be the correct choice as it's negative (numerator xy<0) (high speed train needs less time to cover the distance than regular train, so x<y) but the difference in distances can not be negative as high speed train would cover greater distance than regular train when they meet (for exact same reason choice D can be eliminated as well). Hope it's clear.
_________________



Intern
Joined: 21 Feb 2010
Posts: 3

Re: Highspeed train
[#permalink]
Show Tags
21 May 2010, 05:45
The problem can be solved very easily with the concepts of Relative velocity.. if we assume that slow moving tarin is at stand still and high speed train is moving with the speed : x+y time taken to travel the distance : z/(x+y) diffrence in the distance tarvelled = distance traveled by high speed tarin  distance traveled by slow moving train = x[z/(x+y)]  y[z/(x+y)] = z(xy)/(x+y) hope it will be helpfull..



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10629
Location: Pune, India

Re: It takes the highspeed train x hours to travel the z miles
[#permalink]
Show Tags
07 Feb 2012, 08:03
joyseychow wrote: It takes the highspeed 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 highspeed 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 highspeed 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 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(yx)/(x+y) more than regular train.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Joined: 28 Jul 2011
Posts: 152

Re: It takes the highspeed train x hours to travel the z miles
[#permalink]
Show Tags
01 Apr 2012, 07:17
Vote for A
S * T = DIST
Given
High ST  xhrs to travel Z miles Regular ST = yhrs to travel z miles
therefore HST speed = z/x RST speed = z/y
Trains start at same time = t
Travel towards each other [z/x + z/y] * t = z t = t / [z/x + z/y] t = zxy / z(x+y) t = xy / (x+y)
Dist covered by HST = zt/x Dist covered by RST = zt/y
How much more did HST travel = zt/x  zt/y = zt(yz) / xy = z [xy / (x+y)] * [(yz) / xy] = z (yz) / (x+y)



Director
Joined: 22 Mar 2011
Posts: 576
WE: Science (Education)

Re: MGMAT
[#permalink]
Show Tags
26 Sep 2012, 11:43
Bunuel wrote: ishanand wrote: It takes the highspeed 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 highspeed 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 highspeed 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) Merging similar topics. Please ask if anything remains unclear. 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 highspeed 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.
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Intern
Joined: 19 Aug 2012
Posts: 4
Location: United States
Concentration: Strategy, Marketing
GMAT Date: 11202012
GPA: 3.19
WE: Information Technology (Computer Software)

Re: Man Cat 4 #12High speed train v. Regular Train
[#permalink]
Show Tags
15 Oct 2012, 15:48
gurpreetsingh wrote: joyseychow wrote: It takes the highspeed 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 highspeed 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 highspeed train have traveled than the regular train when the two trains pass each other? (A) z(y – x)/x + y > Contender (B) z(x – y)/x + y > y>x as B is slow train thus time taken by B > than by A> Wrong (C) z(x + y)/y – x > Distance between them is Z. This is greater than Z. Not possible (D) xy(x – y)/x + y > Wrong ..same reason as B (E) xy(y – x)/x + y >The dimensions of this expression are not of Distance. Wrong
Though I had calculated this during my CAT, the best way to solve is as stated above. Awesome method! Should stick on to this.. And I tried by cooking up few simple values, calculated the answer required and substituted it in the equations given. The one which satisfies the values would be my option! I tried with the following values.. x = 3 y = 2 z = 60 ( a value divisible by both x and y, aid for simple calculation) the time when both the trains would meet = z /(x+y) [Relative Speed theory] = 12 mins Distance traveled by Train A = 36 miles Distance traveled by Train B = 24 miles Difference = 12 miles (the actual answer expected while substituting the given values in equation) Substituting the values of x,y and z in option A, => z(xy)/(x+y) => 60(32)/(3+2) => 12  equates the value expected. Hence answer is option A. Though this method takes a considerable time of explanation, it takes less than a minute to solve this way. But the choice of assumption values must be small and should make calculations easier.



Senior Manager
Joined: 13 Aug 2012
Posts: 386
Concentration: Marketing, Finance
GPA: 3.23

Re: It takes the highspeed train x hours to travel the z miles
[#permalink]
Show Tags
01 Dec 2012, 19:49
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 highspeed train minus regular train:
\((\frac{z}{x})(\frac{xy}{x+y})(\frac{z}{y})(\frac{xy}{x+y})\)
Answer: \(z(\frac{yx}{x+y})\)



Manager
Joined: 07 Feb 2011
Posts: 89

Re: Manhattan CAT math question
[#permalink]
Show Tags
28 Mar 2013, 12:17
Bunuel wrote: Prax wrote: Hi,
I have another doubt:
It takes the highspeed 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 highspeed 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 highspeed train have traveled than the regular train when the two trains pass each other?
z(y – x)/x + y
z(x – y)/x + y z(x + y)/y – x
xy(x – y)/x + y
xy(y – x)/x + y
Please help me with this question. It takes the highspeed train x hours to travel the z miles > rate of highspeed train is \(rate_{highspeed}=\frac{distance}{time}=\frac{z}{x}\); It takes the regular train y hours to travel the same distance > rate of regular train is \(rate_{regular}=\frac{distance}{time}=\frac{z}{y}\); Time in which they meet is \(time=\frac{distance}{combinedrate}=\frac{z}{\frac{z}{x}+\frac{z}{y}}=\frac{xy}{x+y}\). Difference in distances covered: {Time}*{Rate of highspeed train}  {Time}{Rate of regular train} > \(\frac{xy}{x+y}*\frac{z}{x}\frac{xy}{x+y}*\frac{z}{y}=\frac{z(yx)}{x+y}\). Answer: A. Hey can someone help me. So something I sort of missed was why we combine the rate (understood this) but then to figure out the time the trains cross each other to simply divide the distance z by the combined rates? The distance z is the total distance for A to B. Which is different for what we're looking for, no? Aren't we looking for the time (and hence corresponding) distance where A and B cross each other? That is not the same as z (to me) but should be shorter than z



Math Expert
Joined: 02 Sep 2009
Posts: 64951

Re: Manhattan CAT math question
[#permalink]
Show Tags
28 Mar 2013, 12:28
manimgoindowndown wrote: Bunuel wrote: Prax wrote: Hi,
I have another doubt:
It takes the highspeed 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 highspeed 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 highspeed train have traveled than the regular train when the two trains pass each other?
z(y – x)/x + y
z(x – y)/x + y z(x + y)/y – x
xy(x – y)/x + y
xy(y – x)/x + y
Please help me with this question. It takes the highspeed train x hours to travel the z miles > rate of highspeed train is \(rate_{highspeed}=\frac{distance}{time}=\frac{z}{x}\); It takes the regular train y hours to travel the same distance > rate of regular train is \(rate_{regular}=\frac{distance}{time}=\frac{z}{y}\); Time in which they meet is \(time=\frac{distance}{combinedrate}=\frac{z}{\frac{z}{x}+\frac{z}{y}}=\frac{xy}{x+y}\). Difference in distances covered: {Time}*{Rate of highspeed train}  {Time}{Rate of regular train} > \(\frac{xy}{x+y}*\frac{z}{x}\frac{xy}{x+y}*\frac{z}{y}=\frac{z(yx)}{x+y}\). Answer: A. Hey Bunuel so something I sort of missed was why we combine the rate (understood this) but then to figure out the time the trains cross each other to simply divide the distance z by the combined rates? The distance z is the total distance for A to B. Which is different for what we're looking for, no? Aren't we looking for the time (and hence corresponding) distance where A and B cross each other? 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?
_________________



Intern
Joined: 18 Nov 2011
Posts: 29
Concentration: Strategy, Marketing
GMAT Date: 06182013
GPA: 3.98

Re: Manhattan CAT math question
[#permalink]
Show Tags
28 Mar 2013, 15:05
Bunuel wrote: Hi,
Time in which they meet is \(time=\frac{distance}{combinedrate}=\frac{z}{\frac{z}{x}+\frac{z}{y}}=\frac{xy}{x+y}\).
I must be making a dumb algebra error somewhere in this step. How did you convert the fraction like this? I keep ending up with somehting slightly different



Math Expert
Joined: 02 Sep 2009
Posts: 64951

Re: Manhattan CAT math question
[#permalink]
Show Tags
29 Mar 2013, 02:01
hitman5532 wrote: Bunuel wrote: Hi,
Time in which they meet is \(time=\frac{distance}{combinedrate}=\frac{z}{\frac{z}{x}+\frac{z}{y}}=\frac{xy}{x+y}\).
I must be making a dumb algebra error somewhere in this step. How did you convert the fraction like this? I keep ending up with somehting slightly different \(\frac{z}{\frac{z}{x}+\frac{z}{y}}=\frac{z}{\frac{zy+zx}{xy}}=\frac{xyz}{zy+zx}=\frac{xy}{x+y}\).
_________________



Director
Joined: 17 Dec 2012
Posts: 631
Location: India

Re: It takes the highspeed train x hours to travel the z miles
[#permalink]
Show Tags
29 Mar 2013, 02:54
joyseychow wrote: It takes the highspeed 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 highspeed 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 highspeed 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 Let us call the trains as H and R resp. Given: Distance = z Time taken by H = x Time taken by R = y Question: When the trains meet how much more distance has H traveled than R? For that we need to calculate the speed of both the trains and the time taken for them to meet. Deductions: Speed of H = z/x Speed of R= z/y To calculate the time taken for them to meet we need to use the total distance between the two towns and the combined speed as they are moving towards each other. Time taken for the trains to meet= z/(z/x +z/y) = xy/(x+y) Distance traveled by H when the trains meet= time taken to meet* speed of H =xy/(x+y) * z/x =zy/(x+y) Similarly for R , we have the distance = zx/(x+y) The difference is z(yx)/(x+y) Hence choice A.
_________________
Srinivasan Vaidyaraman Magical LogiciansHolistic and Holy Approach



Manager
Joined: 07 Feb 2011
Posts: 89

Re: Manhattan CAT math question
[#permalink]
Show Tags
29 Mar 2013, 05:04
Hey Bunuel so something I sort of missed was why we combine the rate (understood this) but then to figure out the time the trains cross each other to simply divide the distance z by the combined rates?
The distance z is the total distance for A to B. Which is different for what we're looking for, no? Aren't we looking for the time (and hence corresponding) distance where A and B cross each other?[/quote]
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?[/quote]
Combining the rates makes sense to me, they are both moving to each other relatively
Here's what doesn't make sense to me.
The distance between the two starting points of the train is 100 miles (z). If we are trying to find what time they will pass each other, that distance MUST BE less than 100 if both trains have a positive velocity.
This distance is less than the starting points of the train from 100 miles (z)?
So I don't see how we can simple plug in z here. Maybe there's a test assumption that simplifies this situation for us.



Math Expert
Joined: 02 Sep 2009
Posts: 64951

Re: Manhattan CAT math question
[#permalink]
Show Tags
29 Mar 2013, 05:43
manimgoindowndown wrote: Hey Bunuel so something I sort of missed was why we combine the rate (understood this) but then to figure out the time the trains cross each other to simply divide the distance z by the combined rates?
The distance z is the total distance for A to B. Which is different for what we're looking for, no? Aren't we looking for the time (and hence corresponding) distance where A and B cross each other? 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?[/quote] Combining the rates makes sense to me, they are both moving to each other relatively Here's what doesn't make sense to me. The distance between the two starting points of the train is 100 miles (z). If we are trying to find what time they will pass each other, that distance MUST BE less than 100 if both trains have a positive velocity. This distance is less than the starting points of the train from 100 miles (z)? So I don't see how we can simple plug in z here. Maybe there's a test assumption that simplifies this situation for us.[/quote] "Pass" here means "meet".
_________________



Intern
Joined: 02 May 2013
Posts: 18
Concentration: International Business, Technology
WE: Engineering (Aerospace and Defense)

Re: It takes the highspeed train x hours to travel the z miles
[#permalink]
Show Tags
29 Jul 2013, 18:16
speed of high speed train, Vh=z/x speed of regular train, Vr=z/y Let p be the distance covered by high speed train when both trains met Time taken to meet both trains =px/z=(zp)y/z p=zy/(x+y) but required is how much more distance covered by high speed train than regular train i.e required = 2pz = zy/(xy) z =z(yx)/x+y




Re: It takes the highspeed train x hours to travel the z miles
[#permalink]
29 Jul 2013, 18:16



Go to page
1 2 3
Next
[ 41 posts ]

