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

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

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19 May 2009, 23:36
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 07 Feb 2012, 07:34, edited 1 time in total.
Edited the question and added the OA

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Re: Manhattan CAT math question [#permalink]

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21 May 2010, 04:10
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Prax wrote:
Hi,

I have another doubt:

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?

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

It takes the high-speed train x hours to travel the z miles --> rate of high-speed train is $$rate_{high-speed}=\frac{distance}{time}=\frac{z}{x}$$;

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 $$time=\frac{distance}{combined-rate}=\frac{z}{\frac{z}{x}+\frac{z}{y}}=\frac{xy}{x+y}$$.

Difference in distances covered: {Time}*{Rate of high-speed train} - {Time}{Rate of regular train} --> $$\frac{xy}{x+y}*\frac{z}{x}-\frac{xy}{x+y}*\frac{z}{y}=\frac{z(y-x)}{x+y}$$.

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Re: Man Cat 4 #12-High speed train v. Regular Train [#permalink]

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20 May 2009, 00:04
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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( y-x)/ x +y

Ans A

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Re: Man Cat 4 #12-High speed train v. Regular Train [#permalink]

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22 Oct 2010, 20:38
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joyseychow 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 -> 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.
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21 May 2010, 06:45
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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(x-y)/(x+y)

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

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07 Feb 2012, 09:03
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joyseychow 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

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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17408 [5], given: 232 Math Expert Joined: 02 Sep 2009 Posts: 41913 Kudos [?]: 129536 [3], given: 12201 Re: Manhattan CAT math question [#permalink] ### Show Tags 21 May 2010, 04:56 3 This post received KUDOS Expert's post 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 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. You can solve this way too, but you made a mistake in calculation: $$time=\frac{distance}{rate}=\frac{d}{\frac{z}{x}}=\frac{z-d}{\frac{z}{y}}$$ --> $$dx=zy-dy$$ (not d/x = (z-d)/y) --> $$d=\frac{zy}{x+y}$$. Difference $$2d-z=\frac{2zy}{x+y}-z=\frac{z(y-x)}{x+y}$$. Answer: A. You could spot that answer B can not be the correct choice as it's negative (numerator x-y<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. _________________ Kudos [?]: 129536 [3], given: 12201 Director Joined: 22 Mar 2011 Posts: 610 Kudos [?]: 1060 [2], given: 43 WE: Science (Education) Re: MGMAT [#permalink] ### Show Tags 26 Sep 2012, 12:43 2 This post received KUDOS Bunuel wrote: ishanand 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) 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 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. _________________ PhD in Applied Mathematics Love GMAT Quant questions and running. Kudos [?]: 1060 [2], given: 43 Senior Manager Joined: 13 Aug 2012 Posts: 458 Kudos [?]: 542 [2], given: 11 Concentration: Marketing, Finance GPA: 3.23 Re: It takes the high-speed train x hours to travel the z miles [#permalink] ### Show Tags 01 Dec 2012, 20:49 2 This post received KUDOS 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: $$z(\frac{y-x}{x+y})$$ _________________ Impossible is nothing to God. Kudos [?]: 542 [2], given: 11 Math Expert Joined: 02 Sep 2009 Posts: 41913 Kudos [?]: 129536 [1], given: 12201 Re: Manhattan CAT math question [#permalink] ### Show Tags 28 Mar 2013, 13:28 1 This post received KUDOS Expert's post manimgoindowndown wrote: Bunuel wrote: Prax wrote: Hi, I have another doubt: 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? 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 high-speed train x hours to travel the z miles --> rate of high-speed train is $$rate_{high-speed}=\frac{distance}{time}=\frac{z}{x}$$; 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 $$time=\frac{distance}{combined-rate}=\frac{z}{\frac{z}{x}+\frac{z}{y}}=\frac{xy}{x+y}$$. Difference in distances covered: {Time}*{Rate of high-speed train} - {Time}{Rate of regular train} --> $$\frac{xy}{x+y}*\frac{z}{x}-\frac{xy}{x+y}*\frac{z}{y}=\frac{z(y-x)}{x+y}$$. 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? _________________ Kudos [?]: 129536 [1], given: 12201 Director Joined: 17 Dec 2012 Posts: 608 Kudos [?]: 521 [1], given: 16 Location: India Re: It takes the high-speed train x hours to travel the z miles [#permalink] ### Show Tags 29 Mar 2013, 03:54 1 This post received KUDOS Expert's post joyseychow 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 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(y-x)/(x+y) Hence choice A. _________________ Srinivasan Vaidyaraman Sravna http://www.sravnatestprep.com/regularcourse.php Pay After Use Standardized Approaches Kudos [?]: 521 [1], given: 16 SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1853 Kudos [?]: 2635 [1], given: 193 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Re: It takes the high-speed train x hours to travel the z miles [#permalink] ### Show Tags 19 Jun 2014, 21:44 1 This post received KUDOS Refer diagram below: Let the trains meet at point P Say the fast train has travelled distance "a" from point A, so the slow train travels distance "z-a" from point B We require to find the difference = a - (z-a) = 2a - z ......... (1) Time taken by high speed train = Time taken by slow train (To meet at point P) Setting up the equation $$\frac{a}{(\frac{z}{x})} = \frac{z-a}{(\frac{z}{y})}$$ $$a = \frac{yz}{x+y}$$ Placing value of a in equation (1) $$= \frac{2yz}{x+y} - z$$ $$= \frac{z(y-z)}{x+y}$$ Answer = A Attachments tr.jpg [ 15.71 KiB | Viewed 1659 times ] _________________ Kindly press "+1 Kudos" to appreciate Kudos [?]: 2635 [1], given: 193 Intern Joined: 24 Jul 2013 Posts: 14 Kudos [?]: 16 [1], given: 9 It takes the high-speed train x hours to travel the z miles [#permalink] ### Show Tags 10 Aug 2015, 05:04 1 This post received KUDOS There's an even faster way to do this. If x = y, z should be 0. This eliminates C. If y>>>>x, x/y should approach 0 and answer should approach z. D and E don't mention of z, so eliminated. Out of A & B, A approaches z, while B approaches -z. Hence, A is the answer. Kudos [?]: 16 [1], given: 9 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7677 Kudos [?]: 17408 [1], given: 232 Location: Pune, India Re: It takes the high-speed train x hours to travel the z miles [#permalink] ### Show Tags 24 Apr 2017, 01:14 1 This post received KUDOS Expert's post mbaprep2016 wrote: Bunuel wrote: Prax wrote: Hi, I have another doubt: 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? 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 high-speed train x hours to travel the z miles --> rate of high-speed train is $$rate_{high-speed}=\frac{distance}{time}=\frac{z}{x}$$; 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 $$time=\frac{distance}{combined-rate}=\frac{z}{\frac{z}{x}+\frac{z}{y}}=\frac{xy}{x+y}$$. Difference in distances covered: {Time}*{Rate of high-speed train} - {Time}{Rate of regular train} --> $$\frac{xy}{x+y}*\frac{z}{x}-\frac{xy}{x+y}*\frac{z}{y}=\frac{z(y-x)}{x+y}$$. Answer: A. Bunuel VeritasPrepKarishma Can anybody please tell me what's wrong with my approach High speed train x miles/hour slow train at y miles/hour combined speed x+y total time when they meed would be z/x+y distance covered by FAST train will be x(z/x+y) by slow train would be y(z/x+y) Difference will be x(z/x+y) - y(z/x+y) I am getting B as answer help me please Note that x hours is the time taken by the high speed train, not its speed. Similarly, y hours is the time taken by the regular train, not its speed. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Man Cat 4 #12-High speed train v. Regular Train [#permalink]

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29 Nov 2009, 14:26
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

-----
(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).
-----
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Re: Manhattan CAT math question [#permalink]

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21 May 2010, 04:30
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 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.

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Re: Manhattan CAT math question [#permalink]

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07 Feb 2012, 07:14
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 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.

If it takes x hours to travel z miles, then it takes d(x/z) hours to travel d miles. But you have wrongly calculated by directly considering d/x.
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]

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01 Apr 2012, 08:17
Vote for A

S * T = DIST

Given

High ST - x-hrs to travel Z miles
Regular ST = y-hrs 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(y-z) / xy
= z [xy / (x+y)] * [(y-z) / xy]
= z (y-z) / (x+y)

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Re: Man Cat 4 #12-High speed train v. Regular Train [#permalink]

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15 Oct 2012, 16:48
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gurpreetsingh wrote:
joyseychow 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 -> 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(x-y)/(x+y)
=> 60(3-2)/(3+2)
=> 12 -- equates the value expected.

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.
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Re: Manhattan CAT math question [#permalink]

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28 Mar 2013, 13:17
Bunuel wrote:
Prax wrote:
Hi,

I have another doubt:

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?

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

It takes the high-speed train x hours to travel the z miles --> rate of high-speed train is $$rate_{high-speed}=\frac{distance}{time}=\frac{z}{x}$$;

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 $$time=\frac{distance}{combined-rate}=\frac{z}{\frac{z}{x}+\frac{z}{y}}=\frac{xy}{x+y}$$.

Difference in distances covered: {Time}*{Rate of high-speed train} - {Time}{Rate of regular train} --> $$\frac{xy}{x+y}*\frac{z}{x}-\frac{xy}{x+y}*\frac{z}{y}=\frac{z(y-x)}{x+y}$$.

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
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Re: Manhattan CAT math question   [#permalink] 28 Mar 2013, 13:17

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