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Manager  Joined: 04 Dec 2008
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It takes the high-speed train x hours to travel the z miles  [#permalink]

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Question Stats: 55% (03:00) correct 45% (02:59) wrong based on 856 sessions

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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

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 V
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Re: Manhattan CAT math question  [#permalink]

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46
30
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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14
7
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
##### General Discussion
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Re: Man Cat 4 #12-High speed train v. Regular Train  [#permalink]

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1
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).
-----
Benjiboo
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Re: Manhattan CAT math question  [#permalink]

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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.
Math Expert V
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Posts: 64951
It takes the high-speed train x hours to travel the z miles  [#permalink]

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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 $$2d-z=\frac{2zy}{x+y}-z=\frac{z(y-x)}{x+y}$$.

You could spot that answer B cannot 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.
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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)

hope it will be helpfull.. Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
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Re: It takes the high-speed train x hours to travel the z miles  [#permalink]

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

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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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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)

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.

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

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

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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: $$z(\frac{y-x}{x+y})$$
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Re: Manhattan CAT math question  [#permalink]

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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]

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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

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 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?
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Re: Manhattan CAT math question  [#permalink]

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Bunuel wrote:
Hi,

Time in which they meet is $$time=\frac{distance}{combined-rate}=\frac{z}{\frac{z}{x}+\frac{z}{y}}=\frac{xy}{x+y}$$.

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
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Posts: 64951
Re: Manhattan CAT math question  [#permalink]

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hitman5532 wrote:
Bunuel wrote:
Hi,

Time in which they meet is $$time=\frac{distance}{combined-rate}=\frac{z}{\frac{z}{x}+\frac{z}{y}}=\frac{xy}{x+y}$$.

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

$$\frac{z}{\frac{z}{x}+\frac{z}{y}}=\frac{z}{\frac{zy+zx}{xy}}=\frac{xyz}{zy+zx}=\frac{xy}{x+y}$$.
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Re: It takes the high-speed train x hours to travel the z miles  [#permalink]

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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

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.
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Re: Manhattan CAT math question  [#permalink]

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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.
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Re: Manhattan CAT math question  [#permalink]

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

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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=(z-p)y/z
p=zy/(x+y)

but required is how much more distance covered by high speed train than regular train
i.e required = 2p-z = zy/(x-y) -z
=z(y-x)/x+y Re: It takes the high-speed train x hours to travel the z miles   [#permalink] 29 Jul 2013, 18:16

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