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Manhattan Test Problem - Why time is taken as constant?

Expert replies
by bubbliiiiiiii » Mon Jun 27, 2011 8:15 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?


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

Solution:

Since the trains traveled the z miles in x and y hours, their speeds can be represented as z/x and z/y respectively.

We can again use an RTD chart to evaluate how far each train travels when they move toward each other starting at opposite ends. Instead of using another variable d here, let's express the two distances in terms of their respective rates and times.



High-speed Regular Total

R z/x z/y

T t t

D zt/x zt/y z


Since the two distances sum to the total when the two trains meet, we can set up the following equation:

zt/x + zt/y = z divide both sides of the equation by z
t/x + t/y = 1 multiply both sides of the equation by xy
ty + tx = xy factor out a t on the left side
t(x + y) = xy divide both sides by x + y
t =
xy

x + y


To find how much further the high-speed train went in this time:
(ratehigh × time) – (ratereg × time)
(ratehigh – ratereg) × time

zy – zx xy
------- x ------
xy x+y


z(y – x)
-------
x + y




The correct answer is A.

Question: Why time is taken as constant?
Regards,

Pranay
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Source: — Problem Solving |

by bubbliiiiiiii » Mon Jun 27, 2011 8:20 am
Also, can someone help me with the working of work-rate related problems. I usually feel difficult while solving these problems, which mostly fall in 700-800 difficult level.
Regards,

Pranay
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by Frankenstein » Mon Jun 27, 2011 8:24 am
Hi,
Both the trains start at the same time. So, they take same time by the time they meet.
Let me know if you are referring to some other time.
If you feel the solution is hard to comprehend, please specify. Will try to make it easier.
Cheers!

Things are not what they appear to be... nor are they otherwise
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by bubbliiiiiiii » Mon Jun 27, 2011 9:53 am
So, they take same time by the time they meet.
Hi Frank,

Thanks for the response dude.

I am still not clear with the sentence above. Can you put it in other way please.

I am Ok with the rest of the problem working.
Regards,

Pranay
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by Frankenstein » Mon Jun 27, 2011 9:57 am
Hi,
Let's say both the trains started at 9:00A.M.
They meet each other at some point say 11:00A.M.
Now both trains have traveled for same time(2 hours) right?
I hope you are referring to the time each has taken right? If not please specify which time you are referring to.
Cheers!

Things are not what they appear to be... nor are they otherwise
Join the discussion

by bubbliiiiiiii » Mon Jun 27, 2011 10:06 am
Since the two trains are travelling at different speeds what I understand is that the slow moving train would need more time to cover the same distance that is travelled by fast moving train.

However, we can simply answer the question saying that when two elements among three in formula speed = distance/time, are variables the third has to be treated as constant. But, I want to understand, why is it constant?

Hope this makes the question more clear.
Regards,

Pranay
Join the discussion

by GMATGuruNY » Mon Jun 27, 2011 10:39 am
bubbliiiiiiii 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?


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

Solution:

Since the trains traveled the z miles in x and y hours, their speeds can be represented as z/x and z/y respectively.

We can again use an RTD chart to evaluate how far each train travels when they move toward each other starting at opposite ends. Instead of using another variable d here, let's express the two distances in terms of their respective rates and times.



High-speed Regular Total

R z/x z/y

T t t

D zt/x zt/y z


Since the two distances sum to the total when the two trains meet, we can set up the following equation:

zt/x + zt/y = z divide both sides of the equation by z
t/x + t/y = 1 multiply both sides of the equation by xy
ty + tx = xy factor out a t on the left side
t(x + y) = xy divide both sides by x + y
t =
xy

x + y


To find how much further the high-speed train went in this time:
(ratehigh × time) – (ratereg × time)
(ratehigh – ratereg) × time

zy – zx xy
------- x ------
xy x+y


z(y – x)
-------
x + y

The correct answer is A.
Plugging in values seems much easier.

Let x=2 miles per hour.
Let y=3 miles per hour.
Combined rate for x+y = 2+3 = 5 miles per hour.
Let z=10 miles.
Time for the trains to meet = d/r = 10/5 = 2 hours.
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.

Now we plug x=2, y=3 and z=10 into the answers to see which yields our target of 2.

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

The correct answer is A.
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by Frankenstein » Mon Jun 27, 2011 10:47 am
bubbliiiiiiii wrote:Since the two trains are travelling at different speeds what I understand is that the slow moving train would need more time to cover the same distance that is travelled by fast moving train.

However, we can simply answer the question saying that when two elements among three in formula speed = distance/time, are variables the third has to be treated as constant. But, I want to understand, why is it constant?

Hope this makes the question more clear.
Hi,
Okay got your point. Here 't' is not the time taken for each train to travel the entire distance between two stations. Here 't' is the time taken by each of the trains to reach from their respective sources to their point of intersection.
Cheers!

Things are not what they appear to be... nor are they otherwise
Join the discussion

by bubbliiiiiiii » Mon Jun 27, 2011 7:19 pm
Hey Frank,

Thanks for your inputs. Yesterday night I got the point why t is constant and understood the very obvious concept.

@Mitch, Thanks for the alternate solution.

Can anyone also help me on work-rate related problems?
Regards,

Pranay
Join the discussion