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Trains.

by goyalsau » Wed Sep 29, 2010 1:43 am
I want to know if this can be solved other than by putting values.
I got it wrong in the first attempt.



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
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by kmittal82 » Wed Sep 29, 2010 2:24 am
If two objects are moving towards each other, you can add their velocities.

Speed of high speed train = z/x
Speed of normal train = z/y

total speed = z/x + z/y = z(x+y)/xy

Time before they meet = Distance/speed = z/[(z(x+y)/xy] = xy/(x+y)

Distance travelled by high speed train in this time = (z/x) * xy(x+y) = yz/(x+y)
Distance travelled by normal train in this time = (z/y) * xy(x+y) = xz/(x+y)

Thus, extra distance travelled by fast train = yz/(x+y) - xz/(x=y) = z[(x-y)/(x+y)]

Should be (B)
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by goyalsau » Wed Sep 29, 2010 2:41 am
official Asnwer is A
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by kmittal82 » Wed Sep 29, 2010 3:16 am
Oh.. sorry, I did a minor mistake in my last calculation:

yz/(x+y) - xz/(x+y) = z[(x-y)/(x+y)]

is incorrect, it should be

yz/(x+y) - xz/(x+y) = z[(y-x)/(x+y)], same as A.
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by gmatmachoman » Wed Sep 29, 2010 3:29 am
goyalsau wrote:official Asnwer is A
HST speed V1 = Z/X

LST Speed V2 = Z/Y

Meeting point time = Distance travelled / Sum of speeds of 2 entities { memorise this..will help u a lot}
= Z/ ( V1 +V2)


Upon solving : t = XY / (X+Y)

Now Distance travelled by HST( H1 )= V1 * t
= (Z/ X ) * (XY / (X+Y)
=ZY / ( X+Y)
Distance travelled by LST (H2) = V 2 * t

= Z/Y * (XY / (X+Y)
= Z X/(X+Y)

H1- H2

= z(y - x)
---------
x + y

So u got A!!
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by goyalsau » Wed Sep 29, 2010 3:34 am
Great work Guys,
But Guys there are too many X and Y in the solution.

I did it by plugging in numbering and then eliminating the choices but even then because of the tedious calculation i got it wrong at the first attempt. So i am looking for a way which is not really lengthy.
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by GMATGuruNY » Wed Sep 29, 2010 11:47 am
goyalsau wrote:I want to know if this can be solved other than by putting values.
I got it wrong in the first attempt.

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
We can plug in values for x, y and z.

Let the high speed rate = 3 miles per hour.
Let the regular rate = 2 miles per hour.
Combined rate for the trains = 2+3 = 5 miles per hour.
Let z = 30 miles.
Time for the high speed train to travel 30 miles = x = 30/3 = 10 hours.
Time for the regular train to travel 30 miles = y = 30/2 = 15 hours.
Time for the trains to meet = 30/5 = 6 hours.
Distance traveled by the high speed train in 6 hours = r*t = 3*6 = 18 miles.
Distance traveled by the regular train in 6 hours = r*t = 2*6 = 12 miles.
Distance for the high speed train - distance for the regular train = 18-12 = 6. This is our target.

Now we plug x=10, y=15, and z=30 into the answers to see which yields our target of 6.

Only answer choice A works:
z(y - x)/(x+y) = 30*(15-10)/(10+15) = 6.

The correct answer is A.
Last edited by GMATGuruNY on Sun Oct 23, 2011 2:15 am, edited 1 time in total.
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by goyalsau » Wed Sep 29, 2010 4:55 pm
GMATGuruNY wrote:
Plugging in values makes this problem a snap:

Let x = 2/hr
Let y= 3/hr
Combined rate for x+y = 2+3 = 5/hr
z=10
Time for trains to meet = d/r = 10/5 = 2hrs.
Distance for x = r*t = 2*2 = 4
Distance for y = r*t = 2*3 = 6
Distance for y - distance for x = 6-4 = 2. This is our target.

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

Total time to solve was less than 30 seconds.
Thanks Guru,
I solved the same way but i take too many big values.
The values you assumed for x and y are easy to calculate.
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by frank1 » Wed Sep 29, 2010 8:58 pm
Some books say it Smart numbers ....
sometime for eg 12....it is factor of both 2...3...4...6...
so right number selection does help in overall process...
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by send2dar » Wed Jul 24, 2013 4:40 pm
I found a very interesting and efficient way of arriving at an answer to this.

As z is the only distance number among x, y, and z, the answer should contain z. Rules out options D and E.
As high-speed train would take lesser time than regular time, x should be less than y. Further, as our answer should be positive, it should not contain x - y term (x - y is negative). Rules out B.
As we are looking at the difference in distances travelled, the numerator should be of the form p - q. Rules out C.

Only Option A is left, which is our answer. In this way, without solving the problem, one can get to the correct answer.

Hope this helps!

If anybody finds loopholes in the above logic, please do let me know.
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by dhirajdas53 » Wed Jul 24, 2013 7:23 pm
The unit of answer should be miles. Hence remove D & E.

For high speed train, the time taken should be lesser than the regular train.
Hence, x is less then Y.

When you look at the option, B gives negative value. Hence remove this.

Carefully, look at the option A and C. C gives value more than Z, which is the total distance between A & B and cannot be true.

Thus, the answer should be A.
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