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# Time and Distance problem

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Manager
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Time and Distance problem [#permalink]

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06 Aug 2009, 02:06
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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

Can any one help me in solving this problem?
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Senior Manager
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Re: Time and Distance problem [#permalink]

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06 Aug 2009, 03:22
z= A-B
z/x= time that speedy train needs.
z/y= time that regular train needs.

z/(x+y) = time that they need to pass each other.

in that time, speedy will go x.z/(x+y) miles and regular will go y.z/(x+y) miles.

So difference is

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

answer is B.

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Senior Manager
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Re: Time and Distance problem [#permalink]

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06 Aug 2009, 05:25
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Ans is a) z(y – x)/x + y

Solution

Fast train speed = z/x
Regular train speed = z/y

Let d be distance traveled by the faster train
So, (z-d) is the distance is the regular train.

Time taken will be the same when they meet

Equating time

Time taken by faster train - dx/z
Time taken by regular train - ((z-d)*y)/z

so solving for d we get

d= zy/(x+y)

Difference in distance traveled by faster over regular train = 2d-z

Solving for 2d-z , where d= zy/(x+y)

we get
z(y – x)/x + y
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Director
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Re: Time and Distance problem [#permalink]

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06 Aug 2009, 05:54
Hi, Maly
plese note that z/x and z/y are respective speeds of both trains.

The time of the meeting of both trains is z/(z/x+z/y) or x*y/(x+y)

required difference in distance is speed of fast train *t -speed slower train *t

or (z/x)*[(x*y)/x+y)] -(z/y)*[(x*y)/x+y)] after canceling out we get

[z*(y-x)]/(x+y)

regards

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Manager
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Location: Washington DC
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Re: Time and Distance problem [#permalink]

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06 Aug 2009, 19:27
thanks guys.....simplertrader.....your explanation is right....
OA is a.....initially i tried doing in the way ....similar to maliey.....and got answer b.....but didnt understand why the OA was A.

thank you all once again in helping me understand.
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Senior Manager
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Location: Vagabond
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WE 2: Banking
Re: Time and Distance problem [#permalink]

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06 Aug 2009, 19:40
mokri

B says - z(x – y)/x + y

Now just think about it logically

if X is the time taken by the fast train - so it should be lower than y (slow train)

So x-y = negative

Denominator is positive, so the ans will be negative. This is simply not possible!!

The method suggested by BG is the best!! I redid the whole problem and it saved precious time

hope this helps
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If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Kudos [?]: 96 [0], given: 41

Re: Time and Distance problem   [#permalink] 06 Aug 2009, 19:40
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# Time and Distance problem

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