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Two trains, 100 miles apart, travelling towards each other on parallel

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Two trains, 100 miles apart, travelling towards each other on parallel  [#permalink]

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New post 11 Jul 2019, 20:35
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Question Stats:

66% (02:47) correct 34% (02:19) wrong based on 82 sessions

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Two trains, 100 miles apart, travelling towards each other on parallel tracks pass each other at x hours. If the same two trains were to travel in the same direction, starting 100 miles apart, they would meet each other at y hours. What is the ratio of the speed of the faster train to that of the slower train?

A. \(\frac{y}{x}\)
B. \(\frac{2y}{x+y}\)
C. \(\frac{y + x}{y -x}\)
D. \(\frac{y}{y-x}\)
E. \(\frac{y+x}{x}\)

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Re: Two trains, 100 miles apart, travelling towards each other on parallel  [#permalink]

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New post 12 Jul 2019, 04:57
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DisciplinedPrep wrote:
Two trains, 100 miles apart, travelling towards each other on parallel tracks pass each other at x hours. If the same two trains were to travel in the same direction, starting 100 miles apart, they would meet each other at y hours. What is the ratio of the speed of the faster train to that of the slower train?

A. \(\frac{y}{x}\)
B. \(\frac{2y}{x+y}\)
C. \(\frac{y + x}{y -x}\)
D. \(\frac{y}{y-x}\)
E. \(\frac{y+x}{x}\)


Take smart numbers: Say x = 2 hrs and y = 5 hrs

So when covering the distance together, sum of speeds = 100/2 = 50 mph = s1 + s2

When travelling in same direction, difference of speeds = 100/5 = 20 mph = s1 - s2

Then s1 = 35 and s2 = 15

s1/s2 = 35/15 = 7/3

Only (C) satisfies
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Re: Two trains, 100 miles apart, travelling towards each other on parallel  [#permalink]

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New post 11 Jul 2019, 20:58
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Let's call the faster train A and the slower train B.

In the first scenario, A and B travel a total of 100 miles in X hours. This gives us the following equation:
AX + BX = 100

In the second scenario, B is ahead of A by 100 miles but they eventually meet at some point after Y hours. This gives us the following equation:
AY = 100 + BY

The question asks for the following equation:
A/B = ?

We can arrange the second equation to equal the first:
AY - BY = 100 = AX + BX

Now, we can move the As to one side and the Bs to the other and factor them out on either sides:
AY - AX = BX + BY
A(Y - X) = B(X + Y)

Now simply arrange to get A/B on one side:
A(Y - X)/B = (X + Y)
A/B = (X + Y)/(Y - X)

Therefore the answer is (X + Y)/(Y - X), which is equivalent to (Y + X)/(Y - X)
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Two trains, 100 miles apart, travelling towards each other on parallel  [#permalink]

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New post 12 Jul 2019, 04:18
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The question may be solved also by the use of real numbers instead of X and Y.

Let A have a speed 66 and B - 33.
So they will cover 100 km in 1 hour.
And A will overtake B in 3 hours (if there will be 100 km between them).

So all we need is to find the appropriate proportion which will give us 66/33=2/1

Only C is ok:
(Y+x)/(y-x) = (3+1)/(3-1) =4/2=2/1

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Re: Two trains, 100 miles apart, travelling towards each other on parallel  [#permalink]

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New post 06 Oct 2019, 20:00
DisciplinedPrep wrote:
Two trains, 100 miles apart, travelling towards each other on parallel tracks pass each other at x hours. If the same two trains were to travel in the same direction, starting 100 miles apart, they would meet each other at y hours. What is the ratio of the speed of the faster train to that of the slower train?

A. \(\frac{y}{x}\)
B. \(\frac{2y}{x+y}\)
C. \(\frac{y + x}{y -x}\)
D. \(\frac{y}{y-x}\)
E. \(\frac{y+x}{x}\)



By using concept of relative velocity this question can be solved in real quick time.

Case 1: Trains travelling in opposite direction.
Let speed of trains are S1 and S2.
Hence given time to cross each other =x=100/(S1+S2)

Case 2 : Trains travelling in same direction.

Here time for one train to cross the other=y=100/(S1-S2)

Solving above equation we get
S1/S2=y+x/y-x
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Re: Two trains, 100 miles apart, travelling towards each other on parallel   [#permalink] 06 Oct 2019, 20:00
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