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Two trains running in opposite directions cross a man standing on the

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Two trains running in opposite directions cross a man standing on the  [#permalink]

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New post 29 Nov 2017, 06:35
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A
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Question Stats:

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Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
A. 3:1
B. 1:3
C.3:2
D.2:3
E.5:3
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Re: Two trains running in opposite directions cross a man standing on the  [#permalink]

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New post 10 Dec 2017, 01:31
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Chemerical71 wrote:
Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
A. 3:1
B. 1:3
C.3:2
D.2:3
E.5:3


This question too can be done in weighted average method.
Ans \(\frac{t_{27}}{t_{17}}=\frac{23-17}{27-23}=\frac{3}{2}\)

C...

How?
Let the man stand at their meeting point in the beginning.
After 17 secs, train II would have crossed him and Train I would have another 10secs to go. Train II helps getting down these 10 secs to 6 secs
These 10 secs are covered in the ratio 6:4, so 3:2
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Re: Two trains running in opposite directions cross a man standing on the  [#permalink]

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New post 29 Nov 2017, 07:10
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this is how I did it and worked for me .. using the S=D/T rule we can get the lengths of each train (thinking of it as the distance here)
therefore, Train 1 Let X be speed and distance is then (D=T*S, 27X) and for train 2 it will be 17Y

knowing the time of them combined is 23 then their combined speed would be

S=D/T >> T=D/S .. 23=(27X+17Y)/(x+y)
27x+17y = 23 x+23y

4x=6y
X/Y = 6/4 = 3/2

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Re: Two trains running in opposite directions cross a man standing on the  [#permalink]

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New post 01 Dec 2017, 07:42
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Chemerical71 wrote:
Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
A. 3:1
B. 1:3
C.3:2
D.2:3
E.5:3


We can let r = speed of train 1, s = speed of train 2, t = the length of train 1, and u = the length of train 2. Since speed x time = distance, we have:

r x 27 = t

s x 17 = u

And

(r + s) x 23 = t + u

If we add the first two equations, we have: 27r + 17s = t + u. If we simplify the third equation, we have 23r + 23s = t + u. Since both new equations are equal to t + u, we can equate them:

27r + 17s = 23r + 23s

4r = 6s

r/s = 6/4 = 3/2

Answer: C
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Two trains running in opposite directions cross a man standing on the  [#permalink]

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New post 12 Feb 2019, 13:41
1
Chemerical71 wrote:
Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
A. 3:1
B. 1:3
C.3:2
D.2:3
E.5:3


------->23 seconds----->TRAINS MEET----->4 seconds------>MAN------------------------------->
<----------------------------TRAINS MEET<-----6 seconds<------MAN<--------17 seconds<-------

The TOP train takes 23 seconds to meet the bottom train and another 4 seconds to cross the man, for a total of 27 seconds to cross the man.
The BOTTOM train takes 17 seconds to meet the man and another 6 seconds to meet the top train, for a total of 23 seconds to meet the top train.

The top train takes 4 seconds to travel the distance in blue.
The bottom train takes 6 seconds to travel the distance in blue.
Time and rate have a RECIPROCAL RELATIONSHIP.
Since the time ratio for the two trains = \(\frac{4}{6} = \frac{2}{3}\), the rate ratio for the two trains = \(\frac{3}{2}\).


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Re: Two trains running in opposite directions cross a man standing on the  [#permalink]

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New post 13 Feb 2019, 05:54
GMATGuruNY wrote:
Chemerical71 wrote:
Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
A. 3:1
B. 1:3
C.3:2
D.2:3
E.5:3


------->23 seconds----->TRAINS MEET----->4 seconds------>MAN------------------------------->
<----------------------------TRAINS MEET<-----6 seconds<------MAN<--------17 seconds<-------

The TOP train takes 23 seconds to meet the bottom train and another 4 seconds to cross the man, for a total of 27 seconds to cross the man.
The BOTTOM train takes 17 seconds to meet the man and another 6 seconds to meet the top train, for a total of 23 seconds to meet the top train.

The top train takes 4 seconds to travel the distance in blue.
The bottom train takes 6 seconds to travel the distance in blue.
Time and rate have a RECIPROCAL RELATIONSHIP.
Since the time ratio for the two trains = \(\frac{4}{6} = \frac{2}{3}\), the rate ratio for the two trains = \(\frac{3}{2}\).




Hi GMATGuruNY

Nice solution. But If I start with bottom train in all equation, it will end up choosing the reversed answer D as follows:

The bottom train takes 6 seconds to travel the distance in blue.
The top train takes 4 seconds to travel the distance in blue.
Since the time ratio for the two trains = \(\frac{6}{4} = \frac{3}{2}\), the rate ratio for the two trains = \(\frac{2}{3}\).

What is the trigger to start with top train to up with OA?
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Re: Two trains running in opposite directions cross a man standing on the  [#permalink]

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New post 13 Feb 2019, 08:09
Mo2men wrote:
What is the trigger to start with top train to up with OA?


The times are mentioned in the following order:
27 seconds to cross the man
17 seconds to cross the man
The rate ratio should align with this order.

That said, the GMAT would make the intention crystal clear, perhaps as follows:
Train A, traveling eastward, crosses the man after 27 seconds.
Train B, traveling westward, crosses the man after 17 seconds.
The two trains cross each other after 23 seconds.
What is the ratio of Train A's speed to Train B's speed?

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Re: Two trains running in opposite directions cross a man standing on the  [#permalink]

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New post 15 Feb 2019, 18:43
Chemerical71 wrote:
Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
A. 3:1
B. 1:3
C.3:2
D.2:3
E.5:3


Letting L and M be the lengths of the two trains, and r and s be the respective speeds of the the two trains, we have two distance (length) equations:

27r = L

17s = M

and

23(r + s) = L + M → 23r + 23s = L + M

If we add the first two equations, we have:

27r + 17s = L + M

Subtracting the 3rd equation from the above equation, we have:

4r - 6s = 0

4r = 6s

r/s = 6/4

r/s = 3/2

Answer: C
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Two trains running in opposite directions cross a man standing on the  [#permalink]

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New post 01 Oct 2019, 18:18
Chemerical71 wrote:
Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
A. 3:1
B. 1:3
C.3:2
D.2:3
E.5:3


length of train 1 ; a
length of train 2 ; b
speed of train 1 ; s
speed of train 2 ; t
given
a=27*s and b= 17*t----(1)
also (s+t)*23 = a+b---(2)
add 1 ; a+b = 27s+17t
we can say from 1 &2 ;
23s+23t=27s+17t
6t=4s
s/t =6/4 ; 3/2
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Re: Two trains running in opposite directions cross a man standing on the  [#permalink]

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New post 01 Oct 2019, 22:50
GMATGuruNY wrote:
Chemerical71 wrote:
Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
A. 3:1
B. 1:3
C.3:2
D.2:3
E.5:3


------->23 seconds----->TRAINS MEET----->4 seconds------>MAN------------------------------->
<----------------------------TRAINS MEET<-----6 seconds<------MAN<--------17 seconds<-------

The TOP train takes 23 seconds to meet the bottom train and another 4 seconds to cross the man, for a total of 27 seconds to cross the man.
The BOTTOM train takes 17 seconds to meet the man and another 6 seconds to meet the top train, for a total of 23 seconds to meet the top train.

The top train takes 4 seconds to travel the distance in blue.
The bottom train takes 6 seconds to travel the distance in blue.
Time and rate have a RECIPROCAL RELATIONSHIP.
Since the time ratio for the two trains = \(\frac{4}{6} = \frac{2}{3}\), the rate ratio for the two trains = \(\frac{3}{2}\).



Hi GMATGuruNY,

When I take closer look about your solution, I find you you have related all the 3 times. In highlighted parts, the question says the train coresses each in 23 second while you mention in 23 second they meet. and you deducted/or added that time for both trains. While the wording suggest that I can set up three independent equations as follows:

Letting L and M be the lengths of the two trains, and r and s be the respective speeds of the the two trains, we have two distance (length) equations:

27r = L

17s = M

and

23(r + s) = L + M → 23r + 23s = L + M

What is the difference ? or where did I go wrong?

Thanks in advance for support
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Re: Two trains running in opposite directions cross a man standing on the  [#permalink]

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New post 02 Oct 2019, 04:22
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Mo2men wrote:

Hi GMATGuruNY,

When I take closer look about your solution, I find you you have related all the 3 times. In highlighted parts, the question says the train coresses each in 23 second while you mention in 23 second they meet. and you deducted/or added that time for both trains. While the wording suggest that I can set up three independent equations as follows:

Letting L and M be the lengths of the two trains, and r and s be the respective speeds of the the two trains, we have two distance (length) equations:

27r = L

17s = M


and

23(r + s) = L + M → 23r + 23s = L + M

What is the difference ? or where did I go wrong?

Thanks in advance for support


Your solution is perfect.
If we add together 27r = L and 17s = M, we get:
\(27r+17s=L+M\)

Since 27r+17s = L+M and 23r+23s = L+M, the expressions in blue are EQUAL:
\(27r+17s=23r+23s\)
\(4r=6s\)
\(\frac{r}{s}=\frac{6}{4}=\frac{3}{2}\)
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Re: Two trains running in opposite directions cross a man standing on the  [#permalink]

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New post 18 Oct 2019, 07:57
Chemerical71 wrote:
Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
A. 3:1
B. 1:3
C.3:2
D.2:3
E.5:3



are there more problems like this to practice?? please link if any

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Re: Two trains running in opposite directions cross a man standing on the  [#permalink]

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New post 18 Oct 2019, 08:13
Chemerical71 wrote:
Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
A. 3:1
B. 1:3
C.3:2
D.2:3
E.5:3


Let the speeds of the trains be s1 & s2 and lengths be l1 & l2 respectively

l1/s1 = 27; l1 = 27s1
l2/s2 = 17; l2 = 17s2

(l1+l2)/(s1 + s2) = 23;
l1 + l2 = 23s1 +23s2 = 27s1 + 17s2
4s1 = 6s2
s1/s2 = 6/4 = 3/2

IMO C
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Re: Two trains running in opposite directions cross a man standing on the   [#permalink] 18 Oct 2019, 08:13
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