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

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

C

------->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?

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?

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

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
IMO C

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:
\(27r+17s=23r+23s\)
\(4r=6s\)
\(\frac{r}{s}=\frac{6}{4}=\frac{3}{2}\)

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

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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

Given: 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.

Asked: The ratio of their speeds is:

Let the length of trains be L1 & L2 and speeds be v1 & v2 respectively

L1 = 27v1
L2 = 17v2

L1 + L2 = 23 (v1 + v2)
27v1 + 17v2 = 23v1 + 23v2
4v1 = 6v2
v1/v2 = 3/2

IMO C

1st) Starting Position for the 2 trains to pass each other in opposite directions:

The two trains start nose-to-nose going in opposite directions.

Assume the man is standing right at this nose-to-nose staring point.
(*the star is the man.)


[———— A————] * [———B——-]

We do not know the length of each train, so we will find a constant distance that each has to travel. Over that constant distance, we can find the time taken by each.

2nd) after it takes 17 seconds pass for train B to cover the length of itself and pass the man:

After 17 seconds pass, train B is able to pass the man — i.e., in those 17 seconds, B covers the length of itself = the length of train B.

train A has not covered the length of itself yet. Assume there is D-length remaining to be covered.

Train A still has another 10 seconds to cover this D-length of itself and pass the man (because it takes A a total of 27 seconds).

Speed of train A = D / 10 (length / second)

Concept: when two trains are moving in opposite directions, the Gap distance they have to cover in order to pass each other is done at the Combined Speed of = (Speed of A) + (Speed of B)

We assumed the man was right at the “nose-to-nose start.”

By itself, it would take train A another 10 second to finish covering the remaining length of itself and pass the man.

However, with B’s help, A is able to cover the length of itself and pass the man (and train B) in just 6 more seconds.

This is because it takes 23 seconds in total for the two trains to pass, just 6 seconds after the first 17 seconds during which B covered the length of itself.

So, train B HELPS train A cover this same gap distance in 6 seconds as opposed to 10 seconds

So we can say the combined speed/relative speed of the two trains is:

(Speed of A) + (Speed of B) = D / 6 (length / second)


We can use these two equations to find the relative speeds:

(Sa) = D/10

(Sa) + (Sb) = D/6

(Sb) = (D/6) — (D/10) = (2D / 30) = D/15

Ratio of: (Speed of A) : (Speed of B) = (D/10) : (D/15)

Speed of A : Speed of B = 3 : 2

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