12 Days of Christmas GMAT Competition 2025 - Day 5: A railway employee

Length of first train be l1
Length of 2nd train be l2

Speed of first train be v1
Speed of second train be v2

Given v1 = l1/40
v2 = l2/25

l1 = v1 * 40
l2 = v2 * 25

Now given they cross each other in 30 seconds.
Total distance = l1 + l2
Total relative speed = v1 - (-v2) = v1 + v2

v1 + v2 = (l1 + l2) / 30
30*v1 + 30*v2 = 40*v1 + 25*v2
10*v1 = 5*v2

v1/v2 = 1/2
Correct Answer: C

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20 Dec 2025, 00:31

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When a train passes a stationary person
Length of train ∝ Speed x Time
Train Length 1 = v1 x 40
Train Length 2 = v2 x 25
Time taken to cross = 30 sec
Distance = sum of lengths
(v1+v2)x30 = v1 x 40 + v2 x 25
30v1 +30v2 = 40 v1 + 25 v2
10v1 = 5 v2
v1:v2 = 1:2 , Option C ✅

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20 Dec 2025, 01:14

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Bunuel

ok so we need ratio for both the train speed

First train : - D1 = 40 * V1 ( D1 -- distance of the first train; V1 -- speed of the first train)

Second train : - D2 = 25 * V2

Both train passes each other : - D1 + D2 = 30 * (V1 + V2)

40V1 + 25V2 = 30 V1 + 30V2

V1/V2 =1/2

Answer is C : 1:2

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20 Dec 2025, 04:02

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Train 1 length = l1
Train 2 length = l2

Train 1 speed: v1 = l1/40
Train 2 speed: v2 = l2/25

Using relative motion => Time to meet = Distance between the references / (Speed1 + Speed2); because they are moving in opposite directions
30 = (l1 + l2)/((l1/40)+(l2/25))
On solving => 4l2 = 5l1

l1 = 4/5 * l2

v1 = l2/50
v2 = l2/25

v1<v2

Therefore v1:v2 = 1/2

Answer C

Bunuel

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20 Dec 2025, 04:57

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Bunuel

Train 1

Len : l1
Speed : s1

Train 2

Len : l2
Speed : s2

Given

l1/s1 = 40

l2/s2 = 25

l1+l2/s1+s2 = 30

l1 + l2 = 30s1 + 30s2

40s1 + 25s2 = 30s1 + 30s2

10s1 = 5s2

s1/s2 = 1/2

Option C

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20 Dec 2025, 05:05

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Slower train
Time for passing man = 40 s
Speed = A
Length = 40A

Faster train
Time for passing man = 25 s
Speed B
Length = 25B

Time = Total Distance/Sum of Speed

=> 30 = (40A + 25B)/(A + B)
=> 30A + 30B = 40A + 25B
=> 5B = 10A
=> A/B = 1/2

Option C

Bunuel

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20 Dec 2025, 05:06

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Mind-numbing, to say the least

Anyway, here goes:

We're given a railway employee, standing on a platform, observes two trains travelling in opposite directions on parallel tracks.

Implication: The two trains will be going away from each other; one train going left in front of you will coincide on the parallel tracks with another train going right in front of you; they will both cover a certain distance at their respective speeds (with their RESPECTIVE lengths), and create a monumental challenge for us to calculate

Now, train 1 takes 40 seconds to completely pass the employee, while train 2 takes 25 seconds. Now, if the train that took 25 seconds to pass the employee was two carriages long and only travelling at, say, 20kmph, it would still be slower than, say, the 40-second train (taking longer) which, say, is 8 carriages long, and travelling at double the speed. That's why, the length, speed, distance are still variables, and we can't just calculate things right away.

To clear this logic more directly by assuming values: If a 200-meter-long shorter train travelling 10 meters per second parallel passes a kilometer-long train travelling at 20 meters per second, the longer train will cross the shorter train's front at second 0, cover the 200 meters length of the shorter train at the longer one's 20 meters-per-second pace, but we won't consider the two trains to have passed each other UNTIL the shorter train's made its way to the back of the longer train. However, for every 20 meters the faster / longer train is moving away from the shorter / slower train, the shorter train is moving 10 meters away from its counterpart. Hence, their combined speed here will reach 30m/s.

With all this information, we can form an equation and make our lives simpler.

What are the variables we are given? The variables which are constant to all? The TIME - 25s. 40s. 30s.

Let's use these.

Time = Distance / Speed

We know from the logic above that we need to combine the variables for both trains to answer the question--

Hence:

T(1) + T(2) =

D(1) + D(2)
--------------
S(1) + S(2)

(We further know that Distance = S*T)

HENCE:

=> 30 =

40*S(1) + 25*S(2)
---------------------
S(1) + S(2)

We move the denominator to the LHS.

=> 30S(1) + 30(S2) = 40S(1) + 25S(2)

(bringing like terms together --

=> 5(S1) = 10(S2)

That's a ratio of 1:2, which is the answer.

-----

A railway employee is standing on a platform and observes two trains traveling in opposite directions on parallel tracks. He observes that the first train takes 40 seconds to completely pass the employee (from the moment the locomotive reaches him until the last carriage clears him), while the second takes 25 seconds to completely pass him. If the two trains take 30 seconds to completely pass each other (from the moment the fronts meet to the moment the rears separate), what is the ratio of the speed of the slower train to that of the faster train?

A. 1:4
B. 1:3
C. 1:2
D. 2:3
E. 3:4

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20 Dec 2025, 05:56

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A railway employee is standing on a platform and observes two trains traveling in opposite directions on parallel tracks. He observes that the first train takes 40 seconds to completely pass the employee (from the moment the locomotive reaches him until the last carriage clears him), while the second takes 25 seconds to completely pass him. If the two trains take 30 seconds to completely pass each other (from the moment the fronts meet to the moment the rears separate), what is the ratio of the speed of the slower train to that of the faster train?

Let, the speed of the trains are v1 & v2
Let, the length of the trains are L1 and L2.
Time = Length of train/ speed of train
For first train, L1/v1=40 or L1=40v1
For second train, L2/v2=25 or L2=25v2

Since the trains are moving in opposite directions, their relative speed = v1+v2
Time taken by train to pass each other = (L1+L2)/(v1+v2) = 30
40v1+25v2= 30v1+30v2
10v1=5v2
v1/v2=1/2
Slower:Faster = 1:2

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20 Dec 2025, 06:15

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Answer C

time equals length of train over velocity t1=L1/v1=40sec t2=L2/V2 =25 sec

two train time equals L1+L2 / v1 + v2 = 30 sec
40sec x v1 + 25sec x v2 / v1 +v2 = 30 sec
40 v1 + 25 v2 = 30 v1 + 30 v2
10 V1 = 5 V2
2 V1 : 1 V2 2:1

Bunuel

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20 Dec 2025, 06:30

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A railway employee is standing on a platform and observes two trains traveling in opposite directions on parallel tracks. He observes that the first train takes 40 seconds to completely pass the employee (from the moment the locomotive reaches him until the last carriage clears him), while the second takes 25 seconds to completely pass him. If the two trains take 30 seconds to completely pass each other (from the moment the fronts meet to the moment the rears separate), what is the ratio of the speed of the slower train to that of the faster train?

A. 1:4
B. 1:3
C. 1:2
D. 2:3
E. 3:4

Let's consider length and speed of trains as l1, l2, s1 & s2

s1 = l1/40 ..............(1)

s2 = l2/25 ..............(2)

s1+s2 = (l1+l2)/30 ...........(3)

Putting l1 & l2 from 1st and 2nd equation into 3rd

30(s1+s2)= 40s1 + 25s2

5s2=10s1 ---> s1:s2 = 1:2 , since s1<s2 ......(Answer C)

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20 Dec 2025, 06:51

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From what we are given, let

Speed of train 1 = v1
Speed of train 2 = v2
Length of train 1 = x1
Length of train 2 = x2

We know that,
Time = Length/Speed

So for,
Train 1 => x1 = 40v1
Train 2 => x2 = 25v2

We know that the 2 trains pass each other
=> (x1+x2 )/ v1+v2 = 30
=> (40v1 + 25v2) / v1+v2 = 30
=> 2v1 = v2

Since we need ratio of slowest train to fastest train
v1:v2 = 1:2

C. 1:2

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20 Dec 2025, 07:10

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Let the slower train be A and faster train be B

Let d1 = distance covered by Train A and s1 = speed of train A
d2 = distance covered by Train B and s2 = speed of train B

d1 = 40*s1
d2 = 25*s2

Since trains are moving in opposite direction, the relative speed = s1+s2

Relative distance = d1+d2

And d1+d2/s1+s2 = 30

d1+d2 = 30s1 + 30s2

substitute the value of d1 and d2 from above

40s1 + 25s2 = 30s1+30s2
10s1 = 5s2
s1/s2 = 1/2

Option C

Bunuel

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20 Dec 2025, 07:12

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Speed of the train 1 = v1, length = l1
Speed of train 2 = v2, length = l2

So,
L1 = 40v1, L2=25v2

Using 30 seconds taken to pass each other => l1+l2/v1+v3 = 30 => (40v1+25v2) / v1+v2 = 30

Upon solving, v2 = 2v1
Ration slower/faster = 1/2 =>C

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20 Dec 2025, 07:18

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s1: speed of 1st train
s2: speed of 2nd train
l1: length of 1st train
l2: length of 2nd train

l1/s1 = 40 -> l1 = 40*s1
l2/s2 = 25 -> l2 = 25*s2

completely pass each other -> add lengths and speeds
(l1+l2)/(s1+s2) = 30
l1+l2 = 30*s1+30*s2

40*s1+25*s2 = 30*s1+30*s2
10*s1 = 5*s2

s1/s2 = 5/10 = 1/2
s1:s2 = 1:2

IMO C

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20 Dec 2025, 07:23

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Let X1, V1, X2, V2 be the speeds and the length of the trains 1 and 2.
Hence we can deduct the following facts from the problem:

I) V1 = X1/40
II) V2 = X2/25
III) V1+V2 = (X1 +X2)/30 (to imagine this last equation, just think about the problem with a view from high, as if you are standing on a tower nearby, you see 2 lines like this:

<----------------------| length X1
|--------------> length X2

hence we substitute the I) and II) in the III) and we easily get:

30V1 + 30V2 = 40 V1 + 25V2 ===> V2/V1 = 1:2

IMO C!

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20 Dec 2025, 07:53

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First train 40 seconds:
L1 = 40 * V1

Second train 25 seconds:
L2 = 25 * V2

Both trains:
L1 + L2 = 30 (V1 + V2)

40 * V1 + 25 * V2 = 30 * V1 + 30 * V2
10 * V1 = 5 * V2
2 * V1 = V2

V1:V2 = 1:2

Answer C

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20 Dec 2025, 07:53

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Let's calculate all the trains' movements from the very head of the train, basically from the Driver. The trains are different in size, with L1 and L2 defining their lengths. Also, we have speeds V1 and V2.

Then, how long does it take for a train to pass the employee? Well, the train needs to cover its whole length, from the head to tail. Therefore, $l_1/v_1 = 40$ seconds and $ l_2/v_2=25 $ seconds; and we can also say that $l_1=40v_1$ and $l_2=25v_2$

Now, what happens when the trains pass each other? The moment their heads meet and Drivers say hi, distance between them is 0. When they pass each other fully, the distance between the Drivers will be exactly the sum of trains' lengths, since their rears will now be 'bumping'. Therefore, they need to cover L1+L2 distance with V1+V2 speed, since they're moving towards one another.
So, $\frac{l_1+l_2}{v_1+v_2}=30$ seconds, which means that $l_1+l_2=30v_1 + 30v_2$

Then, $l_1-30v_1=30v_2-l_2$, and $40v_1-30v_1=30v_2-25v_2$
Finally, $10v_1=5v_2$, and $v_1:v_2=1:2.$

The answer is C.

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20 Dec 2025, 07:59

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The length of first train = velocity * time taken to cross the employee= 40 u, where u is the velocity of the first train.
Similarly length of second train = 25 v where v is the velocity of second train.
Given that it takes 30 seconds for the trains to cross each other.
If trains are traveling in opposite direction, then relative velocity = v + u
Total distance covered = 40u + 25v
This gives time taken, 30 = (40u+25v)/(u+v)
30u + 30v = 40u + 25v
5v = 10u
v = 2u
Ratio of slower train to fast train = 1 : 2

Bunuel