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

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dp1234

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Bunuel

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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Train 1 passes employee in 30 sec, let 1st train speed, v1 and length T1, thus v1=T1/40. Similarly for train 2, v2=T2/25
Also, time taken by two train to pass each other is 30 sec, thus (T1+T2)= 30*(v1+v2). Substitute v1 and v2 from above equations in this, and divide both sides by v2, we get a linear equation in form of variable (v1/v2). On solving we get v1/v2=1/2.

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Length=time x speed
so when 2 trains pass each other, Time= Sum of length/sum of speeds
Lets assume slow train speed is 1 uit ad faster train is X units since we only want the ratio
length of slower T= 40x1=40
Length of faster T= 25 x X =25X
Passing each other = 40+25X/1+x=30
40+25x=30+30x
40-30=30x-25x, 10=5x
x=2 s0 slower is to faster is 1:2

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19 Dec 2025, 11:41

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Let the speed and length of the trains be S1L1 and S2L2
Given that time is D/S In this case we use L/S
For train 1 time = 40 sec L1= 40S1
For train 2 time =25 sec L2 = 25S2
When they pass each other total length is L1+L2 while total speed is S1+S2
Meaning total is L1+L2/S1+S2 = 30.
Replace the lengths with 40S1 and 25S2 and solve for S1:S2
40S1+25S2=30S1+30S2
10S1=5S2
S1/S2= 1/2 or 1;2
Ans C

Bunuel

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Bunuel

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An employee is standing at the platform.

Two trains travelling from opposite directions towards the employee.

Let the speeds of the train be S1 and S2, the length of the respective trains be L1 and L2 respectively.

The distance of employee is considered negligible compared to length of the trains.

Train 1 covers the employee in 40 seconds.

S1 = L1 /40

Train 2 covers the employee in 25 seconds.

S2 = L2/25

When the two trains cross each other , their relative speed is added.

S1 + S2 = (L1+L2)/30

Substitute L1 = 40*S1 , and L2 = 25*S2

S1 + S2 = (40* S1 + 25*S2 ) /30

Solving further, we get

5*S2 = 10 * S1

S1: S2 = 1:2

Option C

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reading this it seems like a unique type of distance/speed problem, where instead of a distance we have to consider the length of the trains.

distance = speed * time
length train 1 = x1 speed train 1 = s1
length train 2 = x2 speed train 2 = s2
from the trains passing the employee we know that
x1 = 40 * s1
x2 = 25 * s2

from the trains passing each other
speed = s1 + s2 (because they are moving in opposite directions)
(x1 + x2) = 30 * (s1 + s2)
40s1 + 40s2 = 30 (s1 + s2)
10s1 = 5s2 -> s2 = 2s1
s1 : s2 = 1 : 2

Answer C

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19 Dec 2025, 13:35

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let lengths and speeds of the slow and fast trains be (l1,s1) and (l2,s2) respectively.
s1=l1/40 and s2=l2/25 (trains passing a static point)
s1/s2=(l1/l2) *5/8--------1.
now trains pass each other when moving towards each other in 30 sec=(l1+l2)/(s1+s2)
=> putting values of s1 and s2 in terms of l1 and l2 and solving we get
50 l1=40 l2
l1/l2=4/5-----2.
solving 1. and 2. we get s1/s2=1:2
so C

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	Train 1	Train 2	Together from opposite direction
Speed	S1	S2	S1+S2 (Relative Speed)
Time	40	25	30
Distance	40*S1	25*S2	30*(S1+S2)

40*S1+25*S2 = 30*(S1+S2)
Solving S1/S2 = 1/2

(C) is the answer

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19 Dec 2025, 14:14

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Bunuel

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Lets seperate the question into two scenarios

TRAIN CROSSING Employee scenario

Time taken by train 1 to cross employee = LENGTH OF TRAIN 1 /SPEED

"" """ """ by train 2 to cross employee = length of train 2 / speed.

hence we will get L1= S1 x 40 and L2=S2 X 25

SECOND SCENARIO - TRAIN CROSSING EACH OTHER SCENARIO

TIME TAKEN TO CROSS EACH OTHER = L1+L2/S1+S2.

We have found L1 and L2 in terms of s1 and s2. substitute that value.

30 = 40 S1+25 S2 / S1+S2.

30 (S1+S2) = 40S1 + 25 S2.

10 S1 = 5 S2

S1/S2 = 5/10 = 1/2

HENCE OPTION C

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19 Dec 2025, 14:47

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Bunuel

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Trick--- when train passes a person it actually covers its own length and when it crosses it other it covers the sum of their lengths
lenths of slower train =d1
lenths of faster train =d2
speed of slower train=d1/40
speed of faster train= d2/25
relative speed= d1/40+d1/25
so time taken to pass eachother=30
(d1+d2)/(d1/25+d2/40)=30
d1/d2=5/4
so ratio=(d1/40)/(d2/25)=1/2
Ans-c

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length of first train (l1) = v1 * 40
length of second train (l2) =v2 * 25

the ends of each train to come together

$v1 + v2 = \frac{((v1*40) + (v2*25))}{30}$

v1*30 + v2*30 = v1*40 + v2*25
v2*5 = v1*10

$\frac{5}{10} =\frac{v1}{v2}$

$\frac{1}{2} =\frac{v1}{v2}$

ans: C

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Suppose trains A & B
If the speed of train -A: x
Speed of train-B:y

Length of train-A= Speed X time = x*40=40x
Length of Train-B= 25y

Given that, two trains together take 30 sec to cross eachother, then
(40x + 25y)/(x+y) = 30
40x + 25y = 30x + 30y
10x=5y
Therefore, x<y
slower by faster= 5/10 = 1/2

Bunuel

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Quote:

Passing the employee can be understood as travelling the whole length of the train.
For the first train which takes 40s to do so, we have: d1 = v1*40
For the second train which takes 25s to do so, we have: d2 = v2*25

The two trains passing each other can be understood as both of them travelling their combined length (calculated from the moment the fronts meet). As they were travelling in opposite direction, this gives us: d1+d2 = (v1+v2)*30
=> v1*40 +v2*25 = v1*30+v2*30
=> v1*10 = v2*5 => v2:v1 = 5:10 = 1:2
Answer: C

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Suppose 1st train length L1 & 2nd L2. So first speed L1/40 & second L2/25. When passes each other length & speed both would add. hence (L1+L2)/(L1/40+L2/25)=30 gives L1/L2=4/5. Since 1st takes more time & length is lesser it would be slower therfore question is asking V1/V2 ---> (L1/40)/(L2/25) = 5L1/8L2= (5/8)*(4/5)= 1:2

Bunuel

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Let the speed of first train be s1 , s2 for second train , l1 be the length of first train , l2 be the length of second train. Following equations will be created:

s1=l1/40 , s2 = l2/25 , (s1+s2) = (l1+l2)/30 [based on relative speed concept]

so , l1/40 +l2/25 = l1/30 + l2/30

from here l2 = 5/4 l1

So , l2 >l1 , which implies s2>s1

then we need to find s1/s2 ,

s1/s2 = (l1/40)/(l2/25) = (l1/l2)*(25/40) = 1/2 which is the answer

Bunuel

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let length and speed be L1,v1, L2,v2
Given: L1=40v1 (slower train) L2=25v2 (faster train)
Acc to ques: 30 = (L1 + L2)/(v1 + v2) = (40v1 + 25v2)/(v1 + v2) => on solving v1/v2 = 1/2.

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Time = Length / Speed
Let L1 & L2 be the length of 2 trains
And, V1 & V2 be the speed of 2 trains
For first train : Time = 40 sec
L1/V1 = 40 --- > L1 = 40V1
Second train : Time = 25 sec
L2/V2 = 25 ---> L2 = 25V2
So when they pass each other: (L1 + L2) / (V1 + V2) = 30
Simplify and cross multiply:
(40V1 + 25 V2) = 30 (V1 + V2)
V1 = 2V2
Ratio (Slow : Fast) = 1/2

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Let L1 be the length of 1st train and x be its speed, L1=40x
Let L2 be the length of 2nd train and y be its speed, L2=25y
When train1 & train2 cross each other
L1+L2=30x+30y=40x+25y
10x=5y
Solving, x:y::1:2
Correct answer is C.

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Lets say the first train is a length, with v1 speed.
the second train is b length, with v2 speed.

The first train takes 40 seconds to pass the employee while the second train takes 25s
=> v1 = a/40, v2 = b/25

When the two train completely pass each other, they have moved the total (a+b) distance:
=> a+b = 30(a/40+b/25) => a/b = 4/5

v1:v2 = (4/5*b/40) : (b/25) = 1:2

Answer: C

Bunuel

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When two objects in towards each other in from opposite directions then their relative speed will be sum of their respective speeds.
If they move apart from each other, then relative speed will be difference of their respective speeds.

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Let L1, S1 be the length and speed of train 1 and L2, S2 for Train 2.

We have L1/S1= 40 and L2/S2 =25. We can neglect the length of the employee as it is negligible

For trains to cross each other completely, L1+L2 has to be traversed. Moreover, since they are travelling in opposite directions, the combined speed is S1+S2

Therefore, L1+L2/ S1+S2 = 30

Substituting the first two equations in the above equation, we have

40S1+25S2 = 30S1+ 30S2

S1/S2 =1/2

Therefore, Option C