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

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Bunuel

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19 Dec 2025, 09:00

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18% (02:37) wrong

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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

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Bunuel

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Bunuel

GMAT Club Official Explanation:

This question can be solved in several ways. Below is arguably the most straightforward and reliable approach.

Let L1 and L2 be the lengths, and S1 and S2 the speeds, of the first and second trains, respectively.

The first train takes 40 seconds to pass the employee, so its length is:
L1 = S1 * 40

The second train takes 25 seconds to pass the employee, so its length is:
L2 = S2 * 25

When the two trains pass each other, they cover the total distance L1 + L2 in 30 seconds at a combined speed of S1 + S2.

So, we have the equation:
L1 + L2 = (S1 + S2) * 30

Substitute the expressions for L1 and L2:
S1 * 40 + S2 * 25 = (S1 + S2) * 30
10 * S1 = 5 * S2
S1 : S2 = 5 : 10 = 1 : 2

Answer: C.

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Let speeds be v1 and v2, lengths be l1 and l2.

Passing the man:
l1/v1 = 40
=> l1=40v1

l2/v2 = 25
=> l2 = 25 v2

Passing each other:
(l1+l2)/ (v1+v2) = 30
substitute:
(40v1+25v2)/(v1+v2) = 30
40v1+25v2 = 30 v1 + 30v2
10v1=5v2

v1/v2 = 1/2

Correct Answer: C

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	Speed	Time	Distance
Train 1	x	40	40x
Train 2	y	25	25y
Total		30	40x+25y

As they are moving in opposite direction their relative speed is x+y
40x+25y/(x+y) = 30
40x + 25y = 30x + 30y
10x = 5y
x/y = 1/2

answer C

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Bunuel

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Speed of Train1=v1; Length=L1
Speed of Train2=v2; Length=L2'
Length=Speed*Time
L1=40*v1; L2=25*v2
Relative Speed=v1+v2
Total Length Passed=40v1+25v2
Time=30s, so
(40v1+25v2)/(v1+v2)=30
10v1=5v2
v1:v2=1:2
Since, v1<v2
Slower:Faster=1:2 (OPTION C)

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Bunuel

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let speed of train a = x/sec , b = y/sec
lenght of train a = 40 * x
length of train b = 25 * y

total distance for which they will be in touch for the time 30 sec = length of a + lenght of b = 30 * (speed of train + speed of train B)
40 x + 25 y. = 30 x + 30 y
10 x = 5 y
x:y = 1:2

answer is 1:2

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19 Dec 2025, 09:25

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l1/v1=40
l2/v2=25
(l1+l2)/(v1+v2) = 30
on solving, we get v1/v2 = 1/2
Hence, Answer is C.

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Let the length of trains be L1 & L2 meters and speeds be s1 & s2 m/s respectively

First train: -
Distance travelled = L1
Time = 40 seconds
Speed = L1/40 = s1; L1 = 40s1

Second train:-
Distance travelled = L2
Time = 25 seconds
Speed = L2/25 = s2; L2 = 25s2

Pass each other: -
Distance travelled = L1 + L2
Time = 30 seconds
Speed = (L1 + L2)/30 = s1 + s2
40s1 + 25s2 = 30s1 + 30s2
10s1 = 5s2
s1/s2 = 1/2

The ratio of speed of slower train to that of faster train = 1:2

IMO C

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Train 1 speed=v1
train 2 speed=v2
train 1 length=l1
train 2 length=l2

time=length/speed
train 1, 40=l1/v1
l1=40v1
train 2, 25=l2/v2
l2=25v2

Time=(l1+l2)/(v1+v2)
30=(40v1+25v2)/(v1+v2)

40v1+25v2=30(v1+v2)
40v1+25v2=30v1+30v2
10v1=5v2
2v1=v2
v1:v2=1:2
v1 slower
v2 faster
C.1:2

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Length of two trains = L1 & L2
Speed of two trains = S1 & S2
T1= time taken by train 1 to pass the employee = 40 sec
T2=time taken by train 2 to pass the employee = 25 sec
t= time taken by trains to pass each other =30 sec t
L1= 40*S1 and L2=25*S2

When two trains travel in their opposite directions, thei relative speed = S1+S2
And total distance covered to pass each other = L1+L2
Distance = Speed*Time
L1+L2= (S1+S2)*30
40S1+25S2 =30S1+30S2
10S1=5S2
S1/S2 = 1/2

C

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Let length of first train be D which takes 40 seconds to cross the man with speed S, then, from Distance = Speed x Time:
D = 40S
Similarly, for second train will be
d = 25s

When both trains cross each other in opposite direction, their total lengths will be total distance travelled and total speed will be addition of both their speeds.

We are given that total time taken to cross each other is 30 seconds.
We know that, Total distance travelled = Total speed x Total time taken
D + d = (S + s) x 30
Since we have to find the ratio of speeds, we will replace values of D and d from above highlighted equations

=> 40S + 25s = 30S + 30s
=> 10S = 5s
=> S/s = 1/2

Hence, option C is the answer

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Bunuel

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Let us use the info provided:

Time = length/speed

First train: 40 = L1/v1 => L1 = 40v1

Second train: 25 = L2/v2 => L2 = 25v2

Time taken for both to pass each other = 30 seconds
When they move in opposite directions, speed is a sum of both = v1 + v2

Time taken for both to pass each other:
30 = (L1 + L2)/(v1 + v2)
30 = (40v1 + 25v2)/(v1 + v2)
30v1 + 30v2 = 40v1 + 25v2
10v1 = 5v2
v1/v2 = 1/2

Option C

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let's assume 1st TRAIN speed V & LengthL, 2nd train speed v & length l, and V:v=K, so,V=Kv
the equations L/V=40,-----(i)
l/v=25,------(ii)
(L+l)/(V+v)=30-----(iii)
Now, Replace V=Kv,
from Eq (iii), we get----(L/v(K+1)) +(l/v(K+1))=30
==>40/(k+1) +25k/(k+1)=30
==>40 +25k=30K+30
==>k=2, so option C

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Train 1 covers its length (D1) in $40$ seconds.

$=>$ speed of Train 1 (S1) $= \frac{D1}{40 }$

Similarly, speed of Train 2 (S2) $= \frac{D2}{25}$

The two trains pass each other in 30 seconds. Which means Train 2 takes 5 seconds longer than it would if it covered its own length.

$=>$ Length of Train 1 $>$ length of Train 2

$=>$ D1$>$D2

To completely pass Train 2, Train 1 must cover its entire length as well. But in 30 seconds it only covers $30*\frac{D1}{40} = \frac{3D1}{4}$

$=>$ Train 2 covers the rest of the distance $= D1/4$

Now, Train 2 covers D2 in 25 seconds and takes 5 seconds to cover the remaining distance which is D1/4.

So, speed of Train 2 $= \frac{D1}{4}*\frac{1}{5} = \frac{D1}{20}$

Therefore, S2$>$S1 and $\frac{S1}{S2} = \frac{D1}{40}*\frac{20}{D1} = \frac{20}{40} = \frac{1}{2}$

The answer is C.

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Let train 1 be of length L1, speed s1 and time 40 sec
train 2 is length L2, speed s2 and time 25 sec

s1=L1/40=> L1=40s1
s2=L2/25= L2=25s2

now when two trains pass each other. The total distance is L1+L2, and the total speed is s1+s2 (as mov in opp directions)

s1+s2= (L1+L2)/ 30
sub L1 and L2

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

hence the answer is C

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So, as per the question, S1= Length of Train-1/ Time taken to cross employee= L1/40
Similarly, S2= Length of Train-2/Time taken to cross employee= L2/25
Now, when both train are moving towards each other so speed is= S1+S2= (L1+L2)/30
(L1+L2)/30= L1/40 + L2/25
On Solving L1/L2= 4/5
Say, L1= 4x and L2= 5x
S1= 4x/40= x/10 & S2= 5x/25= x/5... So, S2> S1
S1/S2= 1/2

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Case 1: (T1)40s ----(T2)30s

Case 2: (T1+2)25s

So, we calculate the gap between
1. T1 - T(1+2): 40-25 = 15 --> higher speed
2. T2 - T(1+2): 30-25=5 --> lower speed

ratio lower/higher = 5/15 = 1/3 (C)

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Ans C
Length of each train w.r.t observer, D1=40S1 , D2=25S2 --(1)
When passing each other 30(S1+S2)=D1+D2 ---(2)
Sub (1) in (2) and simplify to get ratio of S1/S2 as 1/2