Last visit was: 21 Sep 2026, 00:25

It is currently 21 Sep 2026, 00:25

Toolkit

Practice Tests

Quantitative Questions

Problem Solving (PS)

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Go to My Error Log Learn more

Request Expert Reply

Confirm Cancel

Sep 21
Live Online Bootcamp Class with Top GMAT Expert Chris Peckover
07:00 PM EDT
-
10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included!
Sep 23
GMAT 705 in 30 days: Best Study Strategy for Working Professionals
12:30 AM EDT
-
01:30 AM EDT
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more
Sep 24
Stuck at 645 After 3 Attempts: The Problem Wasn't Effort
01:30 AM EDT
-
01:30 AM EDT
Tejas studied hard but kept repeating the same mistakes. After 3 attempts at 645, a focused 4-week strategy helped him identify and fix what was holding him back—leading to a 695 with just 12–15 hours of study per week.
Sep 26
Start your free trial!
06:00 PM PDT
-
07:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!

Back to Forum

Create Topic

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

1 2 3 4

dp1234

Joined: 15 Nov 2021

Last visit: 22 Jul 2026

Posts: 108

Own Kudos:

[1]

Given Kudos: 169

GMAT Focus 1: 665 Q86 V83 DI80 Verified score

Posts: 108

Kudos: 77

[1]

19 Dec 2025, 11:23

Kudos

Bookmarks

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

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

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.

linnet

Joined: 11 Dec 2025

Last visit: 22 Jan 2026

Posts: 81

Own Kudos:

[1]

Given Kudos: 1

Posts: 81

Kudos: 42

[1]

19 Dec 2025, 11:36

Kudos

Bookmarks

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

jkkamau

Joined: 25 May 2020

Last visit: 19 Sep 2026

Posts: 285

Own Kudos:

244

[1]

Given Kudos: 148

Location: Kenya

Schools: Jones '26 (A$$)

GMAT 1: 730 Q50 V46 Verified score

GPA: 3.5

Products:

Schools: Jones '26 (A$$)

GMAT 1: 730 Q50 V46 Verified score

Posts: 285

Kudos: 244

[1]

19 Dec 2025, 11:41

Kudos

Bookmarks

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

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

Dereno

Joined: 22 May 2020

Last visit: 30 Jul 2026

Posts: 1,398

Own Kudos:

1,425

[1]

Given Kudos: 433

Posts: 1,398

Kudos: 1,425

[1]

19 Dec 2025, 11:59

Kudos

Bookmarks

Bunuel

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

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

sitrem

Joined: 19 Nov 2025

Last visit: 24 Feb 2026

Posts: 91

Own Kudos:

[1]

Given Kudos: 238

Posts: 91

Kudos: 84

[1]

19 Dec 2025, 13:31

Kudos

Bookmarks

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

pappal

Joined: 24 Nov 2022

Last visit: 18 Sep 2026

Posts: 385

Own Kudos:

134

[1]

Given Kudos: 120

Products:

Posts: 385

Kudos: 134

[1]

19 Dec 2025, 13:35

Kudos

Bookmarks

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

canopyinthecity

Joined: 12 Jul 2025

Last visit: 20 Sep 2026

Posts: 171

Own Kudos:

110

[1]

Given Kudos: 40

Products:

Posts: 171

Kudos: 110

[1]

19 Dec 2025, 13:38

Kudos

Bookmarks

	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

iBN

Joined: 13 Jan 2025

Last visit: 06 Mar 2026

Posts: 45

Own Kudos:

[1]

Given Kudos: 25

Location: India

WE:Marketing (Consulting)

Products:

Posts: 45

Kudos: 36

[1]

19 Dec 2025, 14:14

Kudos

Bookmarks

Bunuel

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

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

rahumangal

Joined: 20 Nov 2022

Last visit: 17 Jun 2026

Posts: 71

Own Kudos:

[1]

Given Kudos: 316

Location: India

Concentration: Finance, Real Estate

GPA: 3.99

WE:Engineering (Technology)

Products:

Posts: 71

Kudos: 66

[1]

19 Dec 2025, 14:47

Kudos

Bookmarks

Bunuel

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

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

sriharsha4444

Joined: 06 Jun 2018

Last visit: 05 Mar 2026

Posts: 125

Own Kudos:

[1]

Given Kudos: 803

Posts: 125

Kudos: 85

[1]

19 Dec 2025, 14:54

Kudos

Bookmarks

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

Veerenk

Joined: 23 Sep 2024

Last visit: 20 Jun 2026

Posts: 28

Own Kudos:

[1]

Given Kudos: 225

Location: India

Posts: 28

Kudos: 10

[1]

19 Dec 2025, 17:38

Kudos

Bookmarks

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

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

HowCanIDoThis

Joined: 11 Oct 2025

Last visit: 06 Feb 2026

Posts: 19

Own Kudos:

[1]

Given Kudos: 3

Posts: 19

Kudos: 14

[1]

19 Dec 2025, 19:40

Kudos

Bookmarks

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

750rest

Joined: 27 Jul 2022

Last visit: 20 Sep 2026

Posts: 47

Own Kudos:

[1]

Given Kudos: 1,199

Concentration: Marketing, Operations

Products:

Posts: 47

Kudos: 34

[1]

19 Dec 2025, 19:48

Kudos

Bookmarks

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

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

arnab24

Joined: 16 Jan 2024

Last visit: 14 Sep 2026

Posts: 96

Own Kudos:

[1]

Given Kudos: 7

Location: India

Schools: ISB '26

GPA: 8.80

Schools: ISB '26

Posts: 96

Kudos: 81

[1]

19 Dec 2025, 20:50

Kudos

Bookmarks

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

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

gemministorm

Joined: 26 May 2025

Last visit: 19 May 2026

Posts: 143

Own Kudos:

131

[1]

Given Kudos: 56

GMAT Focus 1: 565 Q82 V79 DI73 Verified score

GMAT Focus 2: 605 Q84 V83 DI73 Verified score

Posts: 143

Kudos: 131

[1]

19 Dec 2025, 21:59

Kudos

Bookmarks

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.

officiisestyad

Joined: 25 Oct 2025

Last visit: 14 Sep 2026

Posts: 44

Own Kudos:

[1]

Given Kudos: 37

Location: India

Concentration: Entrepreneurship, Real Estate

GMAT Focus 1: 515 Q78 V79 DI70 Verified score

GMAT 1: 510 Q50 V47 Verified score

GPA: 10

WE:Other (Other)

GMAT Focus 1: 515 Q78 V79 DI70 Verified score

GMAT 1: 510 Q50 V47 Verified score

Posts: 44

Kudos: 37

[1]

19 Dec 2025, 22:00

Kudos

Bookmarks

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

Abhiswarup

Joined: 07 Apr 2024

Last visit: 04 May 2026

Posts: 198

Own Kudos:

172

[1]

Given Kudos: 42

Location: India

Posts: 198

Kudos: 172

[1]

19 Dec 2025, 22:19

Kudos

Bookmarks

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.

truedelulu

Joined: 01 Sep 2025

Last visit: 07 Sep 2026

Posts: 81

Own Kudos:

[1]

Given Kudos: 18

Posts: 81

Kudos: 70

[1]

19 Dec 2025, 22:20

Kudos

Bookmarks

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

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

Sumimasen

Joined: 21 Jan 2024

Last visit: 27 Jul 2026

Posts: 48

Own Kudos:

[1]

Given Kudos: 13

Posts: 48

Kudos: 43

[1]

19 Dec 2025, 22:50

Kudos

Bookmarks

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.

Attachments

PS2.jpeg [ 70.35 KiB | Viewed 399 times ]

harishg

Joined: 18 Dec 2018

Last visit: 30 Aug 2026

Posts: 176

Own Kudos:

174

[1]

Given Kudos: 31

GMAT Focus 1: 695 Q88 V84 DI81 Verified score

Products:

GMAT Focus 1: 695 Q88 V84 DI81 Verified score

Posts: 176

Kudos: 174

[1]

19 Dec 2025, 23:31

Kudos

Bookmarks

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