Last visit was: 02 Sep 2026, 15:38

It is currently 02 Sep 2026, 15:38

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 02
Free full-length test!
08:00 PM PDT
-
09:00 PM PDT
Video explanations + diagnosis of 10 weakest areas + 150+ short videos + study plan!
Sep 03
From ‘GMAT Is Not for Me’ to a 695— Beating Her Target by 50 Points
01:30 AM EDT
-
02:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Aug 27
BIG NEWS! We’ve been nominated, and we’d love your vote!
10:00 AM EDT
-
11:00 AM EDT
We’ve worked incredibly hard to build TTP into the best test prep experience possible, and it would mean a lot to us to win Newsweek’s 2026 Readers’ Choice Award for Best Test Prep. If TTP has helped you, we’d be incredibly grateful for your vote.
Sep 02
615 to 695 in 2 Months Despite a 70+ Hr Work Week
12:30 AM EDT
-
01:30 AM EDT
Thought your job leaves no time for GMAT prep? After multiple GMAT attempts and a demanding consulting job, Kshitij improved to 695 in under 2 months. Learn the mindset shifts, study strategy, and execution changes that made the breakthrough possible.
Sep 02
Save on All Magoosh Study Plans
01:00 PM EDT
-
11:59 PM EDT
Take 30% off all study plans with code COURSEWORK - Expires on Monday 9/7
Sep 04
How to Manage GMAT Test Anxiety Like an Elite Performer
10:00 AM PDT
-
11:00 AM PDT
What does test anxiety really look like on the GMAT? For many test takers, it does not look like a panic attack. Instead, it shows up as brain fog, rereading, negative comparisons, mental noise....
Sep 06
Live Online Class with Top EA Expert Logan Thompson
09:30 AM EDT
-
12:30 PM EST
Top scores are possible when you enroll in a powerful EA course, taught live online + 6 months access to TTP OnDemand video courses included! Perfect class schedule and easy course access for working professionals. Class starts Sundays Sept. 6, 2026.
Sep 07
All GMAT Club Tests are free and open for Labor Day on Monday, September 7th!
12:01 AM PDT
-
11:59 PM PDT
On the 7th, GMAT Club Tests will be Free! Including all the Quizzes, Questions, and Tests. 12 AM - 11:59 PM PDT
Sep 09
Masters' Spotlight Fair
07:00 AM PDT
-
10:00 AM PDT
Meet AdComs and explore top Master’s programs - MiM, MiF, MSc, MSBA and more. - Application Fee Waivers - Free 1-Week of GMAT Club Tests: - Master's Application Toolkit - Grand Prize Giveaway
Sep 15
Live Online Class with Top GMAT Expert Rich Zwelling
07:00 PM EDT
-
09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 21
Live Online Bootcamp Class with Top GMAT Expert Chris Peckover
08:00 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! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST

Back to Forum

Create Topic

GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave

1 2 3 4 5 6 7 8

Shlok02

Joined: 08 Jan 2024

Last visit: 02 Sep 2026

Posts: 65

Own Kudos:

[1]

Given Kudos: 6

Products:

Posts: 65

Kudos: 41

[1]

08 Jul 2026, 11:07

Kudos

Bookmarks

Bunuel

Three buses leave Riverton and travel along the same route toward Lakeview. Bus 1 leaves first and travels at a constant speed of 50 km/h. Bus 2 leaves Riverton 6 hours after Bus 1 and travels at a constant speed of 80 km/h. Bus 3 also leaves Riverton after Bus 1 and travels at a constant speed of 100 km/h. If Bus 2 and Bus 3 catch up to Bus 1 at the same time, how many hours after Bus 1 leaves does Bus 3 leave?

A. 2
B. 6
C. 8
D. 10
E. 16

This question was provided by GMAT Club
for the GMAT World Cup Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.

So we take a few variables in this case
Assuming B1 Leaves now we can denote the following
1. B2 leaves after 6 hrs (+6 hours)
2. B3 leaves after x hours ( x>6)
3. B1, B2 and B3 meet after y hours. We have to find x
At y hours the distances covered by each bus would be the same
Distance of B1 = 50*y
Distance of B2 = 80(y-6)
Distance of B3 = 100 (y-x)
Equating B1 and B3 we get
50y = 80(y-6),
we get y = 16 hours
Hence all 3 buses meet 16 hours after B1 leaves
Equating B2 and B3
80(y-6) = 100(y-x), putting y = 16
we get x= 8
Hence Answer is C (8 hours)

KEETAN11

Joined: 20 Apr 2026

Last visit: 24 Aug 2026

Posts: 18

Own Kudos:

[1]

Given Kudos: 34

Posts: 18

Kudos: 17

[1]

08 Jul 2026, 11:12

Kudos

Bookmarks

Bunuel

This question was provided by GMAT Club
for the GMAT World Cup Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.

let t be the total time in hours that Bus 1 travels before its caught up , p be the time after Bus 1 when Bus 3 leaves
Bus 1: Speed = 50kmph , time = t
Distance = 50t km
Bus 2: Speed = 80kmph, time = t-6
Distance = 80(t-6) = 80t-480
Bus 3: Speed = 100kmph, time = t-p
Distance = 100(t-p)

Since all distances are equal
50t = 80t -480
30t = 480
t = 16
80(t-6) = 100(t-p)
4t-24=5t-5p
t+24 = 5p
40 = 5p, p=8
So 8 hrs
C

Shreyamishra14

Joined: 17 Mar 2026

Last visit: 10 Aug 2026

Posts: 3

Own Kudos:

[1]

Given Kudos: 1

Posts: 3

Kudos: 1

[1]

08 Jul 2026, 11:14

Kudos

Bookmarks

let the time be t which we need to find , and distance where the bus met d, then t = d/50 - d/100 --->eq:1
we also know that bus 2 traveled in 6 hours less time, therefore d/50 - d/80 = 6 , from here , d = 800Km
substituting this value in eq:1
t=800/50 - 800/100 = 16 - 8 = 8 hours --> Ans C

rajdeep212003

Joined: 11 Dec 2024

Last visit: 02 Sep 2026

Posts: 156

Own Kudos:

[1]

Given Kudos: 17

Location: India

Schools: ISB '27 Kellogg INSEAD MiM '26

Products:

Schools: ISB '27 Kellogg INSEAD MiM '26

Posts: 156

Kudos: 78

[1]

08 Jul 2026, 11:39

Kudos

Bookmarks

Here I got option choice D . 8
--- so ,, Here I took the time needed Bus A to travel is t ,, Then the time needed for bus B to travel and meet with A is t - 6 ,, so 50 t = 80 (t - 6)
t = 16 hrs
--- Then I calculated the total distance that is covered by bus a is 50 km/h * 16 hours = 800 km
----- So now I have to Calculate the time Bus C took to travel 800 km --- 800 / 100 km/h = 8 hrs ,, so the difference between a and c travel time 16 -8 = 8 hrs so ,, here is the ans

Parry104

Joined: 30 Oct 2025

Last visit: 02 Sep 2026

Posts: 28

Own Kudos:

[1]

Given Kudos: 3

Location: India

Concentration: Technology, Operations

Schools: ISB '27 IIM NUS '27

GRE 1: Q170 V158 Verified score

GPA: 3.14

Products:

Schools: ISB '27 IIM NUS '27

GRE 1: Q170 V158 Verified score

Posts: 28

Kudos: 26

[1]

08 Jul 2026, 11:40

Kudos

Bookmarks

Bus A = 50km/h
Bus B = 80 Km/h
Bus C = 100km/h

Bus B starts 6 hours after A. Let them catch up after x Hours.

Distance travelled by A = 50(X+6)
Distance Travelled by B = 80(x)
Solving for X = 10 Hours.
Bus B catch's A after 10 hours of travel.==> total travel time for A = 16 Hours.

let bus C start after y hours from Bus A's Start time.
Time Bus C spent travelling = 16-x

Distance Bus A travelled = 50(16) = 800
Distance Bus C travelled = 100(16-y)
Solving for Y = 8Hours.

Answer C

Nsanghavi

Joined: 08 Jul 2025

Last visit: 29 Aug 2026

Posts: 59

Own Kudos:

[1]

Given Kudos: 20

Products:

Posts: 59

Kudos: 46

[1]

08 Jul 2026, 11:52

Kudos

Bookmarks

So the question mentions about the following:
a) Bus 1 having speed of 50 km/h (Slowest of all and the first to depart)
b) Bus 2 having speed of 80 km/h ( Departs 6 hours after Bus 1 leaves)
c) Bus 3 having speed of 100 km/h (In how much time after Bus 1 should it depart so that all 3 Bus meet at a same point ?)
To find this out, I have used the following approach:

I will eliminate Option A and Option B - Since the hours is less than 6 hours - the total time after which Bus 2 departs. If Bus 3 departs at the same time or less time as Bus 2 = then it will overtake Bus 2 and hence they will never be together.

Now, we have 3 Options left - So I just followed a trial and error method-
Option C - If Bus 3 leaves after 8 hours from Bus 1 the total distance convered after 9 hours from Bus 1 departure would be:
Bus 1 - 9*50 = 450 km
Bus 2 - (9-6) *80 = 240 km
Bus 3 - (9-8) * 100 = 100 km

Bus 1, Bus 2 , Bus 3 - Hour details
450, 240, 100 - 10th hour
500, 320, 200 - 11th hour
550, 400, 300 - 12th hour
600, 480, 400 - 13th hour
650, 560, 500 - 14th hour
700, 640, 600 - 15th hour
750, 720, 700 - 16th hour
800, 800, 800 -17th hour

Since I am able to get an answer which matches the condition of the question - I have chosen this option.

aincidunt

Joined: 18 Aug 2025

Last visit: 02 Sep 2026

Posts: 15

Own Kudos:

[1]

Given Kudos: 19

Location: Poland

Schools: INSEAD MiM '27

GMAT Focus 1: 735 Q90 V87 DI83 Verified score

Products:

Schools: INSEAD MiM '27

GMAT Focus 1: 735 Q90 V87 DI83 Verified score

Posts: 15

Kudos: 2

[1]

08 Jul 2026, 11:55

Kudos

Bookmarks

First find when Bus 2 catches up to Bus 1.

50x=80(x-6)
5x=8x-48
3x=48
x=16

Plug into Bus 1 to find total distance in 16 hours.

50(16)=800

Find how long Bus 3 travelled.

800/100=8 hours

16-8=8

Bus 3 waited 8 hours before departing.

C

Bunuel

This question was provided by GMAT Club
for the GMAT World Cup Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.

Stiwari4001

Joined: 25 Jul 2025

Last visit: 02 Sep 2026

Posts: 28

Own Kudos:

[1]

Given Kudos: 17

Posts: 28

Kudos: 23

[1]

08 Jul 2026, 12:04

Kudos

Bookmarks

Bus 1 (B1) leaves first, travels at 50 km/hr
Bus 2 (B2) leaves 6 hrs after Bus 1, travels at 80 km/hr
Bus 3 (B3) leaves after Bus 1 as well, travels at 100 km/hr

Both B2 and B3 catch up with B1 at the same time.
This essentially means that the distance covered for all three buses is the same, at different speeds in different times.
Let, distance traveled for all three buses = D

Although the question only says that B3 leaves after B1, since the speed is greater than B2 and distance is the same, we can safely assume it left even after B2. That immediately tells us that the answer is greater than 6, as B2 left 6 hrs after B1 and traveled the same distance.

For B1-
Let, time = X
D=50*X (since distance = speed*time) ------------(1)

For B2-
D=80*(X-6) (since B2 left 6 hours AFTER B1, so it will travel for 6 hrs lesser)

Since distance is same for both:
50*X=80*(X-6)
50X=80X-480
480=80X-50X
480=30X
X=16

Therefore total distance traveled by all three buses (D):
=50*16 (from (1))
=800 kms.

Now for B3:
D=800 kms. (as we just found)
Speed= 100km/hr
Time traveled = 800/100
=8 hrs.

However, note that the question wants to know how many hours AFTER B1 did B3 leave.

Since B1 traveled for 16 hrs, and B3 traveled for 8 hrs
B3 left after (16-8= 8 hrs) after B1.

Ans: C

Rishm

Joined: 05 May 2024

Last visit: 01 Sep 2026

Posts: 369

Own Kudos:

[1]

Given Kudos: 82

Location: India

Schools: Johnson '27 LBS '27 IESE '27

GPA: 10

WE:Consulting (Consulting)

Products:

Schools: Johnson '27 LBS '27 IESE '27

Posts: 369

Kudos: 78

[1]

08 Jul 2026, 12:18

Kudos

Bookmarks

I did not use any formula here, now notice 2 & 6 can never be an option because they are less than 6 hrs of Bus 2, which has smaller speed. So we have 3 active options 8,10 & 16. Now I used 8 as my answer first and plugged in the numbers, at 8 hrs, 1st bus will have travelled 400km, 2nd at 160 and 3rd at 0, now keep adding them and we see 800km at 16th hour. But notice we want time when it left, hence it is 16-8=8 which is C, our answer

Junaidhh99

Joined: 14 Apr 2026

Last visit: 02 Sep 2026

Posts: 6

Own Kudos:

[1]

Given Kudos: 12

Posts: 6

Kudos: 4

[1]

08 Jul 2026, 12:22

Kudos

Bookmarks

All about relative speeds and times, while keeping a general timeline in mind.

B1 - Bus 1; B2 - Bus 2; B3- Bus 3; t= total time elapsed since bus 1 left at t=0;

At t=0, B1 leaves Rivertown at 50kph

At t=6, B2 leaves Rivertown at 80kph

B2/R |-----------<-D->-----------| B1

D=50*6 = 300kms

Time taken by B2 to catch B1 = 300 / (80 - 50) = 10 hours

in the absolute timeline, this means we see the Bus 1 bois at t=16 hours

Now if B3, left at the same time as B2, it would catch B1 in 300/50 = 6 hours, which would be t=12 hours

Looking at the options, just start plugging it in

Option A - 2 hours later,
t=8,
D=50*8=400kms

Time taken by B3 to catch B1 = 400/(100-50) = 8 hours --> t=8+8= 16 hours!

This is how my brain went about this, would love to see more optimized ways of doing this question

jefferyillman

Joined: 01 Dec 2024

Last visit: 01 Sep 2026

Posts: 112

Own Kudos:

[1]

Given Kudos: 3

Posts: 112

Kudos: 81

[1]

08 Jul 2026, 12:22

Kudos

Bookmarks

Catch up time for bus 2.

Bus 1 travels 50 km/h for t hours, so goes 50t kim.

Bus 2 leaves 6 hours later, so travels t-6 hours at 80 km/h

when they meet 50t = 80 (t-6)
solve for t 30t = 480
t=16
so buses meet 16 hours after bus 1 leaves

Bus 1 travels 50km/h x 16 hours = 800km

Bus 3 travels 100 km/h so 800km/100km/h = 8 hours

Since meeting occurs 16 hours after bus 1 left. Then bus 3 must have left 16-8 = 8hours after bus 1

Answer C 8 hours

kavyamahajan9

Joined: 09 Jun 2021

Last visit: 16 Aug 2026

Posts: 44

Own Kudos:

[1]

Given Kudos: 1

Posts: 44

Kudos: 40

[1]

08 Jul 2026, 12:36

Kudos

Bookmarks

Option C - 8

Bus 1 speed - 50km/h - 50km/t
let time = t

Bus 2 - time = (t-6) - as it left 6 hours after bus 1
Speed =80(t-6) km

So bus 1 and 2 meet so

50t = 80(t-6)
t =16
Bus 1 meets bus 2 after 16 hours

Bus 3
speed =100km/t
so bus 1 total distance when it meets bus 2 = 50 x16 = 800

so Bus 3 meets bus 1 = 800/100 = 8 hours
Since meeting of bus 2 occurred 16 hrs after 1 leaves and bus 3 occurred after 8 hrs
so 3 left by = 16-8 hrs
= 8

Bunuel

This question was provided by GMAT Club
for the GMAT World Cup Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.

GlassChamp

Joined: 05 Jun 2026

Last visit: 23 Aug 2026

Posts: 20

Own Kudos:

[1]

Given Kudos: 1

Location: India

Schools: ISB '27 Kellogg

GPA: 7.7

Schools: ISB '27 Kellogg

Posts: 20

Kudos: 15

[1]

08 Jul 2026, 12:48

Kudos

Bookmarks

Speed of Bus 1 = 50km/hr
Distance travelled by Bus 1 when Bus 2 just started to leave (after 6hrs) = 50*6 = 300km
Speed of Bus 2 = 80km/hr
Using the concept of relative velocity,
Bus 2 has to the relative velocity of 80-50=30km/hr with respect to Bus 1, and Bus 2 has to cover 300km with this speed.
Hence, time taken by Bus 2 to reach Bus 1 = 300/30 = 10hr
Now,
Total time Bus 1 was on journey = 6 + 10 = 16hrs
Total distance covered by Bus 1 = 50* 16 =800km
Bus 3 has to cover this 800km to reach Bus 1.
Time taken by Bus 3 to cover this distance = 800/100 = 8hrs
Hence, Bus 3 will reach after 16-8 = 8hrs of Bus 1 leaving.
Thus, Correct ans : C. 8

princesskitt

Joined: 21 May 2026

Last visit: 01 Sep 2026

Posts: 17

Own Kudos:

[1]

Given Kudos: 7

Products:

Posts: 17

Kudos: 13

[1]

08 Jul 2026, 13:05

Kudos

Bookmarks

3 buses: Bus 1, Bus 2 and Bus 3
We know speed = distance/time and hence distance = speed x time
Speed of Bus 1 = 50km/hr
Speed of Bus 2 = 80km/hr
Speed of Bus 3 = 100km/hr

Bus 1 leaves first at say time t1 and travels a distance of d1. Therefore, 50 = d1 / t1

Now, Bus 2 leaves 6 hours after Bus 1 but Bus 2 is faster than Bus 1 and hence can travel the same distance in a shorter period of time and hence we can write the time of Bus 2 as (t1 - 6). Therefore, 80 = d2 / (t1 - 6).

We know that this bus catches up with Bus 1 at a particular distance.
Hence, to get this exact point where Bus 2 has caught up with Bus 1, we can say that their distance travelled is the same.
Hence, equating d1 and d2, we get 50 x t1 = 80 x (t1 - 6)
Therefore, t1 = 16 hours. This implies that Bus 2 catches up with Bus 1 when Bus 1 has travelled for 16 hours.

This means that Bus 1 has travelled at 50km/hr for 16 hours and hence has covered a total distance of 50 x 16 = 800km.
According to the question, Bus 2 and Bus 3 catch up with Bus 1 at the same time and hence also at the same distance and since Bus 2 caught up with Bus 1 at a distance of 800km, so did Bus 3.
Therefore, to travel 800 kms, bus 3 at its speed of 100km/hr would have to travel for 800 / 100 = 8 hours. But Bus 1 left before Bus 3 and travelled for 16 hours before it was caught up with by Bus 3. Therefore, Bus 3 would have to leave Riverton 16 - 8 = 8 hours after Bus 1.

Hence, final answer: C. 8

mehtyas

Joined: 27 May 2025

Last visit: 02 Sep 2026

Posts: 74

Own Kudos:

[1]

Given Kudos: 823

Location: India

Concentration: Finance, Strategy

Products:

Posts: 74

Kudos: 61

[1]

08 Jul 2026, 13:43

Kudos

Bookmarks

IMO, the answer should be Option C - 8.

B1 travels 50km/h
B2 travels at 80 km/hr and leaves 6 hours later than B1.
B3 travels at 100 km/h and leaves x hours later than B1.

Let "t" be the time in hours after B1 leaves when all buses meet.

B1 travels 50t km.
B2 travels 80 (t-6) km

Since they meet at the same time, then --> 50t = 80(t-6)
Simplifying the equation, we find that t =16 hours

B3 travels 100 (16-x)
So, 100 (16-x) = 50* 16
--> x = 8 (Answer - Option C)

MANASH94

Joined: 25 Jun 2025

Last visit: 02 Sep 2026

Posts: 165

Own Kudos:

121

[1]

Given Kudos: 27

Location: India

Schools: IIM IIM ISB

GMAT Focus 1: 525 Q80 V76 DI72 Verified score

GPA: 2.9

Products:

Schools: IIM IIM ISB

GMAT Focus 1: 525 Q80 V76 DI72 Verified score

Posts: 165

Kudos: 121

[1]

08 Jul 2026, 13:46

Kudos

Bookmarks

Bus 1 leaves at t0 or say 0 hours @50kmph.
Bus 2 Leaves at 6 hours @80kmph, Bus 3 leaves at some x hours which should be more than 6 @100kmph.
And lets say all the buses meets at t hours:

For Bus 1 and Bus 2 meeting we can say that 80(t-6) = 50t
=> t =16

Now at the same time bus 3 also must meet them:
So total distance that bus 3 covered during this time = total distance bus 1 covered = 50 x 16 = 800

Now total time 800km/100kmph of bus 3 should give us the time.
So = 8 hours IMO.
Ans: D.

Bunuel

This question was provided by GMAT Club
for the GMAT World Cup Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.

satanclaus

Joined: 04 Sep 2025

Last visit: 26 Aug 2026

Posts: 10

Own Kudos:

[1]

Given Kudos: 17

Posts: 10

Kudos: 7

[1]

08 Jul 2026, 13:48

Kudos

Bookmarks

By the time bus 2 starts its journey, Bus 1 is 6*50=300 km ahead.
To cover this distance, it would take Bus 2 300/(80-50)=10 hours
In these 10 hours, Bus 2 covers 10*80 = 800 km
To travel the same distance and reach this point at the same time, Bus 3 would take 800/100 = 8 hours.

Ans. C. 8

MG121

Joined: 16 Nov 2022

Last visit: 11 Aug 2026

Posts: 19

Own Kudos:

[1]

Given Kudos: 16

Posts: 19

Kudos: 26

[1]

08 Jul 2026, 13:57

Kudos

Bookmarks

B1: T=0, speed = 50 kph
B2: T=6hrs, speed =80kph
B3: T=x, speed= 100 kph

dist between B1,B2 at T=6hrs ==> 50*6 = 300 km
in order for B2 to catchup with B1 it has to cover 300 km relative distance
time= (relative distance)/(relative speed) ==> 300/ (80-50) ==> 10hrs

final distance when B1,B2 meet = 80*10= 800 kms
B3 will take 800/100 ==> 8hrs to cover this distance ==> x+8=6+10 ==> x=8hrs

Bunuel

This question was provided by GMAT Club
for the GMAT World Cup Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.

priyanka06

Joined: 10 Sep 2025

Last visit: 02 Sep 2026

Posts: 60

Own Kudos:

[1]

Given Kudos: 1

Products:

Posts: 60

Kudos: 35

[1]

08 Jul 2026, 13:58

Kudos

Bookmarks

time for :-
bus1 - 6+x
bus2- x
for catch-up situation they both travel same distance, which is 800 kms.
Bus 3 can travel 800km in 8 hours hence to catch up with bus1 & 2 it leaves 8hrs after bus1

Bunuel

This question was provided by GMAT Club
for the GMAT World Cup Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.

MakB

Joined: 05 May 2025

Last visit: 02 Sep 2026

Posts: 73

Own Kudos:

[1]

Given Kudos: 146

Posts: 73

Kudos: 29

[1]

08 Jul 2026, 14:00

Kudos

Bookmarks

I would love to get any shortcut for this as I took good 5 minutes for this one

Here is what I did,
S1 = D1/T1 ---for bus 1
so, D1= 50T1
S2 = D2/T2 ---for bus 2---let the distance be same as bus 1 so that they can meet
so, D2=D1=80 (T1-6) ---Given it started 6 hours later than bus 1
equating both we get,
50T1= 8 (T1 - 6 ) = 80T1 + 480
30T1 = 480
T1 = 16 hours

Now,
D3= S3 (16- t ) -----t is hours after bus 3 started after to bus 1
Let total distance be equal to the distance travelled by bus 1 in total so that they meet
= 16x50= 800 km
So, 800 = 100 (16-t)
16-t= 8
t= 8 hours
Hence C (phew! )
(PLEASE TELL ME SOME SHORTCUT FOR THIS!)