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

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Zeus_

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08 Jul 2026, 08:27

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I think the answer is C.

Bus 1 = 50km/hr
Bus 2 = 80 km/hr
Bus 3 = 100 km/hr

Bus 1 had a headstart and covered distance of 300 km (50km/hr x 6 hour).

Bus 2 had to cover extra distance of 300 km with relative speed of 30km/hr (80-50) so it took 10 hour to cover that. 6 hour of headstart + 10 hour to cover that distance, so bus 1 travelled 16 hours in total which is the total distance.

Total distance by bus 1 = 16 hour x 50km/hr = 800 km

Bus 3 time = 800km/100 = 8 hr

so Bus 3 should leave after 8 hour to reach the point where all buses meet together

Kinshook

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08 Jul 2026, 08:28

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3 buses leave Riverton and travel along the same route toward Lakeview.

Let Bus 3 leave x hours after Bus 1 leave and let all buses meet t hours after Bus 1 leave.

Bus 1:
Speed = 50 km/h
Time = t hours
Distance = 50t km

Bus 2:
Speed = 80 km/h
Time = t -6 hours
Distance = 80(t-6) km

Bus 3:
Speed = 100 km/h
Time = t - x hours
Distance = 100(t-x) km

Since all 3 buses meet t hours after Bus 1 leave, they travel the same distance.

50t = 80(t-6) = 100(t-x)
50t = 80t - 480
t = 16 hours

50*16 = 100(16-x)
800 = 1600 - 100x
x = 8 hours

IMO C

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08 Jul 2026, 08:29

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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

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B1 would have travelled 300 kms in 6 hrs with 50km/hr speed. Now B2 to catch upto B1 would need Time=300/(80-50)= 10hrs .
Now B3 takes same amount of time to cathup to B1 is what given in the question. So B1 would have travelled 500 kms in 10hrs. Meaning B1 is 500 kms ahead and it would take 10 hrs for B1 to travel 500kms with 50km/hr speed. So B3 left 10 hrs after B1 left Riverton.
Answer: D

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08 Jul 2026, 08:30

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Bunuel

This question was provided by GMAT Club
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IMO, answer is option C. Here is how.

Suppose Bus 1 travelled for time "t", and Bus 3 travelled for time (t-x).
So, distance travelled by bus 1 = 50t
Distance travelled by bus 2 = 80(t-6)
And, distance travelled by bus 3 = 100(t-x)

As, the distance travelled by all buses will be same:
50t = 80(t-6)
t = 16

So, 50 * 16 = 100(16-x)
x = 8 hours.

Arya25

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08 Jul 2026, 08:33

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let time taken by Bus 1 be x , then distance = 50x
for bus2, x+6 is the time taken and so distance = 80(x+6)
since both the distances covered are same, equate them i.e 50x=80x+480
x=16. means time taken by bus1 is 16hrs.
Hence , distance travelled = 50*16=800km

Now, bus3 travels at a speed of 100km/hr.
so 100km in 1 hr, then 800km in 8hrs.
Which means when bus1 covers 800km in 16hrs, bus3 would cover 800km in 8hrs which says that bus3 met bus1 after 16hours so bus3 would have left 8hrs after bus1 left.

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08 Jul 2026, 08:33

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Answer is D - 10 hours.

Bus 1 travels at a constant speed of 50km/hr. Since Bus 2 leaves 6 hours after Bus 1, Bus 1 must have covered 300km in these 6 hours. Now it is said that Bus 2 and 3 catch up to Bus 1 at the same time. So we need to calculate the time at which Bus 2 would have met/overtaken bus 1. Bus 2 is travelling at 80km/hr. To cover this distance of 300 km, the relative speed is (80-50)km/hr which is 30km/hr, since they are both travelling in the same direction. Hence time taken = (300km/30km) which is 10 hours. This is the same time that has been taken by Bus 3 to catch up to Bus 1 and bus 3 is travelling at 100km/hr. Assume Bus 3 left x hours after Bus 1. So Bus 1 would have covered 50x km. This 50x km needs to be covered in 10 hours with a relative speed of (100-50)km/hr which is 50km/hr. Hence x is 10.

Hence, Bus 3 left 10 hours after Bus 1 left.

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08 Jul 2026, 08:34

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B1 = 50 km\h started at t = 0
B2 = 80 km\h started at t =6 h
B3 = 100 km\h started at t=x h
needed x

let D be the distance at which they meet

for B1
D = 50T
assuming T is the time it took for it to meet

for B2
D = 80(T-6)

and for B3
D = 100(T-x)

now all three D are equal

equating 1 and 2
50T = 80T - 480
T= 16

now equating 2 and 3, with T = 16

800 = 100(T-x)
8 = 16 - x
x = 8

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08 Jul 2026, 08:38

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As bus 1 , 2 ,3 reach the same destination ,
let's suppose , they catch up each other in S km away.
Now, supoosing bus 1 took t hr to reach S with a speed of 50km per hour ,
S =50 . t ---------- ( i )
and for bus 2 ,
S =80.(t-6) ------------( ii )[ As, bus 2 took 6 hurs less then bus 1]
similiarly , for bus 3,
S =100.(t-x) [ supposing bus 3 left x hr after bus 1 left]

From ( i ) and ( ii ) ,
50t= 80t-480
or, 30t=480
or , t =16

so , bus 1 took 16 hr , bus 2 took ( 16-6) =10 hrs to reach S=50 . 16 =800 km

For bus 3 ,
800= 100 ( t -x )
or , t-x = 8
or, 16-x =8
or , x =8

so bus 3 left 8 hrs after bus 1 left .

ANS.. OPTION C. 8

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08 Jul 2026, 08:38

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distance travelled by bus 1in 5 hours = (50 * 6) = 300 kms
relative speed concept:

Distance bus 2 needs to catch up on =300 kms
relative speed. = 80 - 50 = 30 kmph

time taken to catch up by bus 2 =300/ 30 = 10 hrs.

Total time travelled by bus 1 when all three caught up = 10 + 6 =. 16 hrs
TOtal distance travelled in that time = 50 * 16 kms

total time taken by bus 3 = (50 *16)/ 100 = 8

So bus 3 started 16- 8 = 8 hrs after bus 1.

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08 Jul 2026, 08:38

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lets say bus1 travels for t hrs. Distance covered = 50t.
bus 2 travels for t-6 hrs. Distance covered = 80 (t-6)
since bus 2 catches up with bus 1 -> 50t = 80 (t-6) -> t=16

lets say bus 3 lets after x hrs. distance covered = 100 (t-x)
since bus 3 catches bus 1, 100 (t-x) = 50t and t=16 from above.
x=8

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08 Jul 2026, 08:39

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Bus 1 is going 50 km an hour and Bus 2 leaves 6 hours later going 80 km per hour.

After 6 hours, Bus 1 has already gone 300 km so to catch up Bus 2 needs to close that distance. To calculate how long it would take to do this I subtracted Bus 1's rate (50 km/h) from Bus 2's rate (80 km/h) to see how much closer Bus 2 is getting to Bus 1 every hour which was 30 km. Then I divided the 300 km distance by 30km to get that it takes 10 hours after Bus 2 leaves for it to catch up.

You can multiple that 10 hours by Bus 2's rate of 80 km/h to see Bus 2 and 1 are 800 km away when they meet. To find how long after Bus 1, Bus 3 left we need to see how long it would take Bus 3 to cover 800 km, so 800 km/ 100 km/h = 8 hours. Bus 1 has been driving for 16 hours (6 hours before Bus 2 left + 10 hours for Bus 2 to catch up), so in order for all the buses to catch up at the same time, Bus 3 has to leave 8 hours after bus 1 16-8 =8.

The answer is C. 8

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08 Jul 2026, 08:39

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Answer: C) 8

Bus 1, 2 and 3 will all drive the same distance so we can set them equal to each other. Let's first find out how far they drive and how long Bus 1 drives.

B1: 50 * t = d
B2: 80 * (t-6) = d
50t = 80t - 480, t = 16 h
d = 50 * 16 = 800 km

For Bus 3: 100 * t = 800, t = 8
Bus 3 drives for 8 hours which is (16 - 8 = 8) 8 hours less than Bus 1 and therefore leaves 8 hours after Bus 1.

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08 Jul 2026, 08:47

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BUS 1 TRAVELS 6*50 KMS=300 KMS IN 6 HOURS BEFORE ANY OF BUS 2 & 3 STARTS

NOW IN NEXT t HOURS( AFTER 6 HOURS TIME) BUS 1 WILL TRAVEL---- 300+50t
BUS 2 DURING THAT PERIOD WILL TRAVEL ---- 80t
as per question all should meet at this time so distance covered by each will be same as beginning point is same.

300+50t=80t---> 30t=300---> t=10 hours
distance travelled by each will be 80*10= 800 kms
so time required by bus-3 to travel this distance will be 800/100=8 hours.

total time taken by bus-1 =6+10= 16 hours
so bus-3 will start 16-8 hours to cover the same distance to meet.
ANS: C

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08 Jul 2026, 08:50

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50*16=80*10=100*8

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08 Jul 2026, 08:55

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Three objects traveling in the same direction and speeds are provided for each. We are being asked for the number of hours after which Bus 1 departs that Bus 3 departs.

Let's take it step by step using the given information.

First, we know that Bus 1 has been traveling for 6 hours at a rate of 50km/hr when Bus 2 departs, meaning there is a distance / gap of 300km to cover.

This gap will decrease every hour by their subtracted rates, so it will decrease by 80 - 50 = 30km/hr. So it takes 10 hours for Bus 2 to catch up to Bus 1. That means that Bus 1 has traveled a total of 10 + 6 = 16 hours at a rate of 50 km/hr = 800 km total.

Since Bus 3 and Bus 2 catch up to Bus 1 at the same time, we know that Bus 3 has traveled 800km at its rate and we can use this to find the time it has spend traveling. 100t=800 = 8 hours. That means it has spent 8 hours moving before it catches up to Bus 1 which has been traveling 16 hours, therefore it must have left 8 hours after Bus 1, and the answer is C.

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08 Jul 2026, 08:56

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initially ,
Bus A started at 50km/hr, after 6hr bus B started at 80km/hr.
distance by Bus A= 50km/hr*6= 300km
relative speed between Bus A and bus B = 80-50= 30km/hr
time take for bus B to catch bus A = 300/30= 10Hr

total distance travelled by Bus A when Bus B catch Bus A = (10+6)hr * 50km/hr = 800km

by this time,
Bus C travelling at 100km/hr , total time needed to cover 800km = 800/100= 8hr

Bus c leaves (16-8)= 8hr after BUs A

option C is correct

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08 Jul 2026, 08:59

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Tried to map the logical sequence of events.

1. Bus 1 left, travelling at 50km/hr. Bus 2 and 3 are still stationary
2. Bus 2 left, 6 hours after Bus 1, at 80 km/h: At this point, Bus 1 would have already 300 kms (50 km/hr * 6 hrs = 300 kms).
3. Bus 3 has also left at some point, travelling at 100 km/hr. After travelling for some time, Bus 3 along with Bus 2, catchup to Bus 1 at the same time.

Now since Bus 2 and 3 catch up at the same time, we need to find this time and that should help us get when Bus 3 left exactly.

To find this time, we can use the Gap Distance Concept.
Relative Speed = (Gap Distance b/w the at the point the bodies) / (Total time for which both bodies move simultaneously, denoted going forward by T)

We know that Bus 1 had already travelled 300 kms by the time Bus 2 started moving (i.e. they both started moving simultaneously)

Since they are moving in the same direction, Relative Speed will be the difference of their speeds i.e. 80 - 50 = 30 km/hr

Thus, the above formulat becomes
30 = 300/T
Thus, T = 10 hours.

So the Buses crossed each other 10 hours after Bus 2 started moving (i.e. 16 hours after Bus 1 started moving)

Now in these 16 hours, Bus 1 would have travelled 16*50 = 800 kms

Hence, Bus 2 and Bus 3 catch up with Bus 1 when Bus 1 has travelled 800 kms.

For Bus 3 to travel 800 kms it would take 800/100 = 8 hours (Time = Distance/Speed)

Thus, Bus 1 travelled for 16 hours, and Bus 3 travelled for 8 hours.

This means Bus 3 left 8 hours after Bus 1.

Answer (C)

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08 Jul 2026, 09:00

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My answer is C.) 8 hours
A covered 50*6= 300 km in 6 hours at speed of 50km/hr
B will catch up A in (Distance in between/Relative speed)= 300/30= 10 hours.
Therefore, meeting point is 80*10= 800kms
Time taken by A to reach meeting point= 10+6=16 hrs
Time taken by C to reach meeting point= 800/100=8 hrs. Hence C leaves 8 hours after A leaves.

Bunuel

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The answer is 8. C.

Bus 1 travels at 50kmph
Distance= speed * time
D1= 50*t

Bus 2 travels at 80 kmph but leaves 6 hours after Bus 1
D2=80 (t-6)

Lets equate these two equation as the distances travelled by both would be same for them to meet.

D1=D2
50t= 80 (t-6)
50t= 80t-48
48=30t
t=16

So, bus 2 meets bus 1 after travelling for 16 hrs.

Now let's calculate the distance.
D1= 50t, replacing the value of t, we get
50*16= 800 kms

So, for bus 3 to meet bus 1 and 2 at 800 kms
it will have to travel at the speed of 80 kms for x hrs.
x= 800/ 100= 8 hrs.

To find the time difference between bus1 and bus 3 leaving from Riverton
time travelled y bus 1- time travelled by bus 3
= 16-8
=8

Hence Bus 3 leaves 8 hrs after bus 1.