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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 03:36
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equating distances Bus1 and Bus2:
50*timeBus1 = 80*timeBus2 = 80*(timeBus1-6)
30*timeBus1 = 480
timeBus1 = 16

total distance:
50*16 = 800 km

Bus3 takes to cover 800 km:
100*timeBus3 = 800
timeBus3 = 8

So Bus3 must delay its departure:
16-8 = 8 hours

Answer C
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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 03:41
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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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Bus 1 gets a 6-hour head start:

Distance covered : 50 * 6 = 300 km

Bus 2 is faster by: 80 - 50 = 30 km/h
Time for Bus 2 to catch Bus 1: 300 / 30 = 10 hours
So they meet: 6 + 10 = 16 hours after Bus 1 leaves.

Bus 3 travels at 100 km/h, so the distance it covers before meeting

Bus 1 is: 50 * 16 = 800 km
800 / 100 = 8 hours

So Bus 3 leaves 8 hours after Bus 1.

Option C
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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 03:42
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Let Bus 2 travels for t1 when bus 3 leaves. They all catch up after t hr bus 3 leaves

80t1 + 80t =100t
t=4t1......eq.1


300+50t1+5t = 100t....use eq.1
6=3t1
t1=2
& t=8

Total time 6+2 = 8hr

Ans C
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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 03:49
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Let y be hours that Bus 1 and Bus 2 have to travel to get to the meet-up place (after Bus 1 leaves 6 hours)
Let z be hours that Bus 3 have to travel to get to the meet-up place

We have 2 equations:
1. 50x6 + 50y = 80y (because after bus 1 travel 6 hours ~300km, bus 1 and bus 2 travel the same time y (hours) until to meet-up place) -> y = 10
2. 80y = 100z (because bus 2 and bus 3 have to travel the same amount of distance to meet-up) --> 80x10 = 100z -> z = 8

The total of time that bus 1 has to travel = 10+6=16 hours
The total of time that bus 3 has to travel = z = 8 hours
--> After 16-8=8 hours, Bus 3 leave -> C
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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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 03:51
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We are given that speed of three busses which leave Riverton and travel on same route towards Lakeview,
their speeds are B1 = 50 kmph; B2 = 80 kmph; B3 = 100 kmph

It is given that B2 leaves Riverton 6 hours after B1 leaves, so by that time B1 would cover = 50 * 6 = 300 Km

The time B2 would take to meet B1 = 300 / relative speed.
Since both the buses are moving in same direction, relative speed = 80 - 50 = 30

So time taken by B2 to meet B1 = 300 / 30 = 10 Hours. So B2 would meet B1 after 10 hours and at the same time B3 would also meet B1.

SO by that incident, B1 would have travel for 6 + 10 = 16 hours and would have cover 16 * 50 = 800 Km. B3 would have covered this distance by 800 / 100 = 8 Hours.

So B3 leaves Riverton (16 - 8 = 8 hours) after B1 leaves Riverton

Hence option C is correct as per my calculation
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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 03:55
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t1: time Bus1
t2: time Bus2
t3: time Bus2

t2 = t1 - 6

50*t1 = 80*t2 = 80*(t1-6)
30*t1 = 480
t1 = 16

They meet at 50*16=800 km

100*t3 = 800
t3 = 8

t1 - t3 = 16 - 8 = 8 hours

The correct answer is C
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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 04:53
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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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speed difference between bus 2 and bus 1 is 30km/hr. so bus 2 will travel additional 30 km every hour. for 300km( distance travelled by bus 1 in 6hrs) , so after 300/30 = 10hrs, bus 2 will be at 800km, bus 1 , 10+6 = 16 hrs, will be at 800km, it will take 8hrs for bus 3 to travel 800km.so it will leave after 8hrs. option C.
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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 05:00
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Bus 1 travels for t hrs @ 50km/hr and Bus 2 travels 6 hrs later that means it travelled 6 hrs less than Bus 1 (t-6) hrs we know the speed 80kms/hr we can equate the distance as 50*t= 80*(t-6) therefore we get 80t-480=50t leads to t =16. Bus 3 also started after Bus 1 but we dont know after how many hours and the differential time is what we want to know 50t=100*(t-a) where a is the no of hrs after which bus 3 left . we know t=16 >> 50*16=800=100*(16-a) we get know a=8hrs
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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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 05:03
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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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This question needs to be done by keeping things simple.
Let bus 1 travels for 't' hours before it got catch up by bus 2 and 3.
To find - y
Distance of all bus would be equal- 50*t = 80*(t-6) = 100*(t-y)
Solving first equation, gives t=16 & distance covered = 800 km
Implies that bus C with speed 100 km/hr would cover it in 8 hrs. Therefore, y = 8. Option C- correct
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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 05:15
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The initial distance between Bus 1 and Bus 2 is 6 hr * 50 km/hr =300 km. Bus 2 travels 80 km/hr - 50 km/hr =30 km/hr faster than Bus 1. So, Bus 2 covers this distance in 300 km / 30 km/hr = 10 hr, Bus 1 traveled 10 hr * 50 km/hr =500 km at this time and 300 km previously, hence a total 0f 800 km. Therefore, Bus 3 took 800 km/ 100 km/hr =8 hrs to cover the distance and left 8 hours after Bus 1.

The answer is C. 8

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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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 05:18
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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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Deriving the given information from the question first:
Bus 1: Speed: 50 km/h; Time: Unknown
Bus 2: Speed: 80 Km/h; Time: 6 hours after B1
Bus 3: Speed: 100 Km/h; Time: Unknown but leaves after Bus 1 and meets it with Bus 2

Now, since all 3 travel the same distance (Same meet point), we can equate each
Let the time taken by Bus 2 be "t"; therefore, time taken by Bus 1 will be "t+6"
Distance by Bus 1 = Distance by Bus 2
50*(t+6) = 80 * t
We get; t = 10
Therefore, Bus 1 and Bus 2 travelled for 16hr and 10hr respectively covering a distance of 800Km (50*16 or 80*10 (speed * time))

Therefore, for Bus 3 to cover 800 km, it will need 8hr at 100 km/h
Finally, the time travelled by Bus 3 is 8 hr, and Bus 1 is 16 hr
Therefore, Bus 3 left 8 hr after Bus 1
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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 05:26
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bus 1 and 2 meet after 300+ 50 t= 80(t) t=10
bus 1 will travel 300+500=800 km
bus 3 will travel 800=100t t=8 so it left 8hrs after b1
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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 05:28
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Let time taken by bus 1 be x.

Distance for bus 1: D= 50*x
Distance for bus 2: D= 80(x-6)
Equate both,
50x= 80(x-6)
30x= 480
x=16,
Distance = 50*16= 800 kms

We know speed of bus 3 is 100km/h.
Time taken by bus 3: 800/100 =8 hrs.

So to reach bus 1 at the end of 16 hrs, bus 3 leaves after 16-8 = 8 hrs.

Hence, option c

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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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 05:43
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t = time Bus1

50t = 80(t-6) = 80t - 480
3t = 48
t = 16

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

IMO C
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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 05:44
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Let the buses meet T hrs after Bus1 leaves.

Bus1 distance = 50T (time * speed)
Bus2 distance as it leaves after 6 hrs after Bus1= 80(T - 6)
Bus3 distance as it leaves after y hrs after Bus1 = 100(T - y)
Since all meet at the same point:

50T = 80(T - 6)
Therefore, T = 16
So, all buses meet 16 hrs after Bus1 leaves.
Now, 50T = 100(16 - y) (since T = 16, and all the buses meet at the same point)
solving we get, x = 8
So, C is the correct option.
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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 05:55
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We know that,
There are 3 buses
Bus 1 leaves first at a constant speed of 50km/hr. ---(1)
Bus 2 leaves 6hrs after Bus 1 leaves at a constant speed of 80km/hr. -- (2)
Bus 3 leaves after bus 1 leaves after a set time at a constant speed of 100km/hr. ----(3)
We need to find how many hrs after bus 1 leaves bus 3 leaves
They all catch up at sometime
=> Let T be the hours after which bus 1 leaves when bus 2 and 3 catch up with it

DIstances travelled by bus 1 = 50T
Distance travelled by bus 2 = 80*(T - 6) (From 2)
=> 80*(T - 6) = 50T
=>30T = 480
=>T = 16

=> Bus 2 catches bus 1 16hrs after bus 1 leaves

We know that bus 3 catches bus 1 at same time
=> Let bus 3 leave A hrs after Bus 1
=> Bus 3 travels for 16 - A hrs
=> 100*(16 - A) = 50*16 (From 3)
=> 16 - A = 8
=> A = 8
=> Bus 3 leaves 8hrs after bus 1 leaves

C. 8
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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 05:56
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Bus123
Speed50 km/h80 km/h100 km/h
Starts @Lets suppose
@ 8am		So it would start
@ 2pm
@2pm300km distance covered

Now, for bus 2 to catch up with bus 1 moving in same direction with distance of 300 km between them it would take
t(80-50) = 300 --> t=10 hrs
And distance covered by each bus would be 800 km (10*80 = 800 km)
And they meet at 12 am midnight (2pm + 10 hrs)

Now bus 3 also catches up with them at same time
For bus 3 to cover 800 km at speed of 100km/h it would require
t*100 = 800 --> 8 hrs
So to meet at midnight the bus 3 would have to start @4pm (12am-8hrs)
Since bus 1 started at 8am, bus 3 started 8 hrs after bus 1

Option C
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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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 05:58
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Bus 1: 50 km/h
Bus 2: 80 km/h
Bus 3: 100 km/h

50 * time = 80 * (time - 6)
3 * time = 48
time = 16

Bus 1: 16 hours
Bus 2: 10 hours

Distance = 16*50 = 800 km

100 * (16 - X) = 800
16 - X = 8
X = 8 hours

The answer is C
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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 06:09
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IMO: C
B1 =50km/hr
B2 = 80km/hr but leaves after 6 hrs
B3 =100 km/hr leaves after t hrs.
now as per the question they all meet at the exact same time, so lets first find the time of meet between B1 and B2 , as that will also be the time of meet for B3.
since B2 leaves after 6 hours, hence in 6 hrs B1 will have covered 50*6 = 300 km.
using s=d/t
relative speed of B1 and B2 = 80-50 = 30 km/hr(since they are moving in the same direction)
30= 300/t ===> t = 10 hours --> i.e 16 hrs for B1
therefore in these 16 hours the distance covered by B1 = 50*16
and since its also meeting B3 here hence the same distance will be travelled by B3
therefore
100*t = 50*16
t =8 hours
now B3 left after 16 -8 = 8 hours.
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GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave 09 Jul 2026, 06:19
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Bus1 ran 50kmh x 6 hrs=300km
Relative speed with Bus2=80-50=30kmh
hence takes 300/30=10 hrs to "catch up"

which means, Bus1 ran for total 10+6=6 hrs

Now Bus 3 starts when we dont know?
But it covers the same distance as Bus1 for 16-x hrs (at double the speed)

So equate the distances=
50x6=100(16-x) will get you x=8 hrs

Answer is C. 8 Hrs.
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