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08 Jul 2026, 14:52
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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
We know that after 6 hours, but one will be 300 km away (50 km/h x 6 hours), and bus 2 is just leaving. Bus 2 travels at 80 km/h, so it will gain 30 km/h on bus one since 80-50 km/h =30. This means it will take take 10 hours for bus 2 to catch up to bus 1, having traveled a total of 800 miles. 50 km/h x16 hrs= 80 km/h x10 hrs.
To travel the same 800 miles, it will take a total 8 hrs since 800 km/100 km per h is 8. Since bus 1 takes a total of 16 hrs, this means bus 3 leaves 8 hrs afterwards. Option c.
A. 2
B. 6
C. 8
D. 10
E. 16
We know that after 6 hours, but one will be 300 km away (50 km/h x 6 hours), and bus 2 is just leaving. Bus 2 travels at 80 km/h, so it will gain 30 km/h on bus one since 80-50 km/h =30. This means it will take take 10 hours for bus 2 to catch up to bus 1, having traveled a total of 800 miles. 50 km/h x16 hrs= 80 km/h x10 hrs.
To travel the same 800 miles, it will take a total 8 hrs since 800 km/100 km per h is 8. Since bus 1 takes a total of 16 hrs, this means bus 3 leaves 8 hrs afterwards. Option c.
⚠️ 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.
08 Jul 2026, 14:57
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B1 travels 300 km before B2 start
B2 is 30km/h faster so it takes 10 hours to cover that B1 head start. Then total time is 10+6 = 16 hours
while total B1 distance = 50 x 16 = 800km
which B3 can cover in 8 hours, so it can give B1 a 8 hour head start
B2 is 30km/h faster so it takes 10 hours to cover that B1 head start. Then total time is 10+6 = 16 hours
while total B1 distance = 50 x 16 = 800km
which B3 can cover in 8 hours, so it can give B1 a 8 hour head start
08 Jul 2026, 15:28
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For convenience purpose, I will call Buses as B1, B2, B3 in the order they are given.
B1 leaves and travels for 6 hours before B2 starts. Hence, by the time B2 leaves, B1 has already travelled 6x50=300.
Now, B2 has to cover 300km in order to catch up with B1.
Using relative speed formula we can find out how much time B2 took before it could catch up to B1.
————-> 300/80-50 = 10 hours.
At this point B3 has also caught up with B1. Now, B1 must have travelled 300km + (10x50)km=800 km
To travel 800km, time taken by B3 is going to be 800/100=8 hours.
And B1 has travelled 16 hours
That means 16-8=8 . B3 left 8 hours after B1 left.
B1 leaves and travels for 6 hours before B2 starts. Hence, by the time B2 leaves, B1 has already travelled 6x50=300.
Now, B2 has to cover 300km in order to catch up with B1.
Using relative speed formula we can find out how much time B2 took before it could catch up to B1.
————-> 300/80-50 = 10 hours.
At this point B3 has also caught up with B1. Now, B1 must have travelled 300km + (10x50)km=800 km
To travel 800km, time taken by B3 is going to be 800/100=8 hours.
And B1 has travelled 16 hours
That means 16-8=8 . B3 left 8 hours after B1 left.
