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

1 2 3 4 5 6 7 8

saibharat5mba

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08 Jul 2026, 09:09

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Distance travelled by bus 1 in 6 hours = 50 kmph * 6 = 300 Km

This 300 km is the distance that bus 2 need to cover to catch up Bus 1.

Time to cover this 300 km = 300km / (80kmph - 50kmph) = 10 hours.

Distance travelled by bus 2 in 10 hours = 80 * 10 = 800 km

Same 800 km needs to be covered by bus 3 to catch bus 1 along with bus 2

Time taken by bus 3 to cover 800 km = 800km / 100kmph = 8 hours

So bus 3 can catch bus 1 along with bus 1 if it starts (10 - 8) = 2 hours after Bus 2.

Hence time after which Bus 3 leaves after bus 1 leave = 6hours + 2hours = 8 hours

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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08 Jul 2026, 09:13

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Let "t" be the number of hours between bus 1 left and Bus 2 & 3 catch it.

Distance travelled by Bus 1 = 50 t
Distance travelled by Bus 2 = 80 (t - 6)

So, 50t = 80 (t-6)

t = 16

Again, let bus 3 left x hours afte bus 1 left,
Then,

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

Answer : C ( 8 )

Bunuel

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08 Jul 2026, 09:15

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The answer is (C), i.e., 8
The LCM of 50 (B1), 80 (B2), 100 (B3) = 800
Bus 1 will reach 800 km in = 800/50 = 16 hours
Bus 3 will reach 800 km in = 800/100 = 8 hours
Therefore, Bus 3 can start after 16-8 = 8 hours to catch Bus 1

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08 Jul 2026, 09:29

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After a certain amount of time taken by each bus, they eventually reach a common point where they are together => They all cover the same distance, so distance is constant and we can equate to Distance = Speed x Time

Let buses be B1, B2, and B3

B1 -> D1 = 50 kmph x (a) hrs
B2 -> D2 = 80(a-6)
B3 -> D3 = 100(a-y)

where a is time taken by first bus and y is no. of hours bus 3 took LESS than bus 1

From the first sentence we know D1 = D2 = D3

Equate B1 and B2 equations => x = 16 hrs
=> D2 = 800km
Use D2 with B3 equation and get y = 8 hrs

So, bus 3 left 8hrs after bus 1 (C)

Bunuel

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08 Jul 2026, 09:31

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This is a typical Speed(S), Time(T), Distance(D) question.
In the given scenario the Distance would be constant as all Buses(B1, B2, B3) meet at the same point. The mechanical way to solve this would=
Write given details
B1: S = 50kmph, T = t (suppose it is 't')
B2: S = 80 kmph, T = t-6 (As it takes 6 hours lesser to reach the same point.
B3: S = 100 kmph, Time(t') = t-x

So to find the actual Distance covered, we could easily do this as:
Distance of B1 = Distance of B2
50*t = 80*(t-6)
On solving, t = 16 hours.
Now we could calculate Distance: D = 50*16 (while solving do not always multiply it because it might be possible that such numbers would have common factors)

B3: Distance = Speed * Time, Distance = 50*16
50* 16 = 100 *(t-x) (t= 16hrs, and also check this: 50 and 100 have common factor 50 now it would make our calculation easier)
On solving x = 8
So B3 starts after 8 hours from B1

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08 Jul 2026, 09:34

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Bus 2 leaves 6hours after B1 leaves. Hence in this 6hrs bus 1 has travelled 6X50= 300kms.
after this 6hrs, B2 tries to catch B1 at a relative speed of 30Kmph ( B2 speed (80) - B1 speed (50)).
Inorder to cover this 300 Kms at 30Kmph relative speed, it would take 300/30 = 10hrs of time and travel distance of 50kmph X 10hrs = 500km additional. Hence distance from the start = 300 + 500 = 800Km.
Hence bus B2 meets B1 after 16hrs from the time B1 started and at a distance of 800Km.
As per question, bus B3 also meets B1 at the same distance.

Now time taken by B3 to cover 800km = 800Km/100Kmph = 8hrs.
Hence Bus B3 should start 16-8hrs= 8hrs after B1 starts to meet B1 and B2 at 800Km of distance.

Hence answer is C 8.

Bunuel

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08 Jul 2026, 09:35

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Let d be the distance travelled, since all three buses start from common point, their distances travelled up to the catching up point will be the same alongside travelling for same time duration.

Bus 1) T = d/50
Bus ) T = d/80

d-80 = d/50 - 6
d = 800 km
Bus 1 took: 800/50 = 16 hrs

Bus 3) T = 800/100 = 8 hrs

Since both buses travelled for same distance and for same time duration, Bus 3 with it's double speed then Bus 1 travelled for half the time. Hence Bus 3 left 8 hours after Bus 1. Option C

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08 Jul 2026, 09:38

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IMO C

Lets assume the Bus 2 catches Bus 1 in t hours.
Then, 50 x t = 80 (t-6), or t = 16
Also, Bus C catches Bus 1 in same 't' time. Assuming it started x hrs after Bus 1
Then, 50 x t = 100 (t-x), putting t=16, we get x=8
Hope this helps.

Bunuel

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08 Jul 2026, 09:42

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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.
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?

_____Speed_____Time_____Distance
1 ___50 _______ t________50t
2____80 ______ t-6 _____ 80(t-6)

Catchy up time for Bus 2
50t=80(t-6)
50t=80t-480
30t=480
t=480/30=16 hours

Bus 3 also catches upto Bus 1 at same t (16 hours). Let, T is number of hours after Bus 1 that Bus 3 departs.
100(16-T)= 50(16)
1600-100T=800
100T=800
T= 8 hours

C

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08 Jul 2026, 09:51

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Bunuel

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B1 travels 300 km in hrs, at this moment, B2 starts travelling. Let's say it takes t time to catch up B1. So, in time t, B1 travels 50t, eventually B2 has to travel d = 300 + 50t. speed of B2 is 80 kmph, which means it travels d distance in time t.
80 = (300+50t)/t
t=10 hrs => total time to catch B1
so in total B1 travelled for 6+10 = 16hrs
d = 300 + 50*10 = 300 + 500 = 800km
Now B3, we don't know when it started, but let's say it takes T time to catch up to B1 and B2.
100kmph = 800 km/T
T = 8hrs
B1 total time - B3 (time took to reach B1) = 16 - 8 = 8 hrs => no. of hours later B3 started after B1.

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08 Jul 2026, 09:54

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R ------------> L
Bus : Time taken by Bus ; Speed
Bus 1 : x hrs ; 50km/h
Bus 2 : x-6 hrs ; 80km/h
Bus 3 : x-y hrs ; 100km/h (Assuming Bus 3 leaves after y hours)

Bus 2 and Bus 3 catch up to Bus 1 ---> Distance of all buses are equal

D=S*t

50*x=80*(x-6)=100*(x-y)

50*x=80*(x-6)
480=30x
x=16 hrs

Therefore,
50*16=100*(16-y)
8=16-y
y=8 hrs

Therefore, Bus 3 will leave after 8 hours

Bunuel

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08 Jul 2026, 09:58

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1. After 6 hours Bus 1 has traveled 300 km.
2. After 6 hours Bus 2 leaves the Riverton with the rate of 80 Km/h
In the 6th hour the distance between Bus 2 and Bus 1 is 300 Km
We need the difference between the rates of Bus 1 and Bus 2 to calculate the catch up time.
Rate of Bus 1: 50Km/H - Rate of Bus 2: 80Km/h = 30Km/H
So Bus 2 takes 300/30 = 10 hour to catch up with Bus 1.
The passage indicates that both Bus 2 and Bus 3 catch up with the Bus 1 at the same time, meaning that as Bus 2 catches the Bus 1 so does the Bus 3.
Let's use the answer choices as it is easier (logically)
B. 6 (Since Bus 3 is faster than Bus 2 we remove the option 6 and 2 hours)
C. 8. (After 8 hours bus 1 will be 400 km away and it takes bus 3 8 hours to catch since the difference between the rates of Bus 1 and Bus 3 is 50 km and 400/50=8. In total it takes 16 hours which is same as the Bus 2)

Answer B
IMO

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08 Jul 2026, 09:59

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Bus2 left 6 hours after Bus1. By that time, Bus1 travelled 6*50kmph = 300 km
To cover this distance, Bus2 will take 300km/speed difference between bus 1 & 2 = 300/(80-50) = 10 hours

That means Bus2 caught up to Bus1 16 hours after Bus1 started the journey.
In these 16 hours, Bus1 travelled a total of 16 hours * 50kmph = 800km

Bus3 also caught up to Bus1 16 hours after Bus1 started the journey. To cover the total distance that Bus1 travelled, Bus3 will take 800km/100kmph = 8hours
So if Bus3 took 16 hours total to catch up to Bus1 for a distance that could be covered in 8 hours. That means Bus3 started 8 hours later than Bus1.

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08 Jul 2026, 10:04

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d/50 = d/80+6 = d/100+x
d is the total distance when they catch up; x is the time after which bus 3 started that bus 1.
solving,
d = 800 and x = 8

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08 Jul 2026, 10:15

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Since the first bus travelled for 6 hours before the other bus 2 starts distance covered by the bus is 6 x 50 = 300km

So the time taken by Bus 2 to catchup with bus 1 would be GAP / differential speed which is 300/(80-30) i.e 10 hours

Now in 10 hours the bus 1 would have travelled another 500 km and the total distance travelled would be 800 km

We are told that bus 3 also caught up with bus 1 in the same time as bus 2 caught up with bus 1. Meaning it also met bus 1 at the point of 800km

Now, Since bus 3 travels at a speed of 100km/h ---> it would have taken bus 3 merely 8 hours cover 800km

But since Bus 3 also met bus 1 after a time interval of 16 hours (i.e time after bus 1 started). It means bus 3 gave a head start of 16-8 i.e 8 hours. So this implies that bus 3 would have started 8 hours AFTER bus 1 started. Hence answer is C

Bunuel

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08 Jul 2026, 10:15

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Lots going on here so take it step at a time.
Diff between bus 1 and 2 is 80-50 = 30km/hr speed.
Not bus 1 has a 6 hour lead on bus 2, so in that time it travels 50x6 = 300km
bus 2 will take 300/30 = 10 hours to make that up plus starts 6 hours later and so 16 hours in total. This means Bus 2 travels 10x80 = 800km
Bus 3 needs to make up 800km, and at 100km/hr will take 8 hours to cover. But Bus 2 meets bus 1 after 16 hours, and bus 3 needs 8 hours to cover that which means bus 3 must have departed after 8 hours.

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08 Jul 2026, 10:24

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Bus 2 leaves 6 hours after B1

In 6 hours, B1 travelled 6*50 = 300 km

B2 needs to fill this gap of 300 km. As both the objects are moving, their relative speed is 80-50 = 30 km/hr

Time taken to fill this gap = 300/30 = 10 hours

In these 10 hours, B1 travels another 500 km and then all three buses meet at 800 km distance away from starting point

Time taken by B3 to travel this 800 km = 800/100= 8 hours

B1 took 16 hours, so B3 left 8 hours after B1

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08 Jul 2026, 10:36

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Bus 2 catches Bus 1 after :
6 + (50*6) /(80-50)
= 6 + 300/30
= 16 hrs

Since Bus 2 & Bus 3 catch Bus 1 at the same time, Bus 3 also catches Bus 1 exactly 16 hrs after Bus 1 leaves

By then, Bus 1 has traveled 50*16 = 800 km

At 100 km/h, Bus 3 needs 800/100 = 8 hrs to cover that distance

So Bus 3 must have left 16-8 = 8 hrs before Bus 1

Ans : C

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08 Jul 2026, 10:42

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Assume it took t hrs, and bus3 started x hrs after bus2. So we have simple equations

50t = 80(t-6) = 100(t-6-x)

solving we get t=16

then we get x = 2

So bus3 started 8hrs after Bus1

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08 Jul 2026, 11:01

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Bunuel

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The answer is option C: 8.

Inference: Since the bus 1, 2 and 3 are "Catching up at the same point," it is safe to infer that they're going to travel the same distance, regardless of rate or time.

Forming three equations:

Let B1, B2, B3 be the three distances the three buses travel and t1, t2 and t3 be the respective times.

$\\
B1 = S (Km/h) * t (h) = 50 * t1\\
B2 = 80 * t2\\
B3 = 100 * t3\\
\\
t1 = t2 - 6 (subtraction because rate is inversely proportional to time; t2 < t1 i.e. t1 travels longer).\\
$

From the inference:

$\\
B1 = B2 = B3\\
\\
=> 50 * t1 = 80 * t2\\
=> 50 * t1 = 80 * (t1 -6)\\
=> 5t1 = 8t1 - 48\\
=> 3t1 = -48\\
=> t1 = 16 Hours\\
$

Using the same inference:

$=> 50 * t1 = 100 * t3$

Substituting t1 for 16,
$\\
=> 50 * 16 = 100 * t3\\
=> 2 t3 = 16 (Dividing both sides by 50)\\
=> t3 = 8 Hours\\
$

Hence, the answer is option C, 8.