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

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Bunuel

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08 Jul 2026, 08:00

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Bunuel

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08 Jul 2026, 09:00

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GMAT Club Official Solution:

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

Bus 2 leaves 6 hours after Bus 1, so by the time Bus 2 starts, Bus 1 has traveled 6 * 50 = 300 km. Thus, Bus 1 has a 300 km lead.

To catch Bus 1, Bus 2 needs:

time = distance/relative speed = 300/(80 - 50) = 10 hours

So Bus 2 catches Bus 1 6 + 10 = 16 hours after Bus 1 leaves.

In 16 hours, Bus 1 travels 16 * 50 = 800 km.

Since Bus 3 catches Bus 1 at the same time, Bus 3 must also travel 800 km before catching Bus 1.

At 100 km/h, Bus 3 needs 800/100 = 8 hours

Therefore, Bus 3 must have left 16 - 8 = 8 hours after Bus 1 left.

Answer: C.

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ankushsambare

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08 Jul 2026, 08:06

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Let meeting time be t hrs
Bus 1: Distance = 50t

Bus 2: 80(t-6) = 50t
30t=480
t=16

Let bus 3 leave after x hrs
100(16-x) = 50*16
1600-100x = 800
100x = 800
x = 8
Answer: C

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08 Jul 2026, 08:09

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We will start by finding when bus 2 catches up with Bus 1.

In 6 hours, bus 1 travels 50 x 6= 300 km

And since bus 2 is 30 km/h faster than bus 1, the relative speed would be 80-50= 30km/hr
Time to catchup (cover 300 km) = 300/30= 10 hours.

Now after 16 hours, bus 1 has travelled 50x16= 800km
bus 3 travels at 100 km/h so to cover 800km, it will take 8 hours.

Since they meet 16 hours after bus 1 started, bus 3 must have started 16-8 hours later

Answer=8

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i am going with option c.
as bus 1 and 2 meet 16 hours later after bus 1 starts at 800km
bus 3 needs 8 hours (800/100).
and as it arrives at 16-hour, 16-8=8.
option a incorrect as it shows time after bus 2 leaves and not bus 1.
option b incorrect it shows departure of bus 2 and not 3
option d incorrect doesn't make up calculation
option e incorrect as it shows bus 1 total time.

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08 Jul 2026, 08:12

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50t=80(t-6)
50t=80t-480
30t=480
t=16
50*16=80(16-x)
800=1600-80x
x=8

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08 Jul 2026, 08:14

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bus	d	s	t
1		50	t+6
2		80	t
3		100	x

50*(t=6)=80*t so t=10

D=50*16=100*x x=8
Time difference B3-B1= 16-8 =8hrs

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08 Jul 2026, 08:15

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At the meeting point, the distance covered by the three buses will be the same.

D1 = D2 = D3
V1 X T1 = V2 X T2 = V3 X T3
50 X T = 80 X (T-6) = 100 X (T-X)
The value of T will be 16 hr
Therefore, X will be 8.

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08 Jul 2026, 08:15

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Let 'h' be the number of hours it takes bus 1 to reach destination.
50*h = 80*(h-6) = 100*(h-t) As Bus 3 leaves after t hours
50h=80*(h-6) As bus 2 leaves after 6 hours
h=16

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

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08 Jul 2026, 08:15

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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

speed bus 1 = 50 kmph
speed of bus 2 = 80 kmph
speed of bus 3 = 100 kmph
bus 2 leaves after 6 hours bus 1 left
time when bus 2 must have catched up with bus 1
let time be x
50x = 80 * (x-6)
50x= 80x-480
30x= 480
x = 16 hours
distance covered by bus 1 in 16 hours will be 16*50 ; 800 km
so time bus 3 catches up with bus 1 = 800/100 ; 8 hours
time ∆ bus 1 & bus 3 ; 16- 8 = 8 hours

OPTION C is correct ; 8 hours

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08 Jul 2026, 08:15

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Here's my solution for this question

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Screenshot_2026-07-08-20-45-00-20_40deb401b9ffe8e1df2f1cc5ba480b12.jpg [ 1.17 MiB | Viewed 1601 times ]

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08 Jul 2026, 08:19

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Divide the problem in sub problems:

Riverton ------- > Lakeview

Bus 1: Started at t = 0, speed (v1) = 50 km/h
Bus 2: Started at t = t - 6 and speed(v2) = 80 km/h
Bus 3: started at t = ?, need to find -> What we need to find and speed (v3) = 100 km/h

Phase 1: consider buses 1 and 2

They catch up at what time ?, if they do catch up, what will be the scenario?, it will be when both have covered the same distance

distance1 = distance2
v1 * t1 = v2 * t2
50 * t = 80 (t - 6)

Solving, we get t = 16 hours, ie, all buses catch up after 16 hours

Phase 2: consider buses 1 and 3 [we can alternatively consider 2&3 also]

distance 1 = distance 3
v1 * t1 = v3 * t3
50 * 16 = 100 * t3

Solving, we get t3 = 8 hours.

After how many hours bus 1 leaves does bus 3 leaves ? -> 16 - 8 = 8 hours

IMO, option (c)

Bunuel

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08 Jul 2026, 08:19

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Head start for Bus 1= 300km
Bus 2 speed gain = 80-50 = 30km/h

Thus, the time to catch up = distance to cover / speed gain = 300/30 = 10hours after bus 2 leaves
Therefore they meet at 800kms.

Bus 3 will reach 800kms at 8 hours. Thus Bus 3, must've left 8 hours after bus 1 did. Option C.

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08 Jul 2026, 08:20

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Three buses, B1, B2, and B3
B1 leaves 6hrs before B2 so it has already travelled 50 x 6 = 300km.
Now, when B2 starts it has to cover the gap of 300 kms between B1 and itself to catch up B1.
So, Total time taken to catch up = 300/(B2-B1) = 300/30 = 10 hrs.

Now, in 10 hrs distance covered by B2 is 10 x 80= 800 kms or we can calculate with B1 as 300 + (10x50) = 800 kms.
So, B3 will require total time to cover this distance = 800/100 = 8 hrs.
Since B3 takes only 8 hrs to cover this distance, it has to start 2 hours later after B2. And 6+2= 8 Hrs later after B1.
So The answer is C.

Bunuel

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08 Jul 2026, 08:22

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Given : 3 buses, Bus 1 leaves first and is travelling at a speed of 50 km / hour.

Bus 2 leaves after 6 hours and is travelling at a speed of 80 km / hour.

Bus 3 is travelling at 100 km / hour.

it is given bus 2 and 3 catch up with bus 1 at the same time. we have speed and time for bus 2, we can use that. bus 1 covers 50*6 = 300 in 6 hours. bus 2 catches up at 300/80-50 = 10 hours. So bus 2 has taken 10 + 6 (since it left after 6 hours) to catch up with bus 1. Bus 3 also takes 16 hours by that logic.

In 16 hours bus 1 has travelled 16*50 = 800 km. Bus 3 speed = 100 km / hour. So it takes 800/100 = 8 hours to travel. Bus 2 took 16 hours so 16-8 = 8.

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08 Jul 2026, 08:23

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A speed= 50 kmph
B speed= 80 kmph
C speed= 100 kmph
Let, hours when B2 catches upto B1 =t

Distance travelled by B1= 50t
Distance travelled by B2= 80(t-6)
When they meet, 50t=80(t-6) ; t=16 hours

Let, B3 leaves x hours after B1.
B3 and B2 catches B1 at same time. B3 travels for 16-x hrs.
Distance travelled by B3= 100(16-x) =50(16)
16-x=16/2
x=8 hours.

C. 8

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08 Jul 2026, 08:24

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Let bus 1 starts at 0800 hrs and travels for x hrs
similarly bus 2 will start at 1400 hrs and travel for (x-6) hrs to meet bus 1
similarly bus 3 will start at (0800+y) hrs and travel for (x-y) hrs to meet bus 1 & 2 both.
thus a/q
distance = speed * time
50*x= 80*(x-6)=100*(x-y)
solving the qbove equation we get x = 16 hrs and y = 8hrs
thus bus 3 leaves after 8hrs after bus 1 left.

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08 Jul 2026, 08:24

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[quote="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

Let d be the distance travelled by all 3 buses and T is time taken by bus 1 to meet bus 2 and 3
This implies bus 2 will take T-6 hrs to meet 1 and 3
This implies bus 3 will take. T- x hrs to meet 1 &2 ( supposing x is time after which bus 3 leaves from riverton after bus 1 leaves.

Applying our formula of speed = distance/ time
Speed of bus 1 = 50km/hr = d/T —-equation 1
Speed of bus 2 = 80 km/hr = d/T-6 — equation 2
Speed of bus 3 = 100 km/hr = d/T-x — equation 3

Now dividing equation 1 by 2
It gives T=16
Now putting this value of T in equation 1; which gives value of d= 800

Now from equation 3
We can determine the value of x = 8 which is our answer

IMO answer option (C)

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08 Jul 2026, 08:24

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Bus 1 D = 50t
Bus 2 D = 80(t-6)

50t=80(t-6)
50t=80t-480
30t/30=480/30
t=16hrs

bus 2 catches bus 1 in 16hrs
meaning50*16=800km

bus 3 leaves??
b3 catch b1 (t=16)
x-no. hrs b3 leaves b1
b3 d=100(16-x)

b3 covers 800km same time
100(16-x)=800
16-x=8
x=8hrs
option C

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08 Jul 2026, 08:26

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Answer(C)

This will use RT=D

Bus 1:
50 kmph
Starts at 0 hrs

Bus 2:
80 Kmph
Starts at 6 hrs

Bus 3:
100 Kmph
Starts at __ hrs. We have to find this.

Bus 1,2,3 meet at the same time so distance will be travelled same by all the bus.

Bus 1 Distance = Bus 2 Distance
Bus 1 Rate x time = Bus 2 Rate x time
$50 (x+6) = 80 (x)$ ----> Here x is supposed to be the total time taken by Bus 2. Because Bus 1 started 6 hours ahead of bus 2, time taken by Bus 1 = x +6.
$x = 10$

Bus 1 => x + 6 = 16 Hrs
Bus 2 => x = 10 hrs

Total Distance by Bus (considered bus 1 for calc)= 50 Kmph x 16 hrs = 800 Kms

Time taken by Bus 3 to cover 800 Kms at 100 Kmph = 8 Hrs.

Bus 1 left 16 Hrs before, hence 16-8 hrs = 8 Hrs. Bus 3 left 8 hrs later than Bus 1.