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09 Jul 2026, 06:29
Kudos
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Let
Bus 1 start at time 0, at 50kmph
Bus 2 leaves 6 hrs later at 80kmph
Bus 3 leaves x hours later at 100kmph
At time t1 when Bus 1 and Bus 2 meet,
Bus 1 travelled for t1 hours
Bus 2 travelled for t1-6 hours
Since they meet, equating distances, 50t1 = 80*(t1-6)
t1 = 16
Since Bus 1 travels for 16 hours at 50kmph, distance covered = 50*16 = 800 km
At 100kmph, Bus 3 needs 800/100 = 8 hrs to cover this.
Since they meet after 16 hrs from Bus 1 start, Bus 3 must have left 16-8 = 8 hrs after Bus 1
Hence answer is 8
Bus 1 start at time 0, at 50kmph
Bus 2 leaves 6 hrs later at 80kmph
Bus 3 leaves x hours later at 100kmph
At time t1 when Bus 1 and Bus 2 meet,
Bus 1 travelled for t1 hours
Bus 2 travelled for t1-6 hours
Since they meet, equating distances, 50t1 = 80*(t1-6)
t1 = 16
Since Bus 1 travels for 16 hours at 50kmph, distance covered = 50*16 = 800 km
At 100kmph, Bus 3 needs 800/100 = 8 hrs to cover this.
Since they meet after 16 hrs from Bus 1 start, Bus 3 must have left 16-8 = 8 hrs after Bus 1
Hence answer is 8
Bunuel
09 Jul 2026, 07:23
Kudos
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let the time at which all three buses meet be 't' hours. Also, the distance they all travel will be the same at the point they all meet.
for bus 1 -> distance travelled will be \(D_1 = 50t\)
for bus 2 -> distance travelled will be \(D_2 = 80(t-6)\)
for bus 3 -> distance travelled will be \(D_3 = 100(t-k)\) assuming bus 3 leaves after 'k' hours
\(D_1 = D_2\)
\(50t = 80(t-6)\)
\(t = 16\)
Now, \(D_1 = D_3\)
\(50(16) = 100(16-k)\)
\(k = 8\)
Answer is 8 hours
for bus 1 -> distance travelled will be \(D_1 = 50t\)
for bus 2 -> distance travelled will be \(D_2 = 80(t-6)\)
for bus 3 -> distance travelled will be \(D_3 = 100(t-k)\) assuming bus 3 leaves after 'k' hours
\(D_1 = D_2\)
\(50t = 80(t-6)\)
\(t = 16\)
Now, \(D_1 = D_3\)
\(50(16) = 100(16-k)\)
\(k = 8\)
Answer is 8 hours
Bunuel
sunshineeee
09 Jul 2026, 07:25
Kudos
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The question asked time (T) when B2, B3, and B1 at the same distance.
Formula used: D=RT or D = (R2-R1) T
D=Distance
R=Rate/Speed
T=Time
So
1. D=RT
50km/h x 6h = 300 km
Bus 1 position when Bus 2 leaves
2. Now, find the distance when B2 catchup Bus 1 along with Bus 3
With R1=50, R2=80, and R3=100, the LCM is 400 then 800.
If they meet at 400km -> not possible because Bus 2 cannot catchup with Bus 1 which already in 300km.
so, it must be 800km
Then for Bus 3
T=D/R = 800km / 100km/h = 8 hour
And Bus 1
800 km : 50 km/h = 16 hour
The difference would be 8 hour (C)
Formula used: D=RT or D = (R2-R1) T
D=Distance
R=Rate/Speed
T=Time
So
1. D=RT
50km/h x 6h = 300 km
Bus 1 position when Bus 2 leaves
2. Now, find the distance when B2 catchup Bus 1 along with Bus 3
With R1=50, R2=80, and R3=100, the LCM is 400 then 800.
If they meet at 400km -> not possible because Bus 2 cannot catchup with Bus 1 which already in 300km.
so, it must be 800km
Then for Bus 3
T=D/R = 800km / 100km/h = 8 hour
And Bus 1
800 km : 50 km/h = 16 hour
The difference would be 8 hour (C)
Bunuel
