Speed - Problem

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Speed - Problem

by Uva@90 » Sat Dec 07, 2013 8:33 pm
A man cycling along the road noticed that every 12 minutes a bus overtakes him and every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at a constant speed, what is the time interval between consecutive buses?
A. 5 minutes
B. 6 minutes
C. 8 minutes
D. 9 minutes
E. 10 minutes

I could not get what is problem states.
I understood that every 12 minutes bus overtakes cyclist. and every 4 minutes once he meets an coming bus.
we need to find the interval between consecutive buses ?

what does consecutive buses mean here ?, whether overtakes or oncoming bus ?
if it is consecutive question itself said it come once in 12 minutes.
if it is oncoming bus, even that in question it is said that it comes once in 4 minutes.

Could please explain in easier way ?

Thanks in advance.

Regards,
Uva.
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by ganeshrkamath » Sun Dec 08, 2013 2:39 am
Uva@90 wrote:A man cycling along the road noticed that every 12 minutes a bus overtakes him and every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at a constant speed, what is the time interval between consecutive buses?
A. 5 minutes
B. 6 minutes
C. 8 minutes
D. 9 minutes
E. 10 minutes
Say bus 1 with speed A overtakes the cyclist with speed C at t0 at point X.
_ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ Y _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ C _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
B2_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ B1_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

And bus 2 with speed A overtakes the cyclist with speed C at t0 + 12 at point Y.
_ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ Y _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ C _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ B2_ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ B1

In 12 minutes, bus 1 has traveled 12A from point X.
At this time, bus 2 and the cyclist are at point y.
So, the distance between bus 1 and bus 2 = D = 12A - (XY) = 12A - 12C

Now, say bus 3 with speed A meets the cyclist with speed C at t1 at point P.
_ _ _ _ _ _ _ P _ _ _ _ _ _ _ _ _ _ Q _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ C _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ B4_
_ _ _ _ _ _ _ B3_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

And bus 4 with speed A meets the cyclist with speed C at t1 + 4 at point Q.
_ _ _ _ _ _ _ P _ _ _ _ _ _ _ _ _ _ Q _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ C _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ B4_ _ _ _ _ _ _ _ _
B3 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

In 4 minutes, bus 3 has traveled 4A from point P.
At this time, bus 4 and the cyclist are at point Q.
So, the distance between bus 3 and bus 4 = D = 4A + (PQ) = 4A + 4C

Equate the 2 distances:
12A - 12C = 4A + 4C
8A = 16C
C = 0.5A

D = 4A + 2A = 6A

Time = distance/speed = 6A/A = 6 minutes

Choose B

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by GMATGuruNY » Sun Dec 08, 2013 4:42 am
Uva@90 wrote:A man cycling along the road noticed that every 12 minutes a bus overtakes him and every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at a constant speed, what is the time interval between consecutive buses?
A. 5 minutes
B. 6 minutes
C. 8 minutes
D. 9 minutes
E. 10 minutes

I could not get what is problem states.
I understood that every 12 minutes bus overtakes cyclist. and every 4 minutes once he meets an coming bus.
we need to find the interval between consecutive buses ?

what does consecutive buses mean here ?, whether overtakes or oncoming bus ?
if it is consecutive question itself said it come once in 12 minutes.
if it is oncoming bus, even that in question it is said that it comes once in 4 minutes.

Could please explain in easier way ?

Thanks in advance.

Regards,
Uva.
The time interval between consecutive buses is equal to how often the buses DEPART from the station: every 5 minutes, every 6 minutes, etc.
All of the buses -- in each direction -- travel at the same uniform speed.
The result is that the distance between consecutive buses is always the same.

Let the distance between consecutive buses = 24 units.
Let b = the rate of each bus and c = the rate of the cyclist.

SAME DIRECTION:
Here, the buses and the cyclist are COMPETING, so we SUBTRACT their rates.
The time needed for the next bus to CATCH UP to the cyclist is 12 minutes.
Thus:
b-c = d/t = 24/12 = 2 units per minute.

OPPOSITE DIRECTIONS:
Here, the buses and the cyclist are WORKING TOGETHER to cover the distance between them, so we ADD their rates.
The time needed for the cyclist and the next oncoming bus to PASS EACH OTHER is 4 minutes.
b+c = d/t = 24/4 = 6 units per minute.

Adding the two equations, we get:
(b-c) + (b+c) = 2+6
2b = 8
b=4 units per minute.

Since the rate of each bus is 4 units per minute and the distance between consecutive buses is 24 units:
The time interval between consecutive buses = d/r = 24/4 = 6 minutes.

The correct answer is B.
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by Uva@90 » Sun Dec 08, 2013 4:43 am
ganeshrkamath wrote:
Uva@90 wrote:A man cycling along the road noticed that every 12 minutes a bus overtakes him and every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at a constant speed, what is the time interval between consecutive buses?
A. 5 minutes
B. 6 minutes
C. 8 minutes
D. 9 minutes
E. 10 minutes
Say bus 1 with speed A overtakes the cyclist with speed C at t0 at point X.
_ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ Y _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ C _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
B2_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ B1_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

And bus 2 with speed A overtakes the cyclist with speed C at t0 + 12 at point Y.
_ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ Y _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ C _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ B2_ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ B1

In 12 minutes, bus 1 has traveled 12A from point X.
At this time, bus 2 and the cyclist are at point y.
So, the distance between bus 1 and bus 2 = D = 12A - (XY) = 12A - 12C

Now, say bus 3 with speed A meets the cyclist with speed C at t1 at point P.
_ _ _ _ _ _ _ P _ _ _ _ _ _ _ _ _ _ Q _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ C _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ B4_
_ _ _ _ _ _ _ B3_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

And bus 4 with speed A meets the cyclist with speed C at t1 + 4 at point Q.
_ _ _ _ _ _ _ P _ _ _ _ _ _ _ _ _ _ Q _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ C _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ B4_ _ _ _ _ _ _ _ _
B3 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

In 4 minutes, bus 3 has traveled 4A from point P.
At this time, bus 4 and the cyclist are at point Q.
So, the distance between bus 3 and bus 4 = D = 4A + (PQ) = 4A + 4C

Equate the 2 distances:
12A - 12C = 4A + 4C
8A = 16C
C = 0.5A

D = 4A + 2A = 6A

Time = distance/speed = 6A/A = 6 minutes

Choose B

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Hi Ganesh,
Your solution is soo clear.
I think this question needs lot of data's to be co-related. You did simply a nice way.
I thank you once again.

Regards,
Uva.
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by Uva@90 » Sun Dec 08, 2013 4:56 am
The time interval between consecutive buses is equal to how often the buses DEPART from the station: every 5 minutes, every 6 minutes, etc.
This is what I failed to understand.

Thanks Mitch.

Regards,
Uva.
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by Mathsbuddy » Mon Dec 09, 2013 12:12 am
Consider relative velocities:
C = cyclist velocity
B = bus velocity
D/T = relative velocity

Towards: C + B = D/4 -> C = D/4 - B
Away: B - C = D/12 -> C = B - D/12

So: D/4 - B = B - D/12
2B = D/4 + D/12 = 2D/6
B = D/6

Hence in the same distance, the time taken is 6.

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by Milovan » Fri Dec 20, 2013 5:16 am
Hi Mitch,

Your explanation is very clear except one thing: how is it possible to assign identical Distance in the amount of 24 to both scenarios: catching up and head to head? Aren't these distances different?
GMATGuruNY wrote:
Uva@90 wrote:A man cycling along the road noticed that every 12 minutes a bus overtakes him and every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at a constant speed, what is the time interval between consecutive buses?
A. 5 minutes
B. 6 minutes
C. 8 minutes
D. 9 minutes
E. 10 minutes

I could not get what is problem states.
I understood that every 12 minutes bus overtakes cyclist. and every 4 minutes once he meets an coming bus.
we need to find the interval between consecutive buses ?

what does consecutive buses mean here ?, whether overtakes or oncoming bus ?
if it is consecutive question itself said it come once in 12 minutes.
if it is oncoming bus, even that in question it is said that it comes once in 4 minutes.

Could please explain in easier way ?

Thanks in advance.

Regards,
Uva.
The time interval between consecutive buses is equal to how often the buses DEPART from the station: every 5 minutes, every 6 minutes, etc.
All of the buses -- in each direction -- travel at the same uniform speed.
The result is that the distance between consecutive buses is always the same.

Let the distance between consecutive buses = 24 units.
Let b = the rate of each bus and c = the rate of the cyclist.

SAME DIRECTION:
Here, the buses and the cyclist are COMPETING, so we SUBTRACT their rates.
The time needed for the next bus to CATCH UP to the cyclist is 12 minutes.
Thus:
b-c = d/t = 24/12 = 2 units per minute.

OPPOSITE DIRECTIONS:
Here, the buses and the cyclist are WORKING TOGETHER to cover the distance between them, so we ADD their rates.
The time needed for the cyclist and the next oncoming bus to PASS EACH OTHER is 4 minutes.
b+c = d/t = 24/4 = 6 units per minute.

Adding the two equations, we get:
(b-c) + (b+c) = 2+6
2b = 8
b=4 units per minute.

Since the rate of each bus is 4 units per minute and the distance between consecutive buses is 24 units:
The time interval between consecutive buses = d/r = 24/4 = 6 minutes.

The correct answer is B.
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by Mathsbuddy » Fri Dec 20, 2013 6:33 am
Relative to the cyclist, the apparent displacement is not.
However, the absolute distance between the buses is uniform, nomatter what direction the buses are going. It could be any distance (24 is just a convenient value), so I called it constant D.

Milovan wrote:Hi Mitch,

Your explanation is very clear except one thing: how is it possible to assign identical Distance in the amount of 24 to both scenarios: catching up and head to head? Aren't these distances different?
GMATGuruNY wrote:
Uva@90 wrote:A man cycling along the road noticed that every 12 minutes a bus overtakes him and every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at a constant speed, what is the time interval between consecutive buses?
A. 5 minutes
B. 6 minutes
C. 8 minutes
D. 9 minutes
E. 10 minutes

I could not get what is problem states.
I understood that every 12 minutes bus overtakes cyclist. and every 4 minutes once he meets an coming bus.
we need to find the interval between consecutive buses ?

what does consecutive buses mean here ?, whether overtakes or oncoming bus ?
if it is consecutive question itself said it come once in 12 minutes.
if it is oncoming bus, even that in question it is said that it comes once in 4 minutes.

Could please explain in easier way ?

Thanks in advance.

Regards,
Uva.
The time interval between consecutive buses is equal to how often the buses DEPART from the station: every 5 minutes, every 6 minutes, etc.
All of the buses -- in each direction -- travel at the same uniform speed.
The result is that the distance between consecutive buses is always the same.

Let the distance between consecutive buses = 24 units.
Let b = the rate of each bus and c = the rate of the cyclist.

SAME DIRECTION:
Here, the buses and the cyclist are COMPETING, so we SUBTRACT their rates.
The time needed for the next bus to CATCH UP to the cyclist is 12 minutes.
Thus:
b-c = d/t = 24/12 = 2 units per minute.

OPPOSITE DIRECTIONS:
Here, the buses and the cyclist are WORKING TOGETHER to cover the distance between them, so we ADD their rates.
The time needed for the cyclist and the next oncoming bus to PASS EACH OTHER is 4 minutes.
b+c = d/t = 24/4 = 6 units per minute.

Adding the two equations, we get:
(b-c) + (b+c) = 2+6
2b = 8
b=4 units per minute.

Since the rate of each bus is 4 units per minute and the distance between consecutive buses is 24 units:
The time interval between consecutive buses = d/r = 24/4 = 6 minutes.

The correct answer is B.