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cyclist

by ritula » Tue Sep 23, 2008 2:57 am
A man cycling along the road noticed that every 12 minutes a bus overtakes him while every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at constant speed, what is the time interval between consecutive buses?

5 minutes
6 minutes
8 minutes
9 minutes
10 minutes
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by Morgoth » Tue Sep 23, 2008 4:24 am
I think this question is not only badly worded but also unclear on what exactly needs to be calculated. I could be wrong, does anybody else thinks the same.

Source pls?
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by 4meonly » Tue Sep 23, 2008 6:50 am
between what consecutive buses?

oncoming? passing?

Agree with Morgoth
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Cant help it

by ritula » Tue Sep 23, 2008 7:07 am
cant help it. I havent framed the question myself. couldnt understand. dats y posted here.
Morgoth wrote:I think this question is not only badly worded but also unclear on what exactly needs to be calculated. I could be wrong, does anybody else thinks the same.

Source pls?
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by niraj_a » Tue Sep 23, 2008 8:14 am
source?
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Re: cyclist

by sudhir3127 » Tue Sep 23, 2008 8:22 am
ritula wrote:A man cycling along the road noticed that every 12 minutes a bus overtakes him while every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at constant speed, what is the time interval between consecutive buses?

5 minutes
6 minutes
8 minutes
9 minutes
10 minutes
i think its B 6.

relative speed formula

2XY/(X+Y)

2*12*4/16
= 6

Hope that helps..
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by stop@800 » Tue Sep 23, 2008 1:11 pm
A very good question:

Let x be speed of cycle and [in km / minutes]
Let y be speed of bus.

In straight direction:
man is moving at x speed and bus is following him at y speed.
Relative speed = (y-x) [y has to be greater than x for bus to catch the cycle]
and after 12 minutes they catch each other so distance traveled is 12*(y-x)


In opposite direction:
man is moving at x speed and bus is coming towards him at y speed.
Relative speed = (y+x)
and after 4 minutes they catch each other so distance traveled is 4*(x+y)

Since time interval between buses is constant and the speed of bus is also constant so
these two distances must be equal.
4*(x+y) = 12*(y-x)

Solve it you will get y = 2*x

Now the time interval between two buses will be = distance traveled / speed of bus
time = (4x+4y)/y or = (12*(y-x)/y)

substitute y = 2x and solve it you will get your answer.
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Re:

by ritula » Tue Sep 23, 2008 8:40 pm
U r a genius
stop@800 wrote:A very good question:

Let x be speed of cycle and [in km / minutes]
Let y be speed of bus.

In straight direction:
man is moving at x speed and bus is following him at y speed.
Relative speed = (y-x) [y has to be greater than x for bus to catch the cycle]
and after 12 minutes they catch each other so distance traveled is 12*(y-x)


In opposite direction:
man is moving at x speed and bus is coming towards him at y speed.
Relative speed = (y+x)
and after 4 minutes they catch each other so distance traveled is 4*(x+y)

Since time interval between buses is constant and the speed of bus is also constant so
these two distances must be equal.
4*(x+y) = 12*(y-x)

Solve it you will get y = 2*x

Now the time interval between two buses will be = distance traveled / speed of bus
time = (4x+4y)/y or = (12*(y-x)/y)

substitute y = 2x and solve it you will get your answer.
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Re: cyclist

by tparekh » Wed Sep 24, 2008 12:01 am
sudhir3127 wrote:
ritula wrote:A man cycling along the road noticed that every 12 minutes a bus overtakes him while every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at constant speed, what is the time interval between consecutive buses?

5 minutes
6 minutes
8 minutes
9 minutes
10 minutes
i think its B 6.

relative speed formula

2XY/(X+Y)

2*12*4/16
= 6

Hope that helps..
Can you explain the relative speed formula?
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Re: cyclist

by stop@800 » Wed Sep 24, 2008 6:27 am
This is actually average speed formula: [when distance is same]

Let say A person moves from point A to point B with x speed and comes back with y speed than average speed will be given by (2xy)/(x+y)

Ex:
A cyclist covered 50 m forward direction in 5 m/s and came back with speed 3 m/s what is his avg speed?

avg = 2*5*3/(5+3) = 30/8 = 3.75

Yes!! It will not be 4. Be careful

The above question if you will solve with formula
avg speed = (total dis)/(total time)
You will get the same answer.

PS: The distance 50m is not needed, I have mentioned just for your better understanding.
tparekh wrote:
sudhir3127 wrote:
ritula wrote:A man cycling along the road noticed that every 12 minutes a bus overtakes him while every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at constant speed, what is the time interval between consecutive buses?

5 minutes
6 minutes
8 minutes
9 minutes
10 minutes
i think its B 6.

relative speed formula

2XY/(X+Y)

2*12*4/16
= 6

Hope that helps..
Can you explain the relative speed formula?
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by niraj_a » Wed Sep 24, 2008 7:49 am
sudhir

like stop@800 said, that definitely is the avg. speed formula, wouldn't apply in this case.
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