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by sudhir3127 » Mon Aug 25, 2008 6:58 am
Two bodies A and B start from opposite ends P and Q of a straight road. They meet at a point 0.6D from P . Find the point of their fourth meeting.

Answer after some discussion..
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by 4meonly » Mon Aug 25, 2008 7:03 am
Are you sure that you posted enough information?
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by sudhir3127 » Mon Aug 25, 2008 7:05 am
4meonly wrote:Are you sure that you posted enough information?
Yes i am ... thats all the Question says !!!
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by pepeprepa » Mon Aug 25, 2008 7:13 am
Couldn't it be 0.2D?
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by sudhir3127 » Mon Aug 25, 2008 7:17 am
pepeprepa wrote:Couldn't it be 0.2D?
its in fact 0.2D
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by 4meonly » Mon Aug 25, 2008 7:22 am
Reasoning please! :D
I've got 0,3D
Last edited by 4meonly on Mon Aug 25, 2008 7:24 am, edited 1 time in total.
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by pepeprepa » Mon Aug 25, 2008 7:23 am
I can't say I used an academic one method, I draw lines with distance 100.
A60 means A is at 60miles, and 60 alone means they meet at 60

First meet:
---------------60----------
We can deduce A speed is 60m/h and B speed is 40m/h

One hour later:
----B20------------A80----

Second meet:
------20--------------------

One hour later:
------A40---------B60-----

Third meet:
--------------------------100

One hour later:
-----A40-------------B60----

Fourth meet:
-----20--------------------
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by 4meonly » Mon Aug 25, 2008 7:26 am
pepeprepa wrote:I can't say I used an academic one method, I draw lines with distance 100.
A60 means A is at 60miles, and 60 alone means they meet at 60

First meet:
---------------60----------
We can deduce A speed is 60m/h and B speed is 40m/h

One hour later:
----B20------------A80----

Second meet:
------20--------------------

One hour later:
------A40---------B60-----

Third meet:
--------------------------100

One hour later:
-----A40-------------B60----

Fourth meet:
-----20--------------------
Ye, i have the same logic. But I made a mistake in Third meet
Can we do it through LCM?
I think there should be a solution throught LCM
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by sudhir3127 » Mon Aug 25, 2008 7:36 am
4meonly wrote:
pepeprepa wrote:I can't say I used an academic one method, I draw lines with distance 100.
A60 means A is at 60miles, and 60 alone means they meet at 60

First meet:
---------------60----------
We can deduce A speed is 60m/h and B speed is 40m/h

One hour later:
----B20------------A80----

Second meet:
------20--------------------

One hour later:
------A40---------B60-----

Third meet:
--------------------------100

One hour later:
-----A40-------------B60----

Fourth meet:
-----20--------------------
Ye, i have the same logic. But I made a mistake in Third meet
Can we do it through LCM?
I think there should be a solution throught LCM
I am not sure how LCMs would work here but there are formula in TSD which might help you..

D + (n-1)2D .......( when 2 bodies are traveling in opposite direction the distance covered by them ...n is the number of trips/ meeting)

Since time is constant...we know the ratio of the speeds

0.6 :0.4
3:2

total distance is D + (4-1)*2D = 7D

we know divide 7D in the ratio of 3:2

therefore A is 4.2 and B is 2.8

thus A having moved a distance of 4.2 will be @ 0.2D from P..

hope this helps..

Thanks pepeprepa.... as usual brilliant !!
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by pepeprepa » Mon Aug 25, 2008 7:46 am
Thanks for posting such brilliant questions Sudhir!
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by rishi235 » Mon Aug 25, 2008 10:11 am
I also got the right answer using pepeprepa's method but took ages to get to the answer
U just gave us a great method sudhir....Thanks
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