TSD

Are you sure that you posted enough information?

4meonly wrote:Are you sure that you posted enough information?

Yes i am ... thats all the Question says !!!

Couldn't it be 0.2D?

pepeprepa wrote:Couldn't it be 0.2D?

its in fact 0.2D

Reasoning please!
I've got 0,3D

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--------------------

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

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 !!

Thanks for posting such brilliant questions Sudhir!

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