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sudhir3127
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TSD
This topic has expert replies
Source: Beat The GMAT — Problem Solving |
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sudhir3127
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Yes i am ... thats all the Question says !!!4meonly wrote:Are you sure that you posted enough information?
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sudhir3127
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pepeprepa
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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--------------------
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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4meonly
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Ye, i have the same logic. But I made a mistake in Third meetpepeprepa 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--------------------
Can we do it through LCM?
I think there should be a solution throught LCM
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sudhir3127
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I am not sure how LCMs would work here but there are formula in TSD which might help you..4meonly wrote:Ye, i have the same logic. But I made a mistake in Third meetpepeprepa 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--------------------
Can we do it through LCM?
I think there should be a solution throught LCM
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 !!
