male -female

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shashank.ism
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Topic: male -female
PostMon Feb 08, 2010 8:14 am

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Consider 4 (dimensionless) flies, 2 males and 2 females. They are situated at the four corners of a square of side 10 meters, the two males at diagonally opposite ends. At any time, each fly tries to reach the male-female fly in front of her-him using the shortest possible route. Since the flies are flying towards another, they will meet each other at a certain time at the center of the square. What is the length of the path that each has traveled at the moment they reach each other?

a) 12.5 mts
b) 10 mts
c) 7.5 mts
d) 5 mts
e) None of these

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PostMon Feb 08, 2010 8:25 am

shashank.ism wrote:
Consider 4 (dimensionless) flies, 2 males and 2 females. They are situated at the four corners of a square of side 10 meters, the two males at diagonally opposite ends. At any time, each fly tries to reach the male-female fly in front of her-him using the shortest possible route. Since the flies are flying towards another, they will meet each other at a certain time at the center of the square. What is the length of the path that each has traveled at the moment they reach each other?

a) 12.5 mts
b) 10 mts
c) 7.5 mts
d) 5 mts
e) None of these
They will fly in an spiral path.
This question is very hard and I think has also been featured in the book "Physics by H. C. Verma".
It requires strong Mathematics.
I will post the visual soln. soon.

By symmetry,they will meet at the centre of the diagonal. i.e.10sqrt(3)/2.
The length of the path will be sqrt(3)/2 x 10sqrt(3)/2 [We are having an angle of <60 sin <60 = sqrt(3)/2.
Answer is 15. E

Shanshank can you post the answer and the solution plz.

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PostMon Feb 08, 2010 1:04 pm

shashank.ism wrote:
Consider 4 (dimensionless) flies, 2 males and 2 females. They are situated at the four corners of a square of side 10 meters, the two males at diagonally opposite ends. At any time, each fly tries to reach the male-female fly in front of her-him using the shortest possible route. Since the flies are flying towards another, they will meet each other at a certain time at the center of the square. What is the length of the path that each has traveled at the moment they reach each other?

a) 12.5 mts
b) 10 mts
c) 7.5 mts
d) 5 mts
e) None of these
The path obviously will not be straight and will be some kind of a curve and by the looks for it requires higher geometry and calculus skills
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PostMon Feb 08, 2010 2:56 pm

The question seems a bit odd, really...

but i am wrong or isn't the distance 1/2 of the diagonal, i.e. 1/2 of the hypethenuse of the triangle with sides 10:10:10*root of 2, i.e, 5*root of 2?
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PostMon Feb 08, 2010 3:10 pm

bln123 wrote:
The question seems a bit odd, really...

but i am wrong or isn't the distance 1/2 of the diagonal, i.e. 1/2 of the hypethenuse of the triangle with sides 10:10:10*root of 2, i.e, 5*root of 2?
In the question,it is written that
Quote:
What is the length of the path that each has traveled at the moment they reach each other?
I think you have confused distance with displacement.
The displacement as you have pointed out will be 5*root of 2.
The distance will be somewhat different and can be found by using derative equation or by symmetry.

Hope,you understand the 2 different concepts mentioned here.

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PostMon Feb 08, 2010 3:11 pm

As the question is worded, it doesn't make any sense; we need to know the direction in which the flies are facing to know which fly is "in front of her-him". If the flies on the leftmost corners are facing each other, for example, they will fly directly towards each other and will never reach the center. If the flies all begin by facing towards the center of the square, the setup doesn't make sense. I'm sure the intention is that each fly is facing in a different direction (one north, one east, one south and one west), though you then get a problem far out of the scope of the GMAT.
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PostMon Feb 08, 2010 3:36 pm

Ian Stewart wrote:
As the question is worded, it doesn't make any sense; we need to know the direction in which the flies are facing to know which fly is "in front of her-him". If the flies on the leftmost corners are facing each other, for example, they will fly directly towards each other and will never reach the center. If the flies all begin by facing towards the center of the square, the setup doesn't make sense. I'm sure the intention is that each fly is facing in a different direction (one north, one east, one south and one west), though you then get a problem far out of the scope of the GMAT.
Hey Ian ,
I do agree with you that the question is out of the scope of GMAT.Actually,the language is quite difficult.
What the question means is this:-
Suppose after 1 sec,each fly has moved 1m[suppose ABCD is the square.I am denoting the fly names as fly M(initially on vertex A) , fly N(initially on vertex B) , fly O(initially on vertex C) and fly P(initially on vertex D)]
Now, fly M is 1m right of vertex A.
fly N is 1m up of vertex B.
fly O is 1m right of vertex C.
fly P is 1m right of vertex D.

So,Now as fly M has to fly towards fly B,it will take an upward path (along the diagonal whose length is sqrt(MB^2 + BN^2).I am denoting the lengths MB and BN w.r.t. the vertex B and the position of the fly M and fly N.]



Like this their path will carry in a spiral direction.[since each one is following the next fly so their angle will change every second]


Well,this is a very high-level advanced problems which you can solve only if you have studied the derivatives in depth.
These type of problem can be found in a russian author's book:-"I.E.Irodov".Have you heard his name??

I had come across such problems while I was doing advanced calculus.(using dy/dx )
I don't think it can come on the GMAT.But if you want to test urself,it's a good problem.(Just for you,not for gmat practice)

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PostThu Feb 11, 2010 10:56 am

Because all flies constantly fly perpendicular to another fly, they all travel the shortest distance to each other, which is 10 meter. All flies make a kind of spiral flight to the center of the square, and during this flight, the flies constantly form a square until they meet in the center. The flies all travel 10 meter
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