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# the championship on boxing

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the championship on boxing [#permalink]  17 Feb 2009, 12:34
Yesterday, studying materials of mathematical Olympiad of 2009 Moscow Lomonosov’s University, I said “this problem is format GMAT and soon the similar problems will meet at examinations.”
Now, I would like to present it for discussion.

Teams A,B,C participated in the championship on boxing. The each boxer from team A boxed exactly with 5 boxers from B and 2 from C, each boxer from B-with 6 from A and 4 from C, and each from C - with 3 from A and m from B. What is m?
A.3
B. 4
C. 5
D. 6
E. 7
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Re: the championship on boxing [#permalink]  18 Feb 2009, 16:41
matematikconsultant wrote:
Yesterday, studying materials of mathematical Olympiad of 2009 Moscow Lomonosov’s University, I said “this problem is format GMAT and soon the similar problems will meet at examinations.”
Now, I would like to present it for discussion.

Teams A, B, C participated in the championship on boxing. The each boxer from team A boxed exactly with 5 boxers from B and 2 from C, each boxer from B-with 6 from A and 4 from C, and each from C - with 3 from A and m from B. What is m?
A.3
B. 4
C. 5
D. 6
E. 7

This problem is in GMAT format but it doesnot mean that it is a GMAT problem. GMAT probllems has some important features such as:

1. It is solvable majority of the test takers. It is not unnecessarily complex or difficult to be solved by test takers..
2. It tests some basic math concepts but not an advanced concepts.
3. It should be solved under a resonable time i.e. 2 minuets.
4. It doesnot require advanced math skills.
5. So on..

Based upon these points above, it is not a GMAT problem for me because it is not solved under 2 minuets, it is too complex, and it tests some advance concepts.

Again, GMAT questions are not too difficult to be answered. If it were not too diffucult, it would have received many responses by now.
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Re: the championship on boxing [#permalink]  18 Feb 2009, 20:43
Thanks, GT!
If a,b,c are number boxers in A,B,C
5a=6b
2a=3c -------> m=5
4b=mc
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Re: the championship on boxing [#permalink]  18 Feb 2009, 21:15
How do you say that 5a = 6b? The question doesnot say that and we do not know how many members are there in A, B or C? There could be 100 boxer in A and 50 in B. Each from A boxed with 5 from B. Each from B boxed with 6 from A.

What is the basis for 5a = 6b?
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Re: the championship on boxing [#permalink]  18 Feb 2009, 21:57
GT,
if each from A have 5 duels against B, then A have 5a duels against B on the other hand B have 6b duels against A ---->
5a=6b
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Re: the championship on boxing [#permalink]  19 Feb 2009, 05:27
5a and 6b are correct but they are not equal. how do you assume that 5a=6b? B could have 5 or 6 or 60 or 100 members.
similarly B could have also any number, so does C.

lets put this way:

a = 10
b = 15
c = 20.

All 10 from a boxed with 5 from b and 2 from c.
All 15 boxed with 6 from a and 4 from c.
All 20 boxed from 3 from a and m from b.

Now how is 5a, which is 50, is equal to 6b, which is 90?

I thought the question is extreamly difficult and needs some advanced concepts but now it looks like a flawed.

Quote:
Teams A,B,C participated in the championship on boxing. The each boxer from team A boxed exactly with 5 boxers from B and 2 from C, each boxer from B-with 6 from A and 4 from C, and each from C - with 3 from A and m from B. What is m?
A.3
B. 4
C. 5
D. 6
E. 7

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Re: the championship on boxing [#permalink]  19 Feb 2009, 06:53
GT!
The equation 5a=6b must always be true because there are only 2 sportsmen
boxing - one from A and other from B, even if some sportsmen have to go to
the ring several times.
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Re: the championship on boxing [#permalink]  19 Feb 2009, 11:03
Not convincing and do not agree with "5a = 6b".

That could be if your question stipulated the same but since the question is silent about it. That is your assumption to solve the question. You cannot say "something has to be" if that is not mentioned in the question.

If the question said that all boxer boxed, then probaly it would much simpler. There could be some boxers in team A who played with boxer from team B and not with boxer from team C and so on....
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Re: the championship on boxing [#permalink]  19 Feb 2009, 12:50
GT!
If I played with you 5 times, how much times did you played with me? Only 5, not 6 or 4!
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Re: the championship on boxing [#permalink]  20 Feb 2009, 13:52
matematikconsultant wrote:
GT!
If I played with you 5 times, how much times did you played with me? Only 5, not 6 or 4!

Correct! Even though I arrived at the three equations it looked like I had 4 unknowns and 3 eq's so unsolvable but

5a=6b
2a=3c
4b=mc

But if we multiply (1) with 2 and (3) with 3

we have 10A= 12B and 12B =3mc

Essentially then 10A=3c (M) means 10A= 2A(M) M=5

But can some one comment on what made these 3 eq's and 4 unknowns solvable? Are there any anomalies to the rule of n unknowns and n equations are needed assuming all equations are different?
Re: the championship on boxing   [#permalink] 20 Feb 2009, 13:52
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# the championship on boxing

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