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# If # represents one of the operations +, -, and *, is k #

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Manager
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If # represents one of the operations +, -, and *, is k # [#permalink]  24 Mar 2008, 12:03
If # represents one of the operations +, -, and *, is k # (l+m) = (k#l) + (k#m) forn all numbers k,l and m?

1) k # l is not equal to 1 # k for some numbers k.

2) # represents subtraction.

Tnx
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Re: sign [#permalink]  24 Mar 2008, 14:12
I am not sure about option 1 as I think it should read "k # l is not equal to l # k for some numbers k" rather than "k # l is not equal to 1 # k for some numbers k". Please correct me if I am wrong.

However Statement 2:
Tells us that # repeasents "-".
SO k # (l+m) = (k#l) + (k#m) => k - (l+m) = (k-l) + (k-m)
=> k -l -m = 2k - l - m, which can be true if k=0 and untrue if k is not equal to true. So this statement alone is not sufficient to answer the question.
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Re: sign [#permalink]  26 Mar 2008, 20:54
Hi, I believe the correct answer is E,
Statement 1 should read'1#k is not equal to k#1' this is true if # is - and when the value of k is negative.
This is insufficient though as if k =0 then the RHS =LHS, if k is greater than or less than zero then the wo two sides are unequal.

statement 2 is also insufficient when # is '-'
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Re: sign [#permalink]  26 Mar 2008, 22:25
My answer is D (both are sufficient), if 1 option is corrected to l # k. Here is my reasoning...
We have to access if
k # (l+m) = (k#l) + (k#m). We also have to identify if # is representing one of the operation- +,-,or *.
Lets look at option 2

2) # represents subtraction
Which means that
k # (l+m) = k#(l+m)= k-(l+m)=k-m-n which is not equal to (k#l) + (k#m), because latter expression will sum up to 2k-l-m. Hence 2 is sufficient. Now the answer choices is limited to B or D

Lets consider statement 1
1) k # l is not equal to l#k for some numbers k.
Now only in multiplication and addition that nxm=mxn (additionally n+m=m+n), which means # stands for subtraction. Hence, by above logic if # represents subtraction, it is sufficient to answer the problem. Hence this statement also works fine.

Therefore D
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Re: sign [#permalink]  27 Mar 2008, 07:23
dushver wrote:
My answer is D (both are sufficient), if 1 option is corrected to l # k. Here is my reasoning...
We have to access if
k # (l+m) = (k#l) + (k#m). We also have to identify if # is representing one of the operation- +,-,or *.
Lets look at option 2

2) # represents subtraction
Which means that
k # (l+m) = k#(l+m)= k-(l+m)=k-m-n which is not equal to (k#l) + (k#m), because latter expression will sum up to 2k-l-m. Hence 2 is sufficient. Now the answer choices is limited to B or D

Lets consider statement 1
1) k # l is not equal to l#k for some numbers k.
Now only in multiplication and addition that nxm=mxn (additionally n+m=m+n), which means # stands for subtraction. Hence, by above logic if # represents subtraction, it is sufficient to answer the problem. Hence this statement also works fine.
Therefore D

First of all If K=0, then k-m-n=2k-l-m, so statement 2 is not sufficient.

Secondly substraction is not the only sign for which k # l is not equal to l#k for some numbers k. Divison is another such operation. Moreover statement is saying "k # l is not equal to 1 # k for some numbers k" and not "k # l is not equal to l#k for some numbers k.", which even I doubted but cannot assume this in GMAT.

It is definitely not D.
Re: sign   [#permalink] 27 Mar 2008, 07:23
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