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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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dushver
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.



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