i needn't have one fix sign if i know that its not * im done, thats what smt1 does for me and so does stmt 2
First of all the question has a mistake by including division operation because of which I got confused.
So now, from statement 1 and statement 2, it's quite clear that @ operation is subtraction. But I am not convinced with the Official Guide-11 explanation.
We are asked whether k-(l+m) = (k-l)+(k-m) for all numbers k,l,m.
But for tuples (k,l,m) = (0, any, any) this equality exists but for other values of k, this equality does not exist. So how can we say that both statements ar e sufficient.
Please respond if I am wrong somewhere.
The Official Answer given is (D) and is correct.
since from both the statements you can say that the operation is "-"
hence you can conclude that
k-(l+m) = (k-l)+(k-m) is NOT TRUE for all numbers answer should be in "Yes" or "No", that's what the question asks.
Statement 1. "k @ 1 is not equal 1 @ k for some numbers k." Well, for all numbers k, k + 1 is the same as 1 + k, and for all numbers k, k*1 is the same as 1*k. Therefore, @ must represent subtraction, and if we know what operation @ represents, we can answer the question. Sufficient.
Statement 2. "@ represents subtraction." Again, if we know what operation @ represents, we can answer the question. Sufficient.
So each statement alone is sufficient to answer the question. The surprising thing is that they each give an answer of "no." This is the only DS in OG11 for which the answer to the question is "no."
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