OG 13, operator problem

[sohailmbaprep]

Q) If @ represents one of the operators +,- and * , is k@(l+m)=(k@l)+(k@m) for all numbers k,l and m ?
1) k@1 is not equal to 1@k for some number k.
2) @ represents substraction.

[neelgandham]

Q) If @ represents one of the operators +,- and * , is k@(l+m)=(k@l)+(k@m) for all numbers k,l and m ?

1)k@1 is not equal to 1@k for some number k.

Let @ be *, then k*1 = 1*k. So, @ can't be *.
Let @ be +, then k+1 = 1+k. So, @ can't be +.
Let @ be -, then k-1 != 1-k(not always).
So, @ is '-'. and the value of k-(l+m) is not equal to (k-l)+(k-m), for all k,l,m.
Statement I is sufficient to answer the question.

2)@ represents substraction.

@ is '-'. and the value of k-(l+m) is not equal to (k-l)+(k-m), for all k,l,m.
Statement II is sufficient to answer the question.

IMO D

[everything's eventual]

Hello Anil, if k = 1, then k-1 = 1-k so IMO we cannot def conclude that @ = '-'.

IMO answer is [spoiler]B)[/spoiler]

[kevincanspain]

Note that (1) tells us that k@1 is not equal to 1@k for some number k. In other words, there is at least one value of k for which k@1 is not equal to 1@k. Since 1-0 is not equal to 0-1, @ being subtraction is consistent with (1). Thus @ could be subtraction.

However, since 1 x k = k x 1 for all values of k, @ being x is not consistent with (1). Thus @ cannot be x (or + for that matter).

Clearly, then, according to (1), @ must be subtraction. It is the only one of the three operations that does not contradict (1).

[SmartAssJun]

Needless to say D