stmnt1: $ is -
==> LHS = a-b + a-c = 2a-b-c
RHS = a-b-c ==> LHS not = RHS answer to the question is NO..hence, SUFF.

stmnt2:m&2 is not equal to 2&m for some numbers m.
this can happen only when & is - as m+2=2+m and 2*m=m*2
which is again stmnt 1 hence SUFF.

answer D

Regards,
Murali.
kudos?

but if a = 0 then LHS = RHS in both (1) and (2), so it does not stand true for "all numbers"

The point here is that the question asks whether \((a@b)+(a@c)=a@(b+c)\) is true FOR ALL NUMBERS a, b, and c?

(1) \(@\) represents subtraction --> the question becomes is \(2a-b-c=a-b-c\), or is \(a=0\)? So \((a@b)+(a@c)=a@(b+c)\) is NOT true for all numbers a, b, and c (so the answer to the question is NO), for this expression to be true \(a\) must equal to zero (so not for all values of \(a\)). Sufficient.

(2) \(m@2\neq{2@}\) --> \(@\) represents subtraction (as it can not be addition or multiplication), so we have the the same info as above. Sufficient.

Answer: D.

Alternately you can see that \((a@b)+(a@c)=a@(b+c)\) to be true FOR ALL NUMBERS a, b, and c then \(@\) must represent multiplication as only for multiplication it's true for all numbers: \(ab+ac=ab+ac\). So the question basically ask whether \(@\) represents multiplication, both (1) and (2) give answer No toth is question.

Hope it's clear.
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yes - thanks Bunuel and Murali - +1 to both...

for some reason i was misinterpreting the expression "all numbers".. this makes sense - the equation is not valid for all values of a,b,c so both are in fact sufficient. thanks.

Statement 1- tells us exact nature of sign( subtraction). therefore,sufficient
Statement 2- 2@m not equal to m@2
take the values m=6,5. and solve and we get what the mean of sign. sufficient
so, Answer is D

Check Arithmetic Operation Functions Questions in Special Questions Directory.
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Is (a&b) + (a&c) = a&(b + c) for all numbers a, b, and c ?

(1) if & represents subtraction, then LHS = (a-b) + (a-c) = 2a-b-c and RHS = a - (b+c) = a-b-c. Now obviously 2a-b-c is NOT equal to a-b-c for all numbers a, b, c. So we get our definite answer as NO for the question stem. Sufficient.

(2) m&2 is NOT equal to 2&m. Now if & is '+', then m+2 = 2+m, for all values of m. So & cannot be '+'. If & is 'x', then also mx2 = 2xm, for all values of m. So & cannot be 'x'. Thus & can only represent subtraction '-'. In which case it becomes same as first statement. Sufficient.

Hence D answer

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