My answer is D...what is the official answer

Hey psm12se,

I am also getting the same answer as sana.noor. ie (D). What's the official answer ?

But here is what I did,

So we have to show that k & (l+m) = (k&l) + (k&m) for all numbers k, l, m

and & can be +,- or *

If you check all these three operations in the question stem, you will find that only subtraction ( -) doesn't work. And the addition and multiplication works. ie.
k+ ( l+m) = (k+l) + (k+m)
k* ( l+m) = (k*l) + ( k*m )

but this operation doesn't work with the subtraction. Moreover, if you know.. + and * operations are linear in nature, so it will always follow the distributive rules.

Ok, now going to the options.

ST.1 :-

1) k&1 is not equal to 1 &k for some numbers k.

As we know only operation which doesn't follow this operation is subtraction( - ) , so it is sufficient.

2) & represents subtraction

It just tells us that it is subtraction operation. So, sufficient.

Since both st1 and st2 are sufficient.

We choose D .

Hope this helps

psm12se wrote:Apologies: Posted this question in wrong forum. This question should have been in DS

If & represents one of the operations +, -, 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 numbers k.
2) & represents subtraction.

Good work guys!

Please notice the importance of the text "for all numbers k,l, and m".

If I change the wording to "for some number k,l, and m", the answer will be (e) because for k=0, the equality holds.

Exactly...nice point Iceman

The Iceman wrote:Good work guys!

Please notice the importance of the text "for all numbers k,l, and m".

If I change the wording to "for some number k,l, and m", the answer will be (e) because for k=0, the equality holds.

thanks for replies