Try this DS problem from Nova prep

is the answer D?

I'd say (A).

(1) 1/3 is in S. Well, we know that the reciprocal will also be in S, so YES, 3 is in S.

(2) 1 is in S. That means that 1/1 is also in S, but 1/1 is just 1, which is already there. So, the set could just be {1}. However, there could be lots of other numbers in S as well, so 3 could be there. Maybe yes, maybe no: insufficient.

However, I suppose that another translation of the rules would require S to contain at least {1,1} based on statement (2), in which case 1+1=2 would be there, in which case 1+2=3 would also be there. I don't think this translation is correct, however.

Stuart Kovinsky wrote:However, I suppose that another translation of the rules would require S to contain at least {1,1} based on statement (2), in which case 1+1=2 would be there, in which case 1+2=3 would also be there. I don't think this translation is correct, however.

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I would prefer to go with the second interpretation tough. And the another translation does not require S to contain at least {1,1}. Nothing is said about x and y. Without any restriction both x and y can refer to the same element in the set.
Let S = {1}
let x be an element in S; the only option is x=1
let y be an element in S; the only option is y=1
Then we can move forward to prove that 3 is present in the set.

IMHO answer is D.

OA is D but I agree with Stuart that the option (B) is open to different interpretations.

I'd say (D)

(1) is straightforward, as if 1/3 is there, then 3 shud be in S
(2) as 1 is part of S, so 1+1=2 also belongs to S, and hence 1+2=3 belongs to S

yes, i agree with you all.

The correct reasoning for B to be true is:

from (B) 1 is in the set so from 1/x we know that 1/1 will also be in the set this makes the set contain atleast {1,1}

now from (2) we know 1+1 =2 will be in the set hence the set will contain atleast {1,1,2}

now from (2) we know 1+2 =3 will be in the set hence the set will contain atleast {1,1,2,3}