Try this DS problem from Nova prep
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Source: Beat The GMAT — Data Sufficiency |
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pavankumar_iitr
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Stuart@KaplanGMAT
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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.
(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.
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simplyjat
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.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.
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.
simplyjat
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netigen
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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}
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}
