Discussion
Is 3 in S?
(1) 
is in S.
(2) 1 is in S.
| (A) | if statement (1) ALONE is sufficient to answer the question but statement (2) alone is not sufficient; |
| (B) | ... |
| (C) | ... |
| (D) | ... |
| (E) | ... |
| (F) | ... |
The solution is
Posted: 02/22/2012 22:57
I don't understand how 2 is sufficient to answer this question. All property II is saying is that the sum of two numbers in the set is included in that set. If you're only given one number, how can you determine whether or not 3 is in the set?
Posted: 02/23/2012 20:42
Justin, thanks for using the app and the post.
According to the author's explanation, in this case, you can apply the property more than once. Since 1 is in S: 1+1 = 2 (in this case x = y, which is not disallowed), 2 is also in S.
Now 1+2=3 (in this case x and y are different), 3 is also in S.
I agree with you though, that this can be confusing. The author should have explicitly said either x cannot be the same as y, or x can be the same as y.
I would have answered A. I will give this feedback to the publisher.
According to the author's explanation, in this case, you can apply the property more than once. Since 1 is in S: 1+1 = 2 (in this case x = y, which is not disallowed), 2 is also in S.
Now 1+2=3 (in this case x and y are different), 3 is also in S.
I agree with you though, that this can be confusing. The author should have explicitly said either x cannot be the same as y, or x can be the same as y.
I would have answered A. I will give this feedback to the publisher.
Posted: 01/22/2013 21:57
What about if the set ends in 1 and starts in a negative integer?