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Easy but trick question?

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Easy but trick question?

by jkwan » Wed Jan 14, 2009 10:10 pm
A set of numbers has the property that for any number t in the set, t + 2 is in the set. If –1 is in the set, which of the following must also be in the set?

I. -3
II. 1
III. 5


A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III

The answer is D... I can't seem to figure out why the answer wouldn't be E? If -1 is in the set, couldn't -1 be the result of T+2, therefore T would be -3? Help please! Thanks
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by sachinkr » Wed Jan 14, 2009 10:57 pm
The question states: if number t is in the set, then number (t+2) will also in the set.
-1 is in set
So the set will have -1, -1+2 = 1, 1+2 =3, 3+2=5 and so on.

-1 will be smallest element in the set.
(-1,1,3,5,7)

-3 could also be in set but it is not definite that it will always be in set.
i.e set could be (-5,-3,-1,1,3,5,and so on), (-3,-1,1,3,5...) or (-1,1,3,5,...)
So -3 is present only in first 2 sets and not in third.

As the question asks which elements must be in the set, the correct answer will be 1 and 5. (Ans: D)
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by jkwan » Wed Jan 14, 2009 10:59 pm
sachinkr wrote:The question states: if number t is in the set, then number (t+2) will also in the set.
-1 is in set
So the set will have -1, -1+2 = 1, 1+2 =3, 3+2=5 and so on.

-1 will be smallest element in the set.
(-1,1,3,5,7)

-3 could also be in set but it is not definite that it will always be in set.
i.e set could be (-5,-3,-1,1,3,5,and so on), (-3,-1,1,3,5...) or (-1,1,3,5,...)
So -3 is present only in first 2 sets and not in third.

As the question asks which elements must be in the set, the correct answer will be 1 and 5. (Ans: D)
Ahh! Awesome..thanks!
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by joostinshu » Thu Jan 29, 2009 7:18 am
If it is given the -1 is in the set then does it not follow that the value preceding -1 must be -3? It states that for any value in the set it is true that t+2 is also in the set. If -1 is in the set then the value before it must be there as well given the construct of the question.

What am I missing here?
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by Ian Stewart » Thu Jan 29, 2009 9:28 am
joostinshu wrote:If it is given the -1 is in the set then does it not follow that the value preceding -1 must be -3? It states that for any value in the set it is true that t+2 is also in the set. If -1 is in the set then the value before it must be there as well given the construct of the question.

What am I missing here?
The set could be:

{-1, 1, 3, 5, 7, 9, 11, ...}

This set agrees with the definition given - for any number t in the set above, t+2 is also in the set. Of course, the set could start at some smaller number - it could also be

{-5, -3, -1, 1, 3, 5, ...}

for example, or it could even consist of all odd integers. But there is no need for -3 to be in the set; the smallest element could be -1.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by nmammeri » Fri Jan 30, 2009 9:30 am
If it is given the -1 is in the set then does it not follow that the value preceding -1 must be -3? It states that for any value in the set it is true that t+2 is also in the set. If -1 is in the set then the value before it must be there as well given the construct of the question.

What am I missing here?
Good question. The statement says t => t+2 this means if t is in the set this implies that t+2 is also in the set and not the vice versa ( fundamental logic definition).

You would be right If it was a logical equivalence i.e. t <=> t+2. which means t => t+2 and t+2 => t at the same time. This question is about implication not logical equivalence.

Hope this will help
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