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Difficult Math Question #32 - Sets

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Difficult Math Question #32 - Sets

by singhpreet1 » Mon Jun 14, 2010 6:05 am
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

I Picked this up from a earlier post back in 2006, though it remains unanswered. i arrived at answer E

the link is https://www.beatthegmat.com/difficult-ma ... -t673.html[/url]

looking forward to valuable inputs...

Thanks.
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by grockit_andrea » Mon Jun 14, 2010 6:27 am
singhpreet1 wrote: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

I Picked this up from a earlier post back in 2006, though it remains unanswered. i arrived at answer E

the link is https://www.beatthegmat.com/difficult-ma ... -t673.html[/url]

looking forward to valuable inputs...

Thanks.
I believe that the answer is D. If -1 is in the set, then so are 1, 3, 5, etc., because each of those is +2 from the previous number t. But the statement doesn't say that t-2 has to be in the set, so there's no reason that -3 would have to be there.
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by singhpreet1 » Mon Jun 14, 2010 6:45 am

I believe that the answer is D. If -1 is in the set, then so are 1, 3, 5, etc., because each of those is +2 from the previous number t. But the statement doesn't say that t-2 has to be in the set, so there's no reason that -3 would have to be there
Hi Andrea...the series does suggest that t+2 is the next no. right? so if we started out from -5, then t+2 is -3. then -1 and 1 and thereon...it doesnt suggest that t cannot be -ve..

if we did take t-2 then we would go from 3-1=1 to t-2= -1 and thereon into -ve values..i agree...but it also doesnt say that that -1 is the first value in the set..it is only a value in the set..so it can be either above or lower right??

what am i missing here...would really appreciate the lack of my understanding.

Thanks in advance.

Preet
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by madhurinarayan » Mon Jun 14, 2010 7:14 am
the problem clearly states that for every t there is a t+2...and not for every t+2 there is a t.....
therefor the set containing 1 and 5 is sufficient to meet the requirement of the set

for example {-1,1,3,5...t+2.....[
this set satisfies the condition
as for everynumber there is a number which is greater than it by 2.
therefor the answer is D
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by outreach » Mon Jun 14, 2010 10:46 am
d looks correct
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by grockit_andrea » Mon Jun 14, 2010 10:52 am
singhpreet1 wrote:

I believe that the answer is D. If -1 is in the set, then so are 1, 3, 5, etc., because each of those is +2 from the previous number t. But the statement doesn't say that t-2 has to be in the set, so there's no reason that -3 would have to be there
Hi Andrea...the series does suggest that t+2 is the next no. right? so if we started out from -5, then t+2 is -3. then -1 and 1 and thereon...it doesnt suggest that t cannot be -ve..

if we did take t-2 then we would go from 3-1=1 to t-2= -1 and thereon into -ve values..i agree...but it also doesnt say that that -1 is the first value in the set..it is only a value in the set..so it can be either above or lower right??

what am i missing here...would really appreciate the lack of my understanding.

Thanks in advance.

Preet
Preet, you're right that the set could be lower (it MUST be higher, because of the t + 2 requirement) and that -3 COULD be in the set. But the question doesn't ask what could be true. It asks what must be true. It is possible, but not necessary, for -3 to be in the set. It is necessary, however, for both 1 and 5 to be there.
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by Testluv » Mon Jun 14, 2010 10:20 pm
Andrea is definitely correct.

From the question, all we know is 2 things:

-1 is in the set
For any number in the set, that number plus 2 is also in the set.

We don't know ANYTHING else.

Now, what can we deduce from the two pieces of info that we do have?

Well, if -1 is there, then according to the rule (the 2nd piece of info), 1 is also there. And, iteratively now, if 1 is there--which it is--then 3 is there. If 3 is there, 5 is there. And so on. Thus, the set certainly contains all positive odd numbers. (The question would have been more interesting if one of the roman numeral statements was "97" or something like that). We don't know whether the set contains odd numbers smaller than -1.

II and III MUST be true. I could be true but also could be false.

Choose D.

In GMAT quant you can only make the following assumptions:

--a line that appears to be straight is in fact straight
--indivisible objects must be positive integers
--the two statements in a DS question will never contradict each other
--the two statemetns in a DS question must be true

That's it; can't make any other assumptions. Can't assume that numbers smaller than -1 are certainly in the set.
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