Data Sufficiency

hey i feel there is something inherently wrong with this question. Let us suppose that p includes 3, we still do not have information whether the set of integers in p are distinct and unique.If they are then 3+ 3 can not result.

The set of P could be this too: [1,2,3,4,5]. this satisfies the statement 1 but does not contain the multiples of 3.

According to me the answer is E.

Also if A is true then Conversely even statement 2 should be true. coz if the set consists of multiples of 3 then any (multiple of 3) - 3 must result in another multiple of 3. so the answer should then be D and not A.


I would like to know the source of this question!

I am not sure if this is a good GMAT question as I see a lot of people having different views and after reading fibonaci's post I am confused .

So here is the original question and official explanation.

If P is a set of integers and 3 is in P, is every positive multiple of 3 in P?

(1) For any integer in P, the sum of 3 and that integer is also in P

(2) for any integer in P, that integer minus 3 is also in P.

The correct answer is A:

To find the answer to this yes-or-no question, try Plugging In to the two statements, starting with the one number that you know is in the set 3. Statement (1) gives us all the positive multiples of 3, which answers the question with a definitive "yes". stetement (2) gives us all the negative multiples of 3, plus 0. The correct answer is A.

Source : Princeton cracking 2009

Hmm..Fibo, you have considered the set p to be [1,2,3,4,5]. If you look at stmt 1, it says 'For any int in P, sum of 3 and that int is also in P".

Let us take the set you have considered - if 1 is present, 1 + 3 should also be present in P. 4 is. Now since 4 is present in P, 4 + 3 should also be present. 7 should be.

Smilarly, if we now consider 3 (since we definitely know from the question stem that it's present in P) 3+3 should be present => 6 should be there. Since 6 is there, 6+3 should be present in P and so on. The set P, then, becomes an infinite series, stretching all the way RIGHT of '0' on the number line, i. e. the positive realm.

Now, lets consider stmt 2:
Since 3 is present, 3-3= 0 should be present, 0-3=-3 should be present, and on it goes till the -ve infinity, but never coming to the positive side of 0.
Of course, it is possible that the series does not start from 3, but from an infinite positive number, for our convenience, lets say 999. Then 999-3=996 should be present, and so on, all the way to negative infinity. But from the Question stem or the stmt 2, we cannot be sure if that a infinite +ve multiple of 3 is present in P or not. So not sufficient.

Makes any sense?

yupp i knew its an indefinite series, but just to put forth my point i laid that set.
Thanks for clarifying. i mis read the question.

positive multiple of 3 include...3,6,9,12,15,......and so on '0' is not a positive multiple of 3
stmt 1

if one no is 3 then other can be 3+3=6,the next is 6+3=9,..and so on.
P=[3, 6, 9, ...]. all these no r +ve multiple of 3
sufficient

stmt 2
if one no is 3 then other can be 3-3=0,then next is 0-3=-3,then next is -3-3=-6....
P=[3, 0, -3,-6 ....]. all these r not +ve multiple of 3
not sufficient


hence A

Question Once again

IF P is a set of integer , is every multiple of 3 in P.

(1) For any integer in P, the sum of 3 and that integer is also in P

( 2) For any integer in , that integer minus 3 is also in P

This is a Yes/No DS question.

From statement(1)
Case 1: Let 6 be an element in set P, then 6+3 is in set P, So result is Yes
Case 2: Let 5 be an element in set P, then 5+3 is in set P, So result is NO

From statement(2)
Case 1: Let 6 be an element in set P, then 6-3 is in set P, So result is Yes
Case 2: Let 5 be an element in set P, then 5-3 is in set P, So result is NO

From statement(1) and statement(2)
Case 1: Let 6 be an element in set P, then 6+3 and 6-3 is in set P, So result is Yes
Case 2: Let 5 be an element in set P, then 5+3 and 5-3 is in set P, So result is NO

Unless we are able to find one element of the set we cant come to a concision.
So the answer choice E is correct.
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