|
Author |
Message |
|
TAGS:
|
|
|
Intern
Joined: 10 Feb 2007
Posts: 3
Followers: 0
Kudos [?]:
0
[0], given: 0
|
I'm flipping through the princeton review 2007 study guide [#permalink]
 10 Feb 2007, 14:32
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
I'm flipping through the princeton review 2007 study guide and come to this question:
Data sufficiency problem
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 answer given is A, but I think it's D, here's why.
Given Statement 1, since 3 is in P, then 3+3=6 is in P, so P = {3,6,9,12......3n}
The answer to the stated question is "Yes" so Statement 1 is sufficient information to answer the question.
Given Statement 2, since 3 is in P, then 3-3 = 0 etc, so P = {3, 0, -3, -6.....-3n}
In which case we can still answer the question "No, all positive multiples of 3 are not in P"
We can still answer the question, and Statement 2 is sufficient information to do that.
Therefore the answer is D.
Tell me if you agree or disagree, and why?
Thanx
Last edited by crankharder on 10 Feb 2007, 14:47, edited 1 time in total.
|
|
|
|
|
|
|
SVP
Joined: 05 Jul 2006
Posts: 1564
Followers: 4
Kudos [?]:
65
[0], given: 37
|
OK , WHY DO U ASSUME IN STATMENT TWO THAT THE SET STARTS WITH 3.......................... Got it 
|
|
|
|
|
|
Intern
Joined: 10 Feb 2007
Posts: 3
Followers: 0
Kudos [?]:
0
[0], given: 0
|
yezz wrote: OK , WHY DO U ASSUME IN STATMENT TWO THAT THE SET STARTS WITH 3.......................... Got it Because it's given in the question "3 is in P"
|
|
|
|
|
|
SVP
Joined: 01 May 2006
Posts: 1837
Followers: 8
Kudos [?]:
73
[0], given: 0
|
crankharder wrote: I'm flipping through the princeton review 2007 study guide and come to this question: Data sufficiency problem 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 answer given is A, but I think it's D, here's why. Given Statement 1, since 3 is in P, then 3+3=6 is in P, so P = {3,6,9,12......3n} The answer to the stated question is "Yes" so Statement 1 is sufficient information to answer the question. Given Statement 2, since 3 is in P, then 3-3 = 0 etc, so P = {3, 0, -3, -6.....-3n}
In which case we can still answer the question "No, all positive multiples of 3 are not in P"
We can still answer the question, and Statement 2 is sufficient information to do that.Therefore the answer is D. Tell me if you agree or disagree, and why? Thanx The bold is wrong  .... U have taken the extrem case in one side... only 3 is a positive multiple of 3 in P.... but what about the extremum case in the other side.... all can be in P ...
Answer A
|
|
|
|
|
|
Intern
Joined: 02 Jan 2007
Posts: 42
Followers: 0
Kudos [?]:
1
[0], given: 0
|
What the other posters are suggesting is that along with what you wrote, this also satisfies statement 2:
3n, 3(n-1), 3(n2),..., 12, 9, 6, 3
Which *does* include all multiple of 3 and, thus, along with your reasoning for statement 2, makes this not sufficient.
_________________
Beginning with the end in mind. Aiming to join the 700+ club.
|
|
|
|
|
|
SVP
Joined: 05 Jul 2006
Posts: 1564
Followers: 4
Kudos [?]:
65
[0], given: 37
|
man it says 3 is in P this doent mean it has to start with 3
|
|
|
|
|
|
Senior Manager
Joined: 04 Jan 2006
Posts: 282
Followers: 1
Kudos [?]:
14
[0], given: 0
|
Can set P in (1) be
P = {-3n, -3(n-1), ..., -9, -6, -3, 0, 3, 6, 9,...3n} ?
P = {..., 3(-3), 3(-2), 3(-1), 3(0), 3(1), 3(2), ...}
If true, wouldn't (1) INSUFF?
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
