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A set of numbers has the property that for any number t in [#permalink]
30 May 2007, 08:30
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
1. 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
-------------------
2. S is a set of integers such that
i) if a is in S, then –a is in S, and
ii) if each of a and b is in S, then ab is in S.
Is –4 in S?
(1) 1 is in S.
(2) 2 is in S.
I chalked out the first one as E, because,
if t = -1; t+2 = 1 (B) and again t+2 = 3 and so on we'll get 5(C)
Again if t+2 = -1 then t = -3 (A)
So, Answer will be (E). Let me know your view.
1. 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
Don't assume that -1 is the first on the set (the stem doesn't say that).
we can say from the stem that "for any number t in the set, t + 2 is in the set" but I think we can't say that "for any number t in the set, t - 2 is in the set" becuase once again it's an assumption !! the GMAT don't want us to assume on PR problems.
so, if -1 is in the set , 1 must be in the set, if 1 is in the set 3 must be in the set and so do 5 ! (but not -3).
Re: Anybody has Solution for these?? [#permalink]
30 May 2007, 17:53
priyankur_saha@ml.com wrote:
1. 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
------------------- 2. S is a set of integers such that i) if a is in S, then –a is in S, and ii) if each of a and b is in S, then ab is in S. Is –4 in S?
(1) 1 is in S. (2) 2 is in S.
I chalked out the first one as E, because, if t = -1; t+2 = 1 (B) and again t+2 = 3 and so on we'll get 5(C) Again if t+2 = -1 then t = -3 (A) So, Answer will be (E). Let me know your view.
For 2 it's B.
Answer should be E and B respectively. What did you not understand?
Re: Anybody has Solution for these?? [#permalink]
31 May 2007, 10:36
LM wrote:
priyankur_saha@ml.com wrote:
1. 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
------------------- 2. S is a set of integers such that i) if a is in S, then –a is in S, and ii) if each of a and b is in S, then ab is in S. Is –4 in S?
(1) 1 is in S. (2) 2 is in S.
I chalked out the first one as E, because, if t = -1; t+2 = 1 (B) and again t+2 = 3 and so on we'll get 5(C) Again if t+2 = -1 then t = -3 (A) So, Answer will be (E). Let me know your view.
For 2 it's B.
Answer should be E and B respectively. What did you not understand?
See this Explanation above
Don't assume that -1 is the first on the set (the stem doesn't say that).
we can say from the stem that "for any number t in the set, t + 2 is in the set" but I think we can't say that "for any number t in the set, t - 2 is in the set" becuase once again it's an assumption !! the GMAT don't want us to assume on PR problems.
so, if -1 is in the set , 1 must be in the set, if 1 is in the set 3 must be in the set and so do 5 ! (but not -3).
I assumed -1 as t+1 and got tha answer choice A also true. Based on the above explanation I shouldn't. Right??
Re: Anybody has Solution for these?? [#permalink]
31 May 2007, 10:56
priyankur_saha@ml.com wrote:
LM wrote:
priyankur_saha@ml.com wrote:
1. 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
------------------- 2. S is a set of integers such that i) if a is in S, then –a is in S, and ii) if each of a and b is in S, then ab is in S. Is –4 in S?
(1) 1 is in S. (2) 2 is in S.
I chalked out the first one as E, because, if t = -1; t+2 = 1 (B) and again t+2 = 3 and so on we'll get 5(C) Again if t+2 = -1 then t = -3 (A) So, Answer will be (E). Let me know your view.
For 2 it's B.
Answer should be E and B respectively. What did you not understand?
See this Explanation above Don't assume that -1 is the first on the set (the stem doesn't say that).
we can say from the stem that "for any number t in the set, t + 2 is in the set" but I think we can't say that "for any number t in the set, t - 2 is in the set" becuase once again it's an assumption !! the GMAT don't want us to assume on PR problems.
so, if -1 is in the set , 1 must be in the set, if 1 is in the set 3 must be in the set and so do 5 ! (but not -3).
I assumed -1 as t+1 and got tha answer choice A also true. Based on the above explanation I shouldn't. Right??
From my understanding of the question, you can't appoint (t+2) as -1. every organ in the set is t, so if (t+2) is in the set (t+2)+2 must be in the set but not (t+2)-2.