BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Is -4 in S? from DS 1000 series

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Data Sufficiency |
Master | Next Rank: 500 Posts
Posts: 127
Joined: Sat Apr 19, 2008 1:02 am
Thanked: 12 times

by jasonc » Sat May 31, 2008 6:17 pm
answer should be b.

given:
S:{a, -a, -a^2, +/-a^3, +/-a^4.....}
(if you dont' understand the above, read (-a) as b and it should make sense)

stem1: 1 is in S => S has at least 1 & -1
INSUFF

stem2: 2 is in S => S has at least 2, -2, -4, +/-2^n where n is any positive integer
SUFF
Top
Legendary Member
Posts: 1161
Joined: Mon May 12, 2008 2:52 am
Location: Sydney
Thanked: 23 times
Followed by:1 members

by mehravikas » Sat May 31, 2008 8:24 pm
jasonc wrote:answer should be b.

given:
S:{a, -a, -a^2, +/-a^3, +/-a^4.....}
(if you dont' understand the above, read (-a) as b and it should make sense)

stem1: 1 is in S => S has at least 1 & -1
INSUFF

stem2: 2 is in S => S has at least 2, -2, -4, +/-2^n where n is any positive integer
SUFF
Your answer is correct, but I am still not able to understand the way you have made a series of values in the set. From what I understood, the set should be something like S = {a, -a, b, ab}

Can you explain further, please?
Top
Master | Next Rank: 500 Posts
Posts: 127
Joined: Sat Apr 19, 2008 1:02 am
Thanked: 12 times

by jasonc » Sat May 31, 2008 9:37 pm
I think the point you may be missing is that the original 2 statements apply to ALL elements of S. Meaning that if 'a' and '-a' are in the set, then 'ab', or in this case 'a' & '-a', must also be in the set.

like I hinted at in my original post, substitue -a in for b.
I beat the GMAT! 760 (Q49/V44)
Top
Legendary Member
Posts: 1161
Joined: Mon May 12, 2008 2:52 am
Location: Sydney
Thanked: 23 times
Followed by:1 members

by mehravikas » Sun Jun 01, 2008 12:09 am
jasonc wrote:I think the point you may be missing is that the original 2 statements apply to ALL elements of S. Meaning that if 'a' and '-a' are in the set, then 'ab', or in this case 'a' & '-a', must also be in the set.

like I hinted at in my original post, substitue -a in for b.
Getting it now :-))

Thanks,

Vikas
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 127
Joined: Tue May 13, 2008 12:28 am
Thanked: 2 times
GMAT Score:710

by airan » Sun Jun 01, 2008 12:28 am
JasonC,
How can we assume -a=b, b can be something else also.
For example:
if a=2, b=1 that means, -2 is also in set but how b=-2 ?

I think it is based on assumption that b=-a, otherwise the answer should be E. Can u pls try to explain it once more(why assume b=-a)
Thanks
Airan
Top
Senior | Next Rank: 100 Posts
Posts: 68
Joined: Thu Feb 14, 2008 4:11 pm
Thanked: 7 times

by zacharyz » Tue Jun 03, 2008 7:17 am
From (2), you know that S contains at least {2}. There is no other guarantee that anything else is in the set... but let's look at the rules:

i) says that if a is in the set, then -a is in the set.
Now you know that -2 is also in the set.

ii) says that if a and b are in the set, then a*b are in the set.
We know that S at least contains {2, -2} now we have an a and a b. In this case, a = 2 and b = -2. Now you know that a*b is in the set, by definition, and therefore -4 is in the set.

So (B) is sufficient.

Other observations, S is going to become an infinite set in 2 as S becomes {2, -2, -4, 4, 8, -8, 16, etc...}

Statement 1 only guaranteed that S is {1, -1}... and then unknowns

Just remember that a and b are not specific numbers, but any of two different variables.
Top
Post new topic Post Reply
• Page 1 of 1