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If a^2 = b, is the value of a between 0 and 1. 1) b is

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If a^2 = b, is the value of a between 0 and 1. 1) b is [#permalink]

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09 Jul 2008, 19:54
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If a^2 = b, is the value of a between 0 and 1.

1) b is between 0 and 1.
2) a> b
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Joined: 30 Apr 2008
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09 Jul 2008, 20:06
It has to be E.

Statement 1) Sure, we know that if 0 < b < 1, then a has to be a fraction (or decimal), like b = 1/4 so A = 1/2 OR -1/2. In the case where a = -1/2, it is not between 0 and 1. INSUFFICIENT.

Statement 2) Tells us the same thing as Statement 1 but in a different way. Insufficient.

Together) Insufficient becuase we're told the same thing with each statement but in different ways. No information we are given tells us if A must be positive or negative.

sachinn wrote:
If a^2 = b, is the value of a between 0 and 1.

1) b is between 0 and 1.
2) a> b

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Current Student Joined: 28 Dec 2004 Posts: 3385 Location: New York City Schools: Wharton'11 HBS'12 Followers: 14 Kudos [?]: 241 [1] , given: 2 Re: DS [#permalink] Show Tags 09 Jul 2008, 20:37 1 This post received KUDOS sachinn wrote: If a^2 = b, is the value of a between 0 and 1. 1) b is between 0 and 1. 2) a> b a^2=+ ..so B is positive.. OK..now if a>b...then sqrt(b)=a.. here we know that sqrt(b) is positive..thus a has to be positive..therefore sufficient B it is.. SVP Joined: 30 Apr 2008 Posts: 1888 Location: Oklahoma City Schools: Hard Knocks Followers: 39 Kudos [?]: 530 [0], given: 32 Re: DS [#permalink] Show Tags 09 Jul 2008, 20:55 I wasn't thinking about b being positive (essentially b = |a|^2, so the answer will never be negative) and if a > b then we know a must be positive. In an indirect way, it tells us that a is indeed positive. Nice catch fresinha. +1 fresinha12 wrote: sachinn wrote: If a^2 = b, is the value of a between 0 and 1. 1) b is between 0 and 1. 2) a> b a^2=+ ..so B is positive.. OK..now if a>b...then sqrt(b)=a.. here we know that sqrt(b) is positive..thus a has to be positive..therefore sufficient B it is.. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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09 Jul 2008, 20:57
Great!!!! thanks for the replies, that was really helpful.
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09 Jul 2008, 21:28
1) alone is NOT sufficient. Because a can be a –ve fraction. Consider a = -1/2, then a^2 = 1/4

2) alone is NOT sufficient. Because 0< b < 1; Consider b = 1/2, a can be an number > 1/2

1) and 2) together is sufficient. Because a > b, a should be +ve. When a is +ve, a should be a fraction itself in order to make a^2 a fraction.

Ans is C
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09 Jul 2008, 22:43
My 2nd trial:

1) alone is NOT Sufficient. Same reason as in my last message.

2) alone is sufficient.
Because a^2 = b, so b > 0 (also because a > b, b should be be 0)
as b > 0, and a > b, then a > 0
a > b = a^2
a > a^2
so a must be less than 1
combine all, 0 < a < 1

So ans is B
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09 Jul 2008, 23:44
sachinn wrote:
If a^2 = b, is the value of a between 0 and 1.

1) b is between 0 and 1.
2) a> b

stat1

b =1/2 ,a =+-sqrt(1/2) ->NS

stat2

a= 1/4, b= 1/16 suff

B
Re: DS   [#permalink] 09 Jul 2008, 23:44
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