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Square Roots DS Question from Kaplan...

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Square Roots DS Question from Kaplan... [#permalink] New post 23 Jan 2006, 16:59
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A
B
C
D
E

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
This question is from Kaplan:

If x is a positive number, is x less than 1 ?

1) x > sqrt(x)
2) -sqrt(x) > -x

My thought process was :

considering statement 2, multiply both sides by negative 1. It gives the same statement as 1.

Now with statement 1, if x = 4, x > 2 which is x > sqrt(4) => x>1
if x = 1/4, then x > -1/2 which is x >sqrt(1\4) => x<1.

So my answer was both of the statements put together are insufficient.

But Kaplan's reasoning was there can't be any number x which is less than 1 and satisfies the condition x > sqrt(x). Obviously Kaplan is assuming that x is a positive integer. But the question says "x is a positive number" which means it can be positive fractional number.

Could some one please clarify whether it is a valid assumption by Kaplan ? Am I missing missing something here ?
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 [#permalink] New post 23 Jan 2006, 17:35
x = 1/4

Statement 1 can't be true since 1/4 is not greater than 1/2.

Statement 2 can't be true since -1/2 is not greater than -1/4.

Answer should be D.
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 [#permalink] New post 23 Jan 2006, 17:51
[quote="lhotseface"]x = 1/4

Statement 1 can't be true since 1/4 is not greater than 1/2.

Statement 2 can't be true since -1/2 is not greater than -1/4.

Answer should be D.[/quote]

But why should we consider only the positive square root (1/2 ?). The problem statement says x is positive it doesn't talk about the positive square root. So in that case 1/4 > -1/2 . And as you have shown 1/4 can't be greater than 1/2 (the other square root). So my little brain says answer should be E. Please correct me if I am wrong
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 [#permalink] New post 23 Jan 2006, 18:21
Choose numbers

From 1) if x = 64 then x > sqrt x, where as if x = 0.64 then x < sqrt x Hence sufficient

From 2) -sqrt(x) > -x hence sqrt(x) < x and the argument from 1) follows. This is also sufficient. Answer should be D.
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 [#permalink] New post 23 Jan 2006, 18:24
On the GMAT, sqrt(4) = 2 and not (2,-2).

thegmatguy wrote:
lhotseface wrote:
x = 1/4

Statement 1 can't be true since 1/4 is not greater than 1/2.

Statement 2 can't be true since -1/2 is not greater than -1/4.

Answer should be D.


But why should we consider only the positive square root (1/2 ?). The problem statement says x is positive it doesn't talk about the positive square root. So in that case 1/4 > -1/2 . And as you have shown 1/4 can't be greater than 1/2 (the other square root). So my little brain says answer should be E. Please correct me if I am wrong
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 [#permalink] New post 23 Jan 2006, 21:48
it is D , got i tried taking x = 0.4 and sqrt(0.4)
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 [#permalink] New post 23 Jan 2006, 22:15
D for me..
a)x=4 => 4>2 so xis not less than 1..This stmt wont suffice with fractional values.

b) x=4=> -2>-4..so x is not less than 1...This stmt wont suffice again with fractional values.
  [#permalink] 23 Jan 2006, 22:15
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