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# DS princeton review drill 11

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DS princeton review drill 11 [#permalink]  16 Aug 2008, 04:08
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(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
ds yes no answer question from princeton review ds drill 11

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If x is a positive number, is x < 1?

(1) 2x < 1
(2) 2x <= 2 (2x is less than and equal to 2)

its gives yes when we plug values less than 1/2 but what if we plug values like 4/5 which is .8 or 3/5 which is .6 it gives no answer so princeton review called statement 1 sufficient despite of the fact it gives sometime positive and sometime negative value.
somebody put more light on it how is that possible ?

Last edited by blover on 04 Sep 2008, 11:35, edited 3 times in total.
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Re: DS princeton review drill 11 [#permalink]  02 Sep 2008, 12:29
what does that symbol denote for the statement (2)?
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Re: DS princeton review drill 11 [#permalink]  02 Sep 2008, 13:55
I'm not sure what the symbol in statement 2 is. But for statement 1 you can rearrange

2x < 1 => x < 1/2

Therefore, x must be less than 1. Suff
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Re: DS princeton review drill 11 [#permalink]  02 Sep 2008, 13:57
A... given that the 'pound symbol' is undefined

1) From 2x<1, we get X<1/2 which is in fact less than 1 (we also know X is postive so 0<x<1/2) -- Suff.

2) ??? This symbol is undefined.. so the answer will either be A or D

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Re: DS princeton review drill 11 [#permalink]  04 Sep 2008, 10:24
Re: DS princeton review drill 11   [#permalink] 04 Sep 2008, 10:24
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