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m20 # 33 [#permalink]
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25 Feb 2009, 20:02
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Question: Is \(X+1<2\)? 1. \((X1)^2 < 1\) 2. \(X^22 < 0\) Source: GMAT Club Tests  hardest GMAT questions ____________________ Would someone please explain to me what the simplest way of solving this problem is? 1) From S1, is there a quick way to come to the conclusion that 0<x<2 2) From S2, is there a quick way to come to the conclusion that 2<X< square root of 2 3) When the question asks if X+1< 2, do we conclude that it is asking if X is either <1 OR <3? Thank you.



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Re: m20 # 33 [#permalink]
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IMO A as (1) is SUFFICIENT from (1), we get 1<x<2 from (2), we get 0<x<1.414 dczuchta wrote: Would someone please explain to me what the simplest way of solving this problem is? 1) From S1, is there a quick way to come to the conclusion that 0<x<2 2) From S2, is there a quick way to come to the conclusion that 2<X< square root of 2 3) When the question asks if X+1< 2, do we conclude that it is asking if X is either <1 OR <3?
Thank you, ____________________ Question:
Is X+1<2?
1) (X1)^2 < 1 2) X^22 < 0
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Re: m20 # 33 [#permalink]
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10 Mar 2009, 21:50
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dczuchta wrote: Would someone please explain to me what the simplest way of solving this problem is? 1) From S1, is there a quick way to come to the conclusion that 0<x<2 2) From S2, is there a quick way to come to the conclusion that 2<X< square root of 2 3) When the question asks if X+1< 2, do we conclude that it is asking if X is either <1 OR >3?
Thank you, ____________________ Question:
Is X+1<2?
1) (X1)^2 < 1 2) X^22 < 0 1) From S1, is there a quick way to come to the conclusion that 0<x<2 (x1)^2 < 1 (x1) < 1 x < 2 (x1) > 1 x > 0 sufff.... 2) From S2, is there a quick way to come to the conclusion that 2<X< square root of 2 x^2  2 < 0 x^2 < 2 x < sqrt2 or x > sqrt2 insuff.......... 3) When the question asks if X+1< 2, do we conclude that it is asking if X is either <1 OR >3? yes, x+1< 2 means 3< x <1. This is how it is solved: If x+1 is +ve, (x+1) < 2. Or, x < 21 Or, x < 1 If x+1 is ve, (x+1) < 2. Or, (x+1) < 2. Or, x1 < 2. Or, 21 < x Or, 3 < x Hence 3 < x < 1. If any typo, thats mine.
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Re: m20 # 33 Kudos? [#permalink]
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12 Mar 2009, 11:03
Thank you. Makes sense. Looking at it now, I'm not even sure what I was confused aboutmaybe a wrong calculation. What happened to the Kudos?; the button seemed to have been deleted.



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Re: m20 # 33 [#permalink]
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30 Mar 2009, 11:10



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Re: m20 # 33 [#permalink]
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28 Apr 2009, 01:29
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maybe I'm having a brainfart, but I think the answer is E
(1) 0<x<2 (2) sqrt(2)<x<sqrt(2)
we need to know if 3<x<1
(1) and (2) tells us nothing that can be definite proof.



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Re: m20 # 33 [#permalink]
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05 Mar 2010, 07:45
Is X+1<2?
1. (X1)^2 < 1 2. X^22 < 0
E.
We have to have 3 < x < 1
1. Consider x=1.1. (1.11)^2 = .1^2 = 0.01 (less than one) falls outside 3 < x < 1 Also consider 0.1. (0.11)^2 = (0.5)^2 = .25 (less than one) falls inside 3 < x < 1
This is insuff because it can fall in and it can fall out of our 3 < x < 1.
2. Consider x = 1.1. 1.1^2  2 = 1.21  2 = (less than 0) falls outside 3 < x < 1 Also Consider x = 0.1. .1^2  2 = 0.01  2 = (less than 0) falls inside 3 < x < 1
This is insuff because it can fall in and it can fall out of our 3 < x < 1.
Also remember you aren't answering "Is X+1<2?". You are answering if 1 or 2 can answer that question.



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Re: m20 # 33 [#permalink]
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05 Mar 2010, 11:58
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is x+1<2 can be rewritten as
x+1<2 or x1<2
x<1 or x>3
1)(x1)^2<1 (x1)<1 x<2 so insufficient
2)x^22<0 x+\sqrt{2} x\sqrt{2} <0 x+\sqrt{2} <0 or x\sqrt{2}<0 x<1.414 or x<1.414 so insufficient
combining both 1 and 2 x can still be 1.2 or 0.5 so both are also insufficient
so ans is E



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Re: m20 # 33 [#permalink]
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05 Mar 2010, 12:35
Ans is E
Q is if 3<X<1
1 gives 0<x<2 2 gives sqroot(2) < X < sqroot(2)



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Re: m20 # 33 [#permalink]
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05 Mar 2010, 21:51
what is OA for me ANS is : E
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Re: m20 # 33 [#permalink]
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08 Mar 2010, 12:38
asimov wrote: maybe I'm having a brainfart, but I think the answer is E
(1) 0<x<2 (2) sqrt(2)<x<sqrt(2)
we need to know if 3<x<1
(1) and (2) tells us nothing that can be definite proof. Absolutely right. So E is the answer.
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Re: m20 # 33 [#permalink]
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13 Mar 2010, 07:08
Yes E is the answer. Both statements 1 and 2 give us that X can be atleast 1 or 0. If X = 1 it fails and if X = 0 it passes the X+1^2 test. Hence data is not sufficient even with both.
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Re: m20 # 33 [#permalink]
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07 Sep 2010, 23:54
Its E.
Question: It becomes 3<x<1
1) Solving 1 says 0<x<2 Insufficient 2) Solving 2 says x< sqrt2 Insufficient



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Re: m20 # 33 [#permalink]
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10 Mar 2011, 08:05
x+1 < 2 x + 1 < 2 or x1 < 2 x < 1 or x > 3 From (1) (x  1)^2 < 1 x^2 2x + 1 < 1 => x(x2) < 0 So 0 < x < 2, so not enough From (2) 1.4 < x < 1.4, so not enough From (1) and (2) also, not enough as x can fall in 1 < x < 1.4, so answer is E.
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Re: m20 # 33 [#permalink]
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13 Mar 2012, 17:22
IMO E.
Both the conditions are not sufficient to conclude.



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Re: m20 # 33 [#permalink]
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16 Apr 2012, 14:43
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Most authors agree that the stem asks if 3<x<1, me too.
However, since it seems that most posts struggle with the first statement, here is my take.
(1) states (x1)2 < 1
If (x1)2 < 1, then most would agree that (x1)2 is also either equal or greater than 0, since squares cannot be negative.
0<= (x1)2 < 1 is what needs to be solved.
(x1)(x1) < 1 x2  2x + 1 < 1 x2  2x < 0 x2 < 2x
At this point we can see that 0 < x <= 1, which makes this statement insufficient as x, per stem cannot be 1.
(2) x < sqrt2 or x > sqrt2
(C) To analyze them together, we have to look at both. 0 < x <= 1 (1.41) sqrt2 < x < sqrt2 (1.41)
Here we can see, that 1 is still part of both, which makes this question E.



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Re: m20 # 33 [#permalink]
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25 Apr 2012, 07:56
dczuchta wrote: Question: Is \(X+1<2\)? 1. \((X1)^2 < 1\) 2. \(X^22 < 0\) Source: GMAT Club Tests  hardest GMAT questions ____________________ Would someone please explain to me what the simplest way of solving this problem is? 1) From S1, is there a quick way to come to the conclusion that 0<x<2 2) From S2, is there a quick way to come to the conclusion that 2<X< square root of 2 3) When the question asks if X+1< 2, do we conclude that it is asking if X is either <1 OR <3? Thank you. let me know if my method is OKay...though it is learnt now only by going through the above observations and my own calculations 1) (x1)sq < 1 (x1) < + or  1 so x1 is between  1 to +1 hence x is between 0 to 2 ( exclusive) so it can take the value 1 , then the solution gives YES if it takes 1.99, then the solution gives NO hence insufficient... A and D are gone 2) xsq  2 < 0 xsq < 2 x is between 1.41 to + 1.41, so this also gives both YES and NO as answer so B is gone too answer should be b/w C and E taking together x would now be between o and 1.41 so here too we get YES and NO for the solution, hence E is the answer
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Re: m20 # 33 [#permalink]
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25 Apr 2012, 09:33
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harshavmrg wrote: let me know if my method is OKay...though it is learnt now only by going through the above observations and my own calculations
1) (x1)sq < 1 (x1) < + or  1 so x1 is between  1 to +1 hence x is between 0 to 2 ( exclusive) so it can take the value 1 , then the solution gives YES if it takes 1.99, then the solution gives NO
hence insufficient... A and D are gone
2) xsq  2 < 0
xsq < 2 x is between 1.41 to + 1.41, so this also gives both YES and NO as answer so B is gone too
answer should be b/w C and E
taking together
x would now be between o and 1.41 so here too we get YES and NO for the solution, hence E is the answer Yes, your solution is correct. Is \(x+1<2\)?Is \(x+1<2\)? > is \(3<x<1\)? (1) \((x1)^2 < 1\) > since both sides of the inequality are nonnegative we can take square root from it: \(x1<1\) > \(0<x<2\). Not sufficient. (2) \(x^22 < 0\) > \(x^2<2\) > again, since both sides of the inequality are nonnegative we can take square root from it: \(x<1.4\) (approximately) > \(1.4<x<1.4\). Not sufficient. (1)+(2) Intersection of the ranges from (1) and (2) is \(0<x<1.4\). So, \(x\) could be in the range \(3<x<1\) (for example if \(x=1\)) as well as out of the range \(3<x<1\) (for example if \(x=1.2\)). Not sufficient. Answer: E.
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Re: m20 # 33 [#permalink]
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15 Mar 2013, 20:52
Complete solution of the problem,
Given in question stem,
\(X+1\)\(<\) \(2\)
If, \((X+1)\)\(< 0\) \((X+1) < 2\) \(X+1 > 2\) \(X > 3\)
If\((X+1) > 0\) \((X+1) > 2\) \(X > 1\)
So we have to check that if any equation proves \(3 < X < 1\), that will be sufficient condition
I  \((X 1)^2 < 1\) \((X 1 ) < 1\) or \((X  1) > 1\)
On solving, \(0 < X < 2\)
This is not sufficient condition, So A and D are incorrect choices.
II  \(X^2 2 < 0\) \(X^2 < 2\)
On solving, \(\sqrt{2} < X < \sqrt{2}\)
This is not sufficient condition, So B and D are incorrect choices.
I and II together,
\(0 < X < \sqrt{2}\)
This is also not sufficient, so Ans is E











