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# m20 # 33

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m20 # 33 [#permalink]

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25 Feb 2009, 19:02
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Question:

Is $$|X+1|<2$$?

1. $$(X-1)^2 < 1$$
2. $$X^2-2 < 0$$

[Reveal] Spoiler: OA
E

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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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28 Feb 2009, 23:02
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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) (X-1)^2 < 1
2) X^2-2 < 0

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Re: m20 # 33 [#permalink]

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10 Mar 2009, 20:50
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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) (X-1)^2 < 1
2) X^2-2 < 0

1) From S1, is there a quick way to come to the conclusion that 0<x<2
(x-1)^2 < 1
(x-1) < 1
x < 2

(x-1) > -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 < 2-1
Or, x < 1

If |x+1| is -ve, -(x+1) < 2.
Or, -(x+1) < 2.
Or, -x-1 < 2.
Or, -2-1 < 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, 10:03
Thank you. Makes sense. Looking at it now, I'm not even sure what I was confused about-maybe 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, 10:10
Kudos system is currently unavailable. We're working on getting it back.
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Re: m20 # 33 [#permalink]

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28 Apr 2009, 00:29
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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, 06:45
Is |X+1|<2?

1. (X-1)^2 < 1
2. X^2-2 < 0

E.

We have to have -3 < x < 1

1. Consider x=1.1. (1.1-1)^2 = .1^2 = 0.01 (less than one) falls outside -3 < x < 1
Also consider 0.1. (0.1-1)^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, 10:58
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is |x+1|<2 can be rewritten as

x+1<2 or -x-1<2

x<1 or x>-3

1)(x-1)^2<1
(x-1)<1
x<2 so insufficient

2)x^2-2<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, 11:35
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, 20:51
what is OA

for me ANS is : E
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Re: m20 # 33 [#permalink]

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08 Mar 2010, 11:38
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, 06:08
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, 22:54
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, 07:05
|x+1| < 2

x + 1 < 2 or -x-1 < 2

x < 1 or x > -3

From (1)

(x - 1)^2 < 1

x^2 -2x + 1 < 1

=> x(x-2) < 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, 16:22
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, 13:43
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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 (x-1)2 < 1

If (x-1)2 < 1, then most would agree that (x-1)2 is also either equal or greater than 0, since squares cannot be negative.

0<= (x-1)2 < 1 is what needs to be solved.

(x-1)(x-1) < 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, 06:56
dczuchta wrote:
Question:

Is $$|X+1|<2$$?

1. $$(X-1)^2 < 1$$
2. $$X^2-2 < 0$$

[Reveal] Spoiler: OA
E

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) (x-1)sq < 1
(x-1) < + or - 1
so x-1 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, 08:33
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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) (x-1)sq < 1
(x-1) < + or - 1
so x-1 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) $$(x-1)^2 < 1$$ --> since both sides of the inequality are non-negative we can take square root from it: $$|x-1|<1$$ --> $$0<x<2$$. Not sufficient.

(2) $$x^2-2 < 0$$ --> $$x^2<2$$ --> again, since both sides of the inequality are non-negative 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, 19:52
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
Re: m20 # 33   [#permalink] 15 Mar 2013, 19:52
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# m20 # 33

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