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m20 # 33 : Retired Discussions [Locked]

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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?

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?

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. _________________

Re: m20 # 33 Kudos? [#permalink]
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.

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. _________________

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 _________________

Regards, Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

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.