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15 Sep 2014, 23:53
Official Solution: First of all: is \(x \gt 1\) means is \(x \lt 1\) \((2, 3, 4, ...)\) or is \(x \gt 1\) \((2, 3, 4, ...)\), so for YES answer \(x\) can be any integer but 1, 0, and 1. (1)\((1  2x)(1 + x) \lt 0\). Rewrite as \((2x  1)(x + 1) \gt 0\) (so that the coefficient of \(x^2\) is positive after expanding): roots are \(x=1\) and \(x=\frac{1}{2}\). \(\gt\)" sign means that the given inequality holds true for: \(x \lt 1\)and \(x \gt \frac{1}{2}\). \(x\) could still equal 1, so not sufficient. (2) \((1  x)(1 + 2x) \lt 0\). Rewrite as \((x  1)(2x + 1) \gt 0\): roots are \(x=\frac{1}{2}\) and \(x=1\). "\(\gt\)" sign means that the given inequality holds true for: \(x \lt  \frac{1}{2}\) and \(x \gt 1\). \(x\) could still equal 1, so not sufficient. (1)+(2) Intersection of the ranges from (1) and (2) is \(x \lt 1\) and \(x \gt 1\). Sufficient. Answer: C
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why do we disregard X< 1/2 for the first solution and X> 1/2 for the second solution? I chose E as although X>1 and X<1 can be derived from both 1 and 2 respectively, the range of 1/2 < x< 1/2 from the information does not align to solution as x can =0



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Re: M1413
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05 May 2015, 07:09
I did by wavy curve method.....Efficient



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Re: M1413
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08 Jun 2015, 05:32
Bunuel wrote: rsamant wrote: why do we disregard X< 1/2 for the first solution and X> 1/2 for the second solution? I chose E as although X>1 and X<1 can be derived from both 1 and 2 respectively, the range of 1/2 < x< 1/2 from the information does not align to solution as x can =0 I don't understand what you mean at all... He means why do we have to consider only extreme ranges in range intersection you explained...while considering (1) and (2), why do we eliminate x>1/2 and x<1/2... because if we donot do so we get a range possible between x<1 and x <1/2 and also x>1/2 and x > 1 so these in bet ranges are confusing.



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Re: M1413
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19 Jun 2015, 10:08
Bunuel wrote: Official Solution:
First of all: is \(x \gt 1\) means is \(x \lt 1\) \((2, 3, 4, ...)\) or is \(x \gt 1\) \((2, 3, 4, ...)\), so for YES answer \(x\) can be any integer but 1, 0, and 1. (1)\((1  2x)(1 + x) \lt 0\). Rewrite as \((2x  1)(x + 1) \gt 0\) (so that the coefficient of \(x^2\) is positive after expanding): roots are \(x=1\) and \(x=\frac{1}{2}\). \(\gt\)" sign means that the given inequality holds true for: \(x \lt 1\)and \(x \gt \frac{1}{2}\). \(x\) could still equal 1, so not sufficient. (2) \((1  x)(1 + 2x) \lt 0\). Rewrite as \((x  1)(2x + 1) \gt 0\): roots are \(x=\frac{1}{2}\) and \(x=1\). "\(\gt\)" sign means that the given inequality holds true for: \(x \lt  \frac{1}{2}\) and \(x \gt 1\). \(x\) could still equal 1, so not sufficient. (1)+(2) Intersection of the ranges from (1) and (2) is \(x \lt 1\) and \(x \gt 1\). Sufficient.
Answer: C Hello Bunuel, Is there any other method to solve this quickly ?? In a test environment sometimes it's hard to think like this and get answer timely. Thanks



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I have some confusion about this question. From statement 1 taken x<1 but not X>1/2 & from statement 2 taken only X<1/2 but not X>1. Why cannot be considered X<1/2 & X>1/2? would anyone like to clarify my confusion. Thank you.



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Hi Bunuel, I know this has to do something with inequalities changing signs when you go beyond extremes. But I needed a refresher on that. In particular I am stuck with how did you come up with this bit: Quote: ">" sign means that the given inequality holds true for: x<−1x<−1and x>12x>12. xx could still equal 1, so not sufficient. Is there any post which I can refer to solidify my concept of finding valid extreme values from Roots of the equation? Thanks in advance. Vaibhav. Ok after doing a bit of looking around, I found this: http://gmatclub.com/forum/inequalitiestrick91482.html#p804990This serves the purpose. Bunuel, is there any other slick/intuitive way of doing this such that one saves time without drawing the line graph and plotting roots on it during the exam?
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05 Mar 2016, 14:35
Bunuel wrote: Official Solution:
First of all: is \(x \gt 1\) means is \(x \lt 1\) \((2, 3, 4, ...)\) or is \(x \gt 1\) \((2, 3, 4, ...)\), so for YES answer \(x\) can be any integer but 1, 0, and 1. (1)\((1  2x)(1 + x) \lt 0\). Rewrite as \((2x  1)(x + 1) \gt 0\) (so that the coefficient of \(x^2\) is positive after expanding): roots are \(x=1\) and \(x=\frac{1}{2}\). \(\gt\)" sign means that the given inequality holds true for: \(x \lt 1\)and \(x \gt \frac{1}{2}\). \(x\) could still equal 1, so not sufficient. (2) \((1  x)(1 + 2x) \lt 0\). Rewrite as \((x  1)(2x + 1) \gt 0\): roots are \(x=\frac{1}{2}\) and \(x=1\). "\(\gt\)" sign means that the given inequality holds true for: \(x \lt  \frac{1}{2}\) and \(x \gt 1\). \(x\) could still equal 1, so not sufficient. (1)+(2) Intersection of the ranges from (1) and (2) is \(x \lt 1\) and \(x \gt 1\). Sufficient.
Answer: C Hi Experts! I'm confused: In statement (1) when you get X<1 and X>1/2, doesn't this show that X is not >1? When you plot the quadratic equation (2X1)(X+1)>0 > 2X^2+X1 >0, you get the range X<1 and X>1/2, thus X obviously has a value at 1  meaning that X is not greater than 1. Wouldn't this make (1) sufficient? Similarly in statement (2), the range is X<1/2 and X>1, thus again X obviously has a value at 1  meaning that X is not less than 1. Again wouldn't this make (2) sufficient? Thank you!
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I found that the most efficient way, for me, was to plug in. The question asks us to determine if \(x>1\) or \(x<1\), so we just need to test 1,0, and 1 in both statements. Statement 1 can be satisfied with 1 or any positive integer so it is not sufficient. It is worth noting that it can not be satisfied with 1 or 0. Statement 2 can be satisfied with 1 or any negative integer, so it is not sufficient. It is with noting that it can not be satisfied with 0 nor with 1. Combined, we see from stmt 1 that it can not be 1 or 0, and from stmt 2 that It can not be 1 or 0 so x must be either greater than 1 or less than 1. I tried the curve approach but I did it in over 3 minutes so in this case it's not efficient for me. Hope this helps.
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25 Dec 2016, 09:47
happyface101 wrote: Bunuel wrote: Official Solution:
First of all: is \(x \gt 1\) means is \(x \lt 1\) \((2, 3, 4, ...)\) or is \(x \gt 1\) \((2, 3, 4, ...)\), so for YES answer \(x\) can be any integer but 1, 0, and 1. (1)\((1  2x)(1 + x) \lt 0\). Rewrite as \((2x  1)(x + 1) \gt 0\) (so that the coefficient of \(x^2\) is positive after expanding): roots are \(x=1\) and \(x=\frac{1}{2}\). \(\gt\)" sign means that the given inequality holds true for: \(x \lt 1\)and \(x \gt \frac{1}{2}\). \(x\) could still equal 1, so not sufficient. (2) \((1  x)(1 + 2x) \lt 0\). Rewrite as \((x  1)(2x + 1) \gt 0\): roots are \(x=\frac{1}{2}\) and \(x=1\). "\(\gt\)" sign means that the given inequality holds true for: \(x \lt  \frac{1}{2}\) and \(x \gt 1\). \(x\) could still equal 1, so not sufficient. (1)+(2) Intersection of the ranges from (1) and (2) is \(x \lt 1\) and \(x \gt 1\). Sufficient.
Answer: C Hi Experts! I'm confused: In statement (1) when you get X<1 and X>1/2, doesn't this show that X is not >1? When you plot the quadratic equation (2X1)(X+1)>0 > 2X^2+X1 >0, you get the range X<1 and X>1/2, thus X obviously has a value at 1  meaning that X is not greater than 1. Wouldn't this make (1) sufficient? Similarly in statement (2), the range is X<1/2 and X>1, thus again X obviously has a value at 1  meaning that X is not less than 1. Again wouldn't this make (2) sufficient? Thank you! please help me to understand:S1: 2x^2+x1>0 =>2x^2+x>1=>x(2x+1)>1=>x>1 or x>0....how it could be 1>x>1/2??? thnx



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25 Dec 2016, 09:53
gupta87 wrote: happyface101 wrote: Bunuel wrote: Official Solution:
First of all: is \(x \gt 1\) means is \(x \lt 1\) \((2, 3, 4, ...)\) or is \(x \gt 1\) \((2, 3, 4, ...)\), so for YES answer \(x\) can be any integer but 1, 0, and 1. (1)\((1  2x)(1 + x) \lt 0\). Rewrite as \((2x  1)(x + 1) \gt 0\) (so that the coefficient of \(x^2\) is positive after expanding): roots are \(x=1\) and \(x=\frac{1}{2}\). \(\gt\)" sign means that the given inequality holds true for: \(x \lt 1\)and \(x \gt \frac{1}{2}\). \(x\) could still equal 1, so not sufficient. (2) \((1  x)(1 + 2x) \lt 0\). Rewrite as \((x  1)(2x + 1) \gt 0\): roots are \(x=\frac{1}{2}\) and \(x=1\). "\(\gt\)" sign means that the given inequality holds true for: \(x \lt  \frac{1}{2}\) and \(x \gt 1\). \(x\) could still equal 1, so not sufficient. (1)+(2) Intersection of the ranges from (1) and (2) is \(x \lt 1\) and \(x \gt 1\). Sufficient.
Answer: C Hi Experts! I'm confused: In statement (1) when you get X<1 and X>1/2, doesn't this show that X is not >1? When you plot the quadratic equation (2X1)(X+1)>0 > 2X^2+X1 >0, you get the range X<1 and X>1/2, thus X obviously has a value at 1  meaning that X is not greater than 1. Wouldn't this make (1) sufficient? Similarly in statement (2), the range is X<1/2 and X>1, thus again X obviously has a value at 1  meaning that X is not less than 1. Again wouldn't this make (2) sufficient? Thank you! please help me to understand:S1: 2x^2+x1>0 =>2x^2+x>1=>x(2x+1)>1=>x>1 or x>0....how it could be 1>x>1/2??? thnx This is totally wrong. You don't solve quadratic inequality this way. How/why did you conclude that x>1 or x>0 from x(2x+1)>1? By the way x>1 or x>0 does not make sense at all. You need to return to basics and study inequalities: Inequalities Made Easy!Solving Quadratic Inequalities  Graphic ApproachInequality tipsDS Inequalities Problems PS Inequalities Problems 700+ Inequalities problemsHope it helps.
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25 Dec 2016, 10:54
understood. but from S1: 2x^2+x1>0 => (x+1)(2x1)>0 => x>1 0r x>1/2...how do u arrive at x<1 ??



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17 Dec 2017, 03:09
Bunuel wrote: Official Solution:
First of all: is \(x \gt 1\) means is \(x \lt 1\) \((2, 3, 4, ...)\) or is \(x \gt 1\) \((2, 3, 4, ...)\), so for YES answer \(x\) can be any integer but 1, 0, and 1. (1)\((1  2x)(1 + x) \lt 0\). Rewrite as \((2x  1)(x + 1) \gt 0\) (so that the coefficient of \(x^2\) is positive after expanding): roots are \(x=1\) and \(x=\frac{1}{2}\). \(\gt\)" sign means that the given inequality holds true for: \(x \lt 1\)and \(x \gt \frac{1}{2}\). \(x\) could still equal 1, so not sufficient. (2) \((1  x)(1 + 2x) \lt 0\). Rewrite as \((x  1)(2x + 1) \gt 0\): roots are \(x=\frac{1}{2}\) and \(x=1\). "\(\gt\)" sign means that the given inequality holds true for: \(x \lt  \frac{1}{2}\) and \(x \gt 1\). \(x\) could still equal 1, so not sufficient. (1)+(2) Intersection of the ranges from (1) and (2) is \(x \lt 1\) and \(x \gt 1\). Sufficient.
Answer: C Hi Bunuel, My approach is a bit different from yours but uses the same concept. But I'm getting an incorrect answer. Could you please advise which step I'm following wrong? As per question Stem  x >1 X< 1 Choice A ( I didn't multiply by 1 the way you did) I got equation  1< X <1/2  So answer is No for the value of X Choice B ( I didn't multiply by 1 the way you did) Equation : 1/2 < X < 1  So answer is No for the value of X Hence I marked D in GMAT Club test which is incorrect. Could you please advise how we can solve without multiplying by  1? And in which step I'm wrong?



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NeverGiveUp Arpit wrote: Hi Bunuel, My approach is a bit different from yours but uses the same concept. But I'm getting an incorrect answer. Could you please advise which step I'm following wrong? As per question Stem  x >1 X< 1 Choice A ( I didn't multiply by 1 the way you did) I got equation  1< X <1/2  So answer is No for the value of X Choice B ( I didn't multiply by 1 the way you did) Equation : 1/2 < X < 1  So answer is No for the value of X Hence I marked D in GMAT Club test which is incorrect. Could you please advise how we can solve without multiplying by  1? And in which step I'm wrong? Hi @NeverGiveUp Arpit The highlighted portion is incorrect. even if you are not multiplying with 1, then you need to realize that coefficient of x will be negative, so on a number line range will start from negative values. Hence when you plot the roots of the quadratic equation on a number line then the range of x will be in the negative regions because the inequality is negative. Refer below image for clarity  Attachment: inequality.jpg Similarly for statement B basically as per your range 0 is a possible solution. so when you put x=0 in statement 1, you will get 1<0, which is not possible
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17 Dec 2017, 04:10
niks18 wrote: NeverGiveUp Arpit wrote: Hi Bunuel, My approach is a bit different from yours but uses the same concept. But I'm getting an incorrect answer. Could you please advise which step I'm following wrong? As per question Stem  x >1 X< 1 Choice A ( I didn't multiply by 1 the way you did) I got equation  1< X <1/2  So answer is No for the value of X Choice B ( I didn't multiply by 1 the way you did) Equation : 1/2 < X < 1  So answer is No for the value of X Hence I marked D in GMAT Club test which is incorrect. Could you please advise how we can solve without multiplying by  1? And in which step I'm wrong? Hi @NeverGiveUp Arpit The highlighted portion is incorrect. even if you are not multiplying with 1, then you need to realize that coefficient of x will be negative, so on a number line range will start from negative values. Hence when you plot the roots of the quadratic equation on a number line then the range of x will be in the negative regions because the inequality is negative. Refer below image for clarity  Attachment: inequality.jpg Similarly for statement B basically as per your range 0 is a possible solution. so when you put x=0 in statement 1, you will get 1<0, which is not possible Makes sense. Cheers mate.



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Re: M1413
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08 Feb 2018, 01:09
Hello all,
how can we just rewrite (12x)(1+x) as (2x1)(x+1) ? How did we turn around the signs for the first braket?
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