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if x + 1  > 2x  1, which of the following represents the correct
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Updated on: 18 Aug 2018, 11:37
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if x + 1  > 2x  1, which of the following represents the correct range of values of x ? A. x < 0 B. x < 2 C. 2 < x < 0 D. 1 < x < 2 E. 0 < x < 2 Source: ExpertsGlobalchetan2u, Shouldn't we consider the overlapping region of (x<0) and (x<2), and the range would then be x < 0 ? Is the OA given wrong ?
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Originally posted by TaN1213 on 18 Aug 2018, 10:23.
Last edited by TaN1213 on 18 Aug 2018, 11:37, edited 1 time in total.



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if x + 1  > 2x  1, which of the following represents the correct
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Updated on: 18 Aug 2018, 12:28
TaN1213 wrote: if x + 1  > 2x  1, which of the following represents the correct range of values of x ? A. x < 0 B. x < 2 C. 2 < x < 0 D. 1 < x < 2 E. 0 < x < 2 Source: ExpertsGlobalShouldn't we consider the overlapping region of (x<0) and (x<2), and the range would then be x < 0 ? Is the OA given wrong ? Given, x + 1  > 2x  1 Or, x+1 > 2x1 (x+1=x+1,when \((x+1)\geq{0}\) or, \(x\geq{1}\))Or, 2x1< x+1 Or, x< 2 So, \(x\ge \:1\quad \mathrm{and}\quad \:x<2\)(1) Given, x + 1  > 2x  1 Or, (x+1) > 2x1 (x+1=(x+1), when (x+1) < 0 Or, x< 1)Or, 2x1< (x+1) Or, 2x1< x1 Or, 3x < 0 Or, x < 0 So, \(x<1\quad \mathrm{and}\quad \:x<0\)(2) Combining (1) & (2), \(\left(x<1\quad \mathrm{and}\quad \:x<0\right)\quad \mathrm{or}\quad \left(x\ge \:1\quad \mathrm{and}\quad \:x<2\right)\) So, \(x < 2\) Ans. (B)
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Originally posted by PKN on 18 Aug 2018, 10:35.
Last edited by PKN on 18 Aug 2018, 12:28, edited 1 time in total.



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Re: if x + 1  > 2x  1, which of the following represents the correct
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18 Aug 2018, 11:20
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Re: if x + 1  > 2x  1, which of the following represents the correct
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18 Aug 2018, 11:36
PKN wrote: TaN1213 wrote: if x + 1  > 2x  1, which of the following represents the correct range of values of x ? A. x < 0 B. x < 2 C. 2 < x < 0 D. 1 < x < 2 E. 0 < x < 2 Source: ExpertsGlobalShouldn't we consider the overlapping region of (x<0) and (x<2), and the range would then be x < 0 ? Is the OA given wrong ? Given, x + 1  > 2x  1 Or, x+1 > 2x1 (x+1=x+1,when (x+1) > 0) Or, 2x1< x+1 Or, x< 2(1) Given, x + 1  > 2x  1 Or, (x+1) > 2x1 (x+1=(x+1), when (x+1) < 0) Or, 2x1< (x+1) Or, 2x1< x1 Or, 3x < 0 Or, x < 0(2) From (1) & (2), x < 2 Ans. (B) Would you please help me understand why aren't we taking the overlap here and choose x<0 ?
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Re: if x + 1  > 2x  1, which of the following represents the correct
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18 Aug 2018, 20:35
TaN1213 wrote: if x + 1  > 2x  1, which of the following represents the correct range of values of x ? A. x < 0 B. x < 2 C. 2 < x < 0 D. 1 < x < 2 E. 0 < x < 2 Source: ExpertsGlobalchetan2u, Shouldn't we consider the overlapping region of (x<0) and (x<2), and the range would then be x < 0 ? Is the OA given wrong ? Hi... As correctly done , two critical points.. 1) x\(\leq{1}\) (x+1)>2x1........x+1<12x......X<0 So all values below and equal to 1 satisfy the equation 2) x>1 x+1>2x1.......x<2 So all values between 1 and 2 satisfy the equation.. Combined.. All values upto 1 inclusive and all values between 1 and 2 So it turns out all values till 2 And that is why both combined give us x<2
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Re: if x + 1  > 2x  1, which of the following represents the correct
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18 Aug 2018, 21:11
chetan2u wrote: TaN1213 wrote: if x + 1  > 2x  1, which of the following represents the correct range of values of x ? A. x < 0 B. x < 2 C. 2 < x < 0 D. 1 < x < 2 E. 0 < x < 2 Source: ExpertsGlobalchetan2u, Shouldn't we consider the overlapping region of (x<0) and (x<2), and the range would then be x < 0 ? Is the OA given wrong ? Hi... As correctly done , two critical points.. 1) x\(\leq{1}\) (x+1)>2x1........x+1<12x......X<0 So all values below and equal to 1 satisfy the equation 2) x>1 x+1>2x1.......x<2 So all values between 1 and 2 satisfy the equation.. Combined.. All values upto 1 inclusive and all values between 1 and 2 So it turns out all values till 2 And that is why both combined give us x<2 Hi chetan2uWhats wrong with this approach.Please explain.
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Re: if x + 1  > 2x  1, which of the following represents the correct
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18 Aug 2018, 21:31
gmat1393 wrote: chetan2u wrote: TaN1213 wrote: if x + 1  > 2x  1, which of the following represents the correct range of values of x ? A. x < 0 B. x < 2 C. 2 < x < 0 D. 1 < x < 2 E. 0 < x < 2 Source: ExpertsGlobalchetan2u, Shouldn't we consider the overlapping region of (x<0) and (x<2), and the range would then be x < 0 ? Is the OA given wrong ? Hi... As correctly done , two critical points.. 1) x\(\leq{1}\) (x+1)>2x1........x+1<12x......X<0 So all values below and equal to 1 satisfy the equation 2) x>1 x+1>2x1.......x<2 So all values between 1 and 2 satisfy the equation.. Combined.. All values upto 1 inclusive and all values between 1 and 2 So it turns out all values till 2 And that is why both combined give us x<2 Hi chetan2uWhats wrong with this approach.Please explain. Hi.. You square both sides ONLY when you are sure both sides are positive... That is the reason you are getting your values when X is positive. But what about x as negative, 2x1 will be negative too at those values. So REMEMBER, do not square when both sides are not 100% to be positive
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1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
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Re: if x + 1  > 2x  1, which of the following represents the correct
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18 Aug 2018, 21:34
Hi chetan2u.
I consistently get these mod questions and inequalities questions wrong. Could you please guide me as to how to improve and practice(i dont want to practice QUESTION PACK questions for these topics)
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Re: if x + 1  > 2x  1, which of the following represents the correct
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18 Aug 2018, 22:02
chetan2u wrote: TaN1213 wrote: if x + 1  > 2x  1, which of the following represents the correct range of values of x ? A. x < 0 B. x < 2 C. 2 < x < 0 D. 1 < x < 2 E. 0 < x < 2 Source: ExpertsGlobalchetan2u, Shouldn't we consider the overlapping region of (x<0) and (x<2), and the range would then be x < 0 ? Is the OA given wrong ? Hi... As correctly done , two critical points.. 1) x\(\leq{1}\) (x+1)>2x1........x+1<12x......X<0 So all values below and equal to 1 satisfy the equation 2) x>1 x+1>2x1.......x<2 So all values between 1 and 2 satisfy the equation.. Combined.. All values upto 1 inclusive and all values between 1 and 2 So it turns out all values till 2 And that is why both combined give us x<2 Thank you so very much for the crisp explanation.
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