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Re M2430 [#permalink]
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16 Sep 2014, 01:22



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Re: M2430 [#permalink]
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02 Jul 2015, 10:34
Bunuel wrote: Official Solution:
If \(\frac{x}{x} \lt x\) which of the following must be true about \(x\)?
A. \(x \gt 1\) B. \(1 \lt x \lt 0\) or \(x \gt 1\) C. \(x \lt 1\) D. \(x = 1\) E. \(x^2 \gt 1\)
Let's see in which ranges \(\frac{x}{x} \lt x\) holds true. A. If \(x \lt 0\), then \(x=x\), so we'll have that \(\frac{x}{x} \lt x\) or \(1 \lt x\). Since we consider the range when \(x \lt 0\) then \(1 \lt x \lt 0\); B. If \(x \gt 0\), then \(x=x\), so we have that \(\frac{x}{x} \lt x\) or \(1 \lt x\). So, given inequality holds true when \(1 \lt x \lt 0\) or \(x \gt 1\).
Answer: B Hi, I am confused about this: A. If x<0, then x=−x .... I thought that x is always positive, why then x? Could you help me? Thanks a lot. Regards. Luis Navarro



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Re: M2430 [#permalink]
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02 Jul 2015, 10:38
luisnavarro wrote: Bunuel wrote: Official Solution:
If \(\frac{x}{x} \lt x\) which of the following must be true about \(x\)?
A. \(x \gt 1\) B. \(1 \lt x \lt 0\) or \(x \gt 1\) C. \(x \lt 1\) D. \(x = 1\) E. \(x^2 \gt 1\)
Let's see in which ranges \(\frac{x}{x} \lt x\) holds true. A. If \(x \lt 0\), then \(x=x\), so we'll have that \(\frac{x}{x} \lt x\) or \(1 \lt x\). Since we consider the range when \(x \lt 0\) then \(1 \lt x \lt 0\); B. If \(x \gt 0\), then \(x=x\), so we have that \(\frac{x}{x} \lt x\) or \(1 \lt x\). So, given inequality holds true when \(1 \lt x \lt 0\) or \(x \gt 1\).
Answer: B Hi, I am confused about this: A. If x<0, then x=−x .... I thought that x is always positive, why then x? Could you help me? Thanks a lot. Regards. Luis Navarro If x is negative, then x = negative = positive, so x is still positive.
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Re: M2430 [#permalink]
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07 Mar 2016, 09:33
I did this question just by plugging number. It takes a little more time (less than 2 mins anyways), but for me the official solution presented is just too complicated.
A. Let's say x=3/2, than 3/2:3/2<3/2, 1<1.5. This might work. B. We know x>1 works, so we try 1/2:I1/2I<1/2, 1<1/2. Also works, so we eliminate A. C. IxI<1, x<1 or x>1, so we test 1/2, 1/2:1/2<1/2, 1<1/2. Incorrect. Eliminate D. x=+/1, 1/I1I<1, 1<1, Incorrect. Eliminate. E. x>1, x<1, we haven't test x<1, 3/2/I3/2I<3/2 1<3/2. Incorrect. Eliminate.
Finally, B is the answer.



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Re: M2430 [#permalink]
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06 Aug 2016, 07:57
why isn't A true? Although A does not cover entire set for x , its indeed true that x> 1.



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Re: M2430 [#permalink]
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06 Aug 2016, 11:06



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Re: M2430 [#permalink]
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16 Dec 2016, 21:32
Great question. Can you please explain the range of values that \(x^2 \gt 1\) will take? Will it be x>1 or x<1 ? Bunuel wrote: If \(\frac{x}{x} \lt x\) which of the following must be true about \(x\)?
A. \(x \gt 1\) B. \(1 \lt x \lt 0\) or \(x \gt 1\) C. \(x \lt 1\) D. \(x = 1\) E. \(x^2 \gt 1\)



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Re: M2430 [#permalink]
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17 Dec 2016, 03:50



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Hi Bunuel , Could you please explain why are we including zero in this range. Since we consider the range when x<0 then −1<x<0. Thanks Abhinash



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Re: M2430 [#permalink]
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08 Dec 2017, 11:12
abhinashgc wrote: Hi Bunuel , Could you please explain why are we including zero in this range. Since we consider the range when x<0 then −1<x<0. Thanks Abhinash Sorry I got it.Thanks



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Re: M2430 [#permalink]
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30 Jun 2018, 00:19
Hi Bunuel, Your solution is elegant. However, lets say I tried something like whats given below, then how do I get to the same ans? \(x / x < x\) x < x x x( x  1) > 0 case 1: x> 0 OR x 1 > 0 x > 0 OR x> 1 OR x < 1 case 2: x < 0 OR x  1 < 0 x< 0 OR 1 < x < 1 Thanks.



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Re: M2430 [#permalink]
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30 Jun 2018, 00:35
sanjay1810 wrote: Hi Bunuel, Your solution is elegant. However, lets say I tried something like whats given below, then how do I get to the same ans? \(x / x < x\) x < x x x( x  1) > 0 case 1: x> 0 OR x 1 > 0 x > 0 OR x> 1 OR x < 1 case 2: x < 0 OR x  1 < 0 x< 0 OR 1 < x < 1 Thanks. x(x  1) > 0 Case 1: x > 0 AND x  1 > 0. Notice that if x > 0, then x = x, so we'd get x > 0 AND x  1 > 0; x > 0 AND x > 1 So, in this case we have x > 1. Case 2: x < 0 OR x  1 < 0. Notice that if x < 0, then x = x, so we'd get x < 0 AND x  1 < 0; x < 0 AND 1 < x; So, in this case we have 1 < x < 0. As you can see we got the same range as in the solution above.
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New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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