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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x
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05 Feb 2017, 05:34
AmritaSarkar89 wrote: Bunuel wrote: boomtangboy wrote: If x≠0, is x <1? (1) x2<1 (2) x < 1/x If \(x\neq{0}\), is \(x <1\)?Is \(x <1\)? > is \(1<x<1\) (\(x\neq{0}\))? (1) \(x^2<1\) > \(1<x<1\). Sufficient. (2) \(x < \frac{1}{x}\) > since LHS (x) is an absolute value which is always nonnegative then RHS (1/x), must be positive (as \(x < \frac{1}{x}\)), so \(\frac{1}{x}>0\) > \(x>0\). Now, if \(x>0\) then \(x=x\) and we have: \(x<\frac{1}{x}\) > since \(x>0\) then we can safely multiply both parts by it: \(x^2<1\) > \(1<x<1\), but as \(x>0\), the final range is \(0<x<1\). Sufficient. Answer D. Can we solve this by plugging in numbers? I did it that way and got the correct answer. Wondering if that was just by chance or was I correct in my approach Yes you can but you have to then validate with every kind of number i.e positive , negative , decimals , range between 0 to 1 etc.
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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x
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27 Aug 2017, 08:25
Bunuel mikemcgarry IanStewart shashankism Engr2012 Quote: Also if \(x<0\) then we have \(x<\frac{1}{x}\) and now if we cross multiply by negative \(x\) then we should flip the sign: \(x^2>1\) > \(x^2+1<0\) which cannot be true for any real value of \(x\) (the sum of two positive value cannot be less than zero). why am I getting \(x^2\) > 1 for this step? I multiplied both sides of \(x<1/x\) with x
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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x
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01 Sep 2017, 04:20
adkikani wrote: Bunuel mikemcgarry IanStewart shashankism Engr2012 Quote: Also if \(x<0\) then we have \(x<\frac{1}{x}\) and now if we cross multiply by negative \(x\) then we should flip the sign: \(x^2>1\) > \(x^2+1<0\) which cannot be true for any real value of \(x\) (the sum of two positive value cannot be less than zero). why am I getting \(x^2\) > 1 for this step? I multiplied both sides of \(x<1/x\) with x You are making an algebraic mistake. If you do assume that x<0, x = x; multiplying both sides of the inequality x<1/x by x you should get, x*(x) < (1/x)*(x) > x^2 < 1 and this is not possible for any real 'x'. Hence, x<0 can not be true. Additionally, you dont need to flip the signs when you multiply by x as we are assuming that x<0, making x >0 and when you multiply an inequality by a positive quantity, you do not need to flip the signs. Hope this helps.



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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x
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23 Nov 2018, 08:26
Project DS Butler: Day 18: Data Sufficiency (DS35) For DS butler Questions Click HereIf \(x\neq{0}\), is \(x\)< 1? (1) \(x^2\) < 1 (2) \(x\) < \(1/x\)
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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x
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23 Nov 2018, 18:33
adkikani wrote: Project DS Butler: Day 18: Data Sufficiency (DS35) For DS butler Questions Click HereIf \(x\neq{0}\), is \(x\)< 1? (1) \(x^2\) < 1 (2) \(x\) < \(1/x\) If \(x\neq{0}\), is \(x\)< 1? This means 1<x<1 (1) \(x^2\) < 1 \(x^2<1\) means x<1, since only fractions with absolute value less than 1 will have their square less than 1. Sufficient (2) \(x\) < \(1/x\) Since x >0, we can write 1/xx>1 Since x and 1 are positive, x also must be positive for above to be true. And since X is also positive, X=X, and X<1/X can be written as x^2<1 Therefore X is between 0 and 1 Sufficient D
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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x
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23 Nov 2018, 19:03
Hi chetan2uI am not able to understand 2. (2) xx < 1/x1/x Since x >0, we can write 1/xx>1 Since x and 1 are positive, x also must be positive for above to be true. And since X is also positive, X=X, and X<1/X can be written as x^2<1 Therefore X is between 0 and 1 Sufficient Where is mod X given as greater than 0 Posted from my mobile device



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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x
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24 Nov 2018, 00:40
adkikani wrote: Project DS Butler: Day 18: Data Sufficiency (DS35) For DS butler Questions Click HereIf \(x\neq{0}\), is \(x\)< 1? (1) \(x^2\) < 1 (2) \(x\) < \(1/x\) From 1 : we know that x^2=IxI so x^2<1 IxI<1 sufficient From 2: IxI<1/x this can only be possible if we have values of +ve x between 0 to 1 so yes sufficient IMO D
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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x
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24 Nov 2018, 01:16
adkikani wrote: Project DS Butler: Day 18: Data Sufficiency (DS35) For DS butler Questions Click HereIf \(x\neq{0}\), is \(x\)< 1? (1) \(x^2\) < 1 (2) \(x\) < \(1/x\) Statement One: Any number squared is less than positive 1. \(0.9^2\) or \(0.5^2\) it will be still less than one. SUFFICIENT Statement Two: Let X be 0.2 i.e. \(\frac{1}{5}\) than \(\frac{1}{0.2}\) = 5 SUFFICIENT have a great weekend



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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x
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24 Nov 2018, 02:01
saurabh9guptaQuote: Where is mod X given as greater than 0 What chetan2u meant was that since we are given that x is not equal to 0, and modulus of any number can either be positive or 0 (not allowed in this problem) the only sign that x can take is positive. Note: 0 is neither positive nor negative. Does that help, my friend?
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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x
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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x
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24 Nov 2018, 17:09
boomtangboy wrote: If x≠0, is x < 1 ?
(1) x^2 < 1 (2) x < 1/x IxI=x^2 From 1: to be valid value of x has to be <1 i.e can be any value 1>x <1 sufficient from 2: IxI*x<1 this would stand valid only when 1<x<1 i.e x is a fraction value IMO clearly D
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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x
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25 Nov 2018, 15:06
the slow but robust way of doing this is solving it for x>0 and x<0.
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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x
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