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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  [#permalink]

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New post 05 Feb 2017, 05:34
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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 non-negative 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  [#permalink]

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New post 27 Aug 2017, 08:25
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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  [#permalink]

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New post 01 Sep 2017, 04:20
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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  [#permalink]

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New post 23 Nov 2018, 08:26
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Project DS Butler: Day 18: Data Sufficiency (DS35)


For DS butler Questions Click Here



If \(x\neq{0}\), 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  [#permalink]

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New post 23 Nov 2018, 18:33
adkikani wrote:

Project DS Butler: Day 18: Data Sufficiency (DS35)


For DS butler Questions Click Here



If \(x\neq{0}\), 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/|x|x>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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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html


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Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x  [#permalink]

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New post 23 Nov 2018, 19:03
Hi chetan2u

I am not able to understand 2.

(2) |x||x| < 1/x1/x
Since |x| >0, we can write 1/|x|x>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

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Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x  [#permalink]

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New post 24 Nov 2018, 00:40
adkikani wrote:

Project DS Butler: Day 18: Data Sufficiency (DS35)


For DS butler Questions Click Here



If \(x\neq{0}\), 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  [#permalink]

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New post 24 Nov 2018, 01:16
adkikani wrote:

Project DS Butler: Day 18: Data Sufficiency (DS35)


For DS butler Questions Click Here



If \(x\neq{0}\), 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 :grin:


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  [#permalink]

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New post 24 Nov 2018, 02:01
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saurabh9gupta


Quote:
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  [#permalink]

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Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x  [#permalink]

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New post 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  [#permalink]

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New post 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   [#permalink] 25 Nov 2018, 15:06

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