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If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x [#permalink]
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08 Apr 2012, 09:43
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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? [#permalink]
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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.
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Re: If x≠0, is x <1? [#permalink]
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19 Apr 2012, 13:25
My answer was A... Explanation> x<1 means x should be between 1 and 1. from (1) x21<0 => x between 1 and 1. sufficient from (2) if x +ve then same as 1 and sufficient if x ve then x=x x2<1 or x2 > 1 not sufficient so and A Please correct me if wrong
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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x [#permalink]
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19 Apr 2012, 21:44
shikhar wrote: If x≠0, is x <1?
(1) x2<1 (2) x < 1/x Merging similar topics. shikhar wrote: My answer was A... Explanation> x<1 means x should be between 1 and 1.
from (1) x21<0 => x between 1 and 1. sufficient
from (2) if x +ve then same as 1 and sufficient if x ve then x=x x2<1 or x2 > 1 not sufficient
so and A
Please correct me if wrong \(x\) cannot be negative. Refer to the solution above. 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).
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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x [#permalink]
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23 Apr 2012, 20:41
Bunuel wrote: shikhar wrote: If x≠0, is x <1?
(1) x2<1 (2) x < 1/x Merging similar topics. shikhar wrote: My answer was A... Explanation> x<1 means x should be between 1 and 1.
from (1) x21<0 => x between 1 and 1. sufficient
from (2) if x +ve then same as 1 and sufficient if x ve then x=x x2<1 or x2 > 1 not sufficient
so and A
Please correct me if wrong \(x\) cannot be negative. Refer to the solution above. Also if \(x<0\) then we have \(x<\frac{1}{x}\) and now if we cross mulitply 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). Well Explained Bunuel



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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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boomtangboy wrote: If x≠0, is x < 1 ?
(1) x^2 < 1 (2) x < 1/x Given question stem asks if x<1> Is 1<x<1 from St 1 we have x^2<1 > 1<x<1 So Sufficient from St2 we have x<1/x Notice that x is a positive value and for any Integer value x> 1/x This implies X is a fraction. In order to satisfy the above equation let us take some fractional value of x and check what happens to the above equation x= 1/2 so we have 1/2<2 No x=3/4 so we have 3/4< 4/3 Yes x=4/3 so we have 4/3 <3/4  no We see that when fraction is between 0<x<1 then the above equation holds true and hence x is between 1 and 1 Ans should be D.... Bunuel's solution is superb. Saves time
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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x [#permalink]
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If x≠0, is x < 1 ?
is x<1 OR is x<1 x>1
Is 1<x<1?
(1) x^2 < 1 x^2<1 x<1
This tells us exactly what the stem looks for. SUFFICIENT
(2) x < 1/x is x < 1/x OR is x < 1/x is x > 1/x SO 1/x < x < 1/x 1 < x^2 < 1 SUFFICIENT
(I am going to use Bunuel's method only because I think it makes more sense and I believe mine is wrong anyways)
x<1/x if x<1/x then 1/x MUST be positive as it is greater than an absolute value. If 1/x is positive then x must also be positive and therefore x<1/x is actually equal to x<1/x. Because we know that x is positive we can multiply both sides by x to simplify.
(x)* x < 1/x *(x) x^2 < 1 x<1 x<1 or x>1 1<x<1
This tells us exactly what the stem is looking for SUFFICIENT (D)
(also, could someone explain to me how to solve by the method I originally chose in #2...the one where I get the positive and negative case for X)
Thanks!



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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x [#permalink]
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04 Jul 2013, 13:00
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WholeLottaLove wrote: If x≠0, is x < 1 ?
(also, could someone explain to me how to solve by the method I originally chose in #2...the one where I get the positive and negative case for X)
Thanks! I.x>0 \(x<\frac{1}{x}\) As x>0, we can safely crossmultiply \(\to x^2<1 \to x<1.\) II.x<0 \(x<\frac{1}{x}\) multiply both sides by\(x^2\), which is a positive quantity \(\to x^3<x\)\([x\neq{0}]\) or \(x(1+x^2)>0 \to\) \((1+x^2)\) and x have same sign and as\((1+x^2)\) is always positive, thus x>0. However, this goes against our assumption. Thus, x is not negative. Sufficient.
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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x [#permalink]
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07 Jul 2013, 18:54
If x≠0, is x < 1 ?
(1) x^2 < 1 x<1, x>1 1<x<1
Valid X is a number that is greater than 1 (i.e. 3/4, 1/2) and less than 1 (i.e. 1/2, 3/4) either way the absolute value of any of those numbers will be less than 1. SUFFICIENT
(2) x < 1/x Multiply both sides by x x^2<1 Same solution as above. SUFFICIENT
(D)
Is it valid to multiply the absolute value of x on one side (like in number two) to simplify as long as x is positive?



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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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08 Jul 2013, 16:39
It helps a lot. Thanks! Bunuel wrote: WholeLottaLove wrote: If x≠0, is x < 1 ?
(1) x^2 < 1 x<1, x>1 1<x<1
Valid X is a number that is greater than 1 (i.e. 3/4, 1/2) and less than 1 (i.e. 1/2, 3/4) either way the absolute value of any of those numbers will be less than 1. SUFFICIENT
(2) x < 1/x Multiply both sides by x x^2<1 Same solution as above. SUFFICIENT
(D)
Is it valid to multiply the absolute value of x on one side (like in number two) to simplify as long as x is positive? As, since from x < 1/x we concluded that x is positive, then yes we can do that: x < 1/x > x^2 < 1. Hope it 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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21 Jul 2013, 20:38
Bunuel wrote: shikhar wrote: If x≠0, is x <1?
(1) x2<1 (2) x < 1/x Merging similar topics. shikhar wrote: My answer was A... Explanation> x<1 means x should be between 1 and 1.
from (1) x21<0 => x between 1 and 1. sufficient
from (2) if x +ve then same as 1 and sufficient if x ve then x=x x2<1 or x2 > 1 not sufficient
so and A
Please correct me if wrong \(x\) cannot be negative. Refer to the solution above. Also if \(x<0\) then we have \(x<\frac{1}{x}\) and now if we cross mulitply 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). Bunuel, I am not clear with this part Also if x<0 then we have x<\frac{1}{x} and now if we cross mulitply by negative x then we should flip the sign: x^2>1 > x^2+1<0I am getting the below: x < 1/x > when we cross multiply by ve x, we get (x*x ) x2 >( flipping) 1 ( x/x) > X2+1 >0 ?? Pls help...
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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x [#permalink]
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21 Jul 2013, 21:44
Mountain14 wrote: Bunuel wrote: shikhar wrote: If x≠0, is x <1?
(1) x2<1 (2) x < 1/x Merging similar topics. shikhar wrote: My answer was A... Explanation> x<1 means x should be between 1 and 1.
from (1) x21<0 => x between 1 and 1. sufficient
from (2) if x +ve then same as 1 and sufficient if x ve then x=x x2<1 or x2 > 1 not sufficient
so and A
Please correct me if wrong \(x\) cannot be negative. Refer to the solution above. Also if \(x<0\) then we have \(x<\frac{1}{x}\) and now if we cross mulitply 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). Bunuel, I am not clear with this part Also if x<0 then we have x<\frac{1}{x} and now if we cross mulitply by negative x then we should flip the sign: x^2>1 > x^2+1<0I am getting the below: x < 1/x > when we cross multiply by ve x, we get (x*x ) x2 >( flipping) 1 ( x/x) > X2+1 >0 ?? Pls help... When I say "multiply by negative x", I mean multiply by x, which is negative, so simply by 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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22 May 2014, 20:33
If x <1 > 1<x<1, why not x < 1/x > 1/x<x<1/x? If so, how does it lead to x>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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Hi Mahmud, x <1 can be written as 1<x<1 because 1 is a constant BUTx < \(\frac{1}{x}\) cannot be written as 1/x<x<1/x because 1/x is a variableSolving, x <\(\frac{1}{x}\) RHS has to be greater than 0 (As LHS can only be +ve or 0) => \(\frac{1}{x}\) > 0 => x>0 (x cannot be ve or 0 , Because, if x is ve then \(\frac{1}{x}\) is ve if x = 0 then\(\frac{1}{x}\) is not defined)Rgds, Rajat
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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x [#permalink]
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18 Oct 2014, 13:22
X cannot be 0. 1. X^2 < 1 ==> X < 1 since X^2 cannot be negative value so both positive and negative values are possible for X. Sufficient. 2. if X=1 then 1<1 not possible X cannot be 0 so X > 0. X^2 < 1 same as st1. Sufficient. ANSWER: D boomtangboy wrote: 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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Re: If x≠0, is x < 1 ? (1) x^2< 1 (2) x < 1/x [#permalink]
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10 Sep 2016, 08:34
Hi Bunuel!
A question here.
How is St 2 sufficient? You say it yourself that 2) implies that 0<x<1 so that is just part of the range of the question stem ( is 1<x<1 ?) So statement two doesn't cover the negative area of the range. How can it be considered to be sufficient?
Do I make myself clear btw?



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