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If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x

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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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Last edited by Bunuel on 08 Apr 2012, 10:02, edited 1 time in total.
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Re: If x≠0, is |x| <1? [#permalink]

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08 Apr 2012, 10:01
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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 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.

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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)
x2-1<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
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Expert's post
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)
x2-1<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)
x2-1<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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04 Jul 2013, 01:24
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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, 05:16
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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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04 Jul 2013, 12:25
1
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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 cross-multiply $$\to x^2<1 \to |x|<1.$$

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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07 Jul 2013, 23:18
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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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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)
x2-1<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<0

I 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)
x2-1<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<0

I 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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22 May 2014, 23:19
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Hi Mahmud,

|x| <1 can be written as -1<x<1 because 1 is a constant BUT

|x| < $$\frac{1}{x}$$ cannot be written as -1/x<x<1/x because 1/x is a variable

Solving,

|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.

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

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11 Sep 2016, 03:21
iliavko wrote:
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?

The question asks: whether x is between -1 and 1. (2) says that x is between 0 and 1, so the answer to the question is YES.
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Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x [#permalink]

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30 Dec 2016, 08:05
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.

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

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Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x   [#permalink] 30 Dec 2016, 08:05

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