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Absolute value

by Uri » Mon Apr 20, 2009 9:11 am
If x&#8800;0, is |x| <1?
(1) x^2 < 1
(2) |x| < (1/x)

Please explain your logic.

OA: [spoiler](D)[/spoiler]
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I don't understand how [spoiler](2)[/spoiler] can be sufficient. Please help me.
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Re: Absolute value

by iamcste » Mon Apr 20, 2009 9:49 am
Uri wrote:If x&#8800;0, is |x| <1?
(1) x^2 < 1
(2) |x| < (1/x)

Please explain your logic.

OA: [spoiler](D)[/spoiler]
Source: GMAT Plus Book
I don't understand how [spoiler](2)[/spoiler] can be sufficient. Please help me.
Second statment means X has to be between 0 and 1

Qtn Is X between -1 and 1?

Yes, any value of X that suffices qtn 2 is between that range and no other value outside that range suffices qtn 2

In short as I dont have any value of x outside range -1 to 1 that sufficies qtn 2, Statement is sufficient

Did you think that x is not between 0 and -1 hence its not sufficient?

Stmt 1 is sufficient

[spoiler]
Choose D[/spoiler]
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by Uri » Tue Apr 21, 2009 1:21 am
i made a very silly mistake....and i am very much prone to making this mistake.
for (2), i simplified as
x|x|<1
but this will be true only when x is positive

you have done it in back-solving method. although this makes sense, i would like to know whether there is any other way to solve this problem
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by iamcste » Tue Apr 21, 2009 2:12 am
Uri wrote:i made a very silly mistake....and i am very much prone to making this mistake.
for (2), i simplified as
x|x|<1
but this will be true only when x is positive

you have done it in back-solving method. although this makes sense, i would like to know whether there is any other way to solve this problem

try to use test data, -2,-1,-1/2,0,1/2. 1 and 2

see the relationship only for for 1/2, you will see that relationship is true...
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by austin » Tue Apr 21, 2009 3:50 am
If x&#8800;0, is |x| <1?
(1) x^2 < 1
(2) |x| < (1/x)

statement 1: x^2 < 1=> -1<x<1 => |x| < 1 sufficient
statement 2: |x| < (1/x).. for x > 1 and x <-1, this does NOT hold true.. it holds true only when -1<x<1 => |x|<1... sufficient

D

NOTE: DO NOT CROSS-MULTIPLY VARIABLES ...we can cross multiply only when we know the sign of that quantity!!!!
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by iamcste » Tue Apr 21, 2009 4:17 am
austin wrote: statement 2: |x| < (1/x).. for x > 1 and x <-1, this does NOT hold true.. it holds true only when -1<x<1 => |x|<1... sufficient

D

are you sure, that this holds true for value of x between 0 and -1?

I dont think so.
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by austin » Tue Apr 21, 2009 4:29 am
x = 1/2

|x| = 1/2; 1/x = 1/(1/2) = 2

=> |x| < (1/x); satisfies the given condition
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by iamcste » Tue Apr 21, 2009 4:48 am
austin wrote:x = 1/2

|x| = 1/2; 1/x = 1/(1/2) = 2

=> |x| < (1/x); satisfies the given condition

Dude, I am talking about a value between 0 and -1

1/2 is not between 0 and -1

Okie, what I ideally meant is IMO stmt 2 means that x has to be between 0 and 1 and not -1 to 1 as you mentioned

so pls recheck this once
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by austin » Tue Apr 21, 2009 7:21 am
iamcste,

You are right...
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by cubicle_bound_misfit » Tue Apr 21, 2009 7:31 am
FOR STMT 2.

TAKE
X = -1/2

THEN |X| > 1/X

ALSO FOR x> 1 |x| > 1/x

relation is only true

for 0<x<1 for all the values in this range |x| <1 hence Suff
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by maihuna » Tue Apr 21, 2009 10:03 am
1. x^2 < 1 for -1<x<1 => |x| <1 Suff

2. |x| < 1/x
trick here is we dont know sign of x but we do know the sign of |x|

so 1/|x| > x
or 1 > x|x|
so 1 > -x^2 or x^2> -1 or x^2+1>0 true as x^2 will be always positive
and 1>x^2 or x^2<1 that means |x| <1
so sufficient

ans: D
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by schumi_gmat » Tue Apr 21, 2009 10:38 am
@ Cubicle
for 0<x<1 for all the values in this range |x| <1 hence Suff
We have to prove lxl < 1 i.e -1<x<1

I agree that the equation is true for 0<x<1.
But how is it possible that it is sufficient to prove -1<x<1. Isnt it partially true?

Can you explain what I am doing wrong here?
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by ketkoag » Tue Apr 21, 2009 11:42 am
statement 2 doesn't hold true when x is between -1 and 0.
IMO : D
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by iamcste » Thu Apr 23, 2009 10:59 am
maihuna wrote:1. x^2 < 1 for -1<x<1 => |x| <1 Suff

2. |x| < 1/x
trick here is we dont know sign of x but we do know the sign of |x|

so 1/|x| > x
or 1 > x|x|
so 1 > -x^2 or x^2> -1 or x^2+1>0 true as x^2 will be always positive
and 1>x^2 or x^2<1 that means |x| <1
so sufficient

ans: D
Maihuna, I feel this is incorrect. you also move x while moving mod x

Also, Stmt 2 is satisied only between 0 and 1 and NOT between -1 and 1, as you conclude.

Pls let me know if I get you incorrect.
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by Stuart@KaplanGMAT » Thu Apr 23, 2009 2:41 pm
schumi_gmat wrote:@ Cubicle
for 0<x<1 for all the values in this range |x| <1 hence Suff
We have to prove lxl < 1 i.e -1<x<1

I agree that the equation is true for 0<x<1.
But how is it possible that it is sufficient to prove -1<x<1. Isnt it partially true?

Can you explain what I am doing wrong here?
You've misinterpreted the question.

You've read the question as "could x be every number that satisfies |x| < 1?"

However, the question is actually "is |x| < 1?"

In order for a statement to be sufficient, it has to give us a definite "yes" or definite "no" answer to the question.

From (2) we can derive 0<x<1. In other words, x must be a positive fraction.

Is the absolute value of every positive fraction less than 1? Definitely YES! Therefore, (2) is sufficient.

Let's look at a much simpler question to illustrate the conceptual error that you made:

Q: Is x greater than 0?

(1) x = 5

For this question, you probably wouldn't even think twice. If x=5, then it's definitely greater than 0... sufficient.

However, if we follow the line of thinking you employed for the original question, we'd say:

"5 doesn't satisfy every possible value that's greater than 0, therefore (1) isn't sufficient."
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