Is x > 1? : GMAT Data Sufficiency (DS)
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# Is x > 1?

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18 Jun 2012, 14:46
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Is x > 1?

(1) (x+1) (|x| - 1) > 0

(2) |x| < 5
[Reveal] Spoiler: OA

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18 Jun 2012, 15:15
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Good question. +1.

Is x> 1?

(1) (x+1) (|x| - 1) > 0. Consider two cases:

If $$x>0$$ then $$|x|=x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (x - 1) > 0$$ --> $$x^2-1>0$$ --> $$x^2>1$$ --> $$x<-1$$ or $$x>1$$. Since we consider range when $$x>0$$ then we have $$x>1$$ for this case;

If $$x\leq{0}$$ then $$|x|=-x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (-x - 1) > 0$$ --> $$-(x+1) (x+1) > 0$$ --> $$-(x+1)^2>0$$ --> $$(x+1)^2<0$$. Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that $$(x+1) (|x| - 1) > 0$$ holds true only when $$x>1$$. Sufficient.

(2) |x| < 5 --> $$-5<x<5$$. Not sufficient.

Hope it's clear.
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18 Jun 2012, 15:26
Thanks Bunuel. Can you please explain how did you get this?

If $$x>0$$ then $$|x|=x$$
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18 Jun 2012, 15:30
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enigma123 wrote:
Thanks Bunuel. Can you please explain how did you get this?

If $$x>0$$ then $$|x|=x$$

Check this: math-absolute-value-modulus-86462.html

Absolute value properties:

When $$x\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \ expression\leq{0}$$ then $$|some \ expression|={-(some \ expression)}$$. For example: $$|-5|=5=-(-5)$$.

When $$x\geq{0}$$ then $$|x|=x$$, or more generally when $$some \ expression\geq{0}$$ then $$|some \ expression|={some \ expression}$$. For example: $$|5|=5$$.

Hope it helps.
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19 Jun 2012, 11:22
Bunuel wrote:
Good question. +1.

Is x> 1?

(1) (x+1) (|x| - 1) > 0. Consider two cases:

If $$x>0$$ then $$|x|=x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (x - 1) > 0$$ --> $$x^2-1>0$$ --> $$x^2>1$$ --> $$x<-1$$ or $$x>1$$. Since we consider range when $$x>0$$ then we have $$x>1$$ for this case;

If $$x\leq{0}$$ then $$|x|=-x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (-x - 1) > 0$$ --> $$-(x+1) (x+1) > 0$$ --> $$-(x+1)^2>0$$ --> $$(x+1)^2<0$$. Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that $$(x+1) (|x| - 1) > 0$$ holds true only when $$x>1$$. Sufficient.

(2) |x| < 5 --> $$-5<x<5$$. Not sufficient.

Hope it's clear.

Dear Bunuel,
i got (x+1)^2<0 .
and further solved to x+1<0 giving x <-1
and ended up in wrong answer E.
square of a number cannot be negative. but just confused for (x+1)^2 where x is a variable?
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20 Jun 2012, 00:07
kashishh wrote:
Bunuel wrote:
Good question. +1.

Is x> 1?

(1) (x+1) (|x| - 1) > 0. Consider two cases:

If $$x>0$$ then $$|x|=x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (x - 1) > 0$$ --> $$x^2-1>0$$ --> $$x^2>1$$ --> $$x<-1$$ or $$x>1$$. Since we consider range when $$x>0$$ then we have $$x>1$$ for this case;

If $$x\leq{0}$$ then $$|x|=-x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (-x - 1) > 0$$ --> $$-(x+1) (x+1) > 0$$ --> $$-(x+1)^2>0$$ --> $$(x+1)^2<0$$. Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that $$(x+1) (|x| - 1) > 0$$ holds true only when $$x>1$$. Sufficient.

(2) |x| < 5 --> $$-5<x<5$$. Not sufficient.

Hope it's clear.

Dear Bunuel,
i got (x+1)^2<0 .
and further solved to x+1<0 giving x <-1
and ended up in wrong answer E.
square of a number cannot be negative. but just confused for (x+1)^2 where x is a variable?

It doesn't matter that $$x$$ is a variable, it's still some number and so is $$x+1$$. So, $$(x+1)^2$$ is a square of that number and it cannot be negative.

Also your way of solving is not correct: $$(x+1)^2<0$$ does not mean $$x+1<0$$ it means that $$|x+1|<0$$. From that you could deduce the same: since absolute value cannot be negative then this equation has no solution.

Hope it's clear.
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21 Jun 2012, 00:04
Hi Bunuel, I understood your solution but can you please tell me where am I going wrong?

When I see the expression, (x+1)(|x|-1)>0, I immediately think that these two brackets must be either positive or negative.

Hence, if that take both of them are positive, then x>-1 and and x>1 & x<-1

For these two ranges we realize when we start plugging in number that only for x>1 the equation hold true.--------(1)

Similarly, if we take both of them to be negative, then x<-1 and x<1 & x>-1

for x<-1 for example, -2, one of the brackets turn out to be positive, which does not satisfy our initial assumption that both must be negative.

for x<1 ex 0, one bracket is +ve and one is -ve <Not Desired>; for 0, the inequality is not > 0 <not desired>

if we go still less it follows the first case

for x>-1 pretty much follows like the above. Hence nothing desired.--------(2)

Hence, x>1

A is sufficient.
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21 Jun 2012, 01:41
pavanpuneet wrote:
Hi Bunuel, I understood your solution but can you please tell me where am I going wrong?

When I see the expression, (x+1)(|x|-1)>0, I immediately think that these two brackets must be either positive or negative.

Hence, if that take both of them are positive, then x>-1 and and x>1 & x<-1

For these two ranges we realize when we start plugging in number that only for x>1 the equation hold true.--------(1)

Similarly, if we take both of them to be negative, then x<-1 and x<1 & x>-1

for x<-1 for example, -2, one of the brackets turn out to be positive, which does not satisfy our initial assumption that both must be negative.

for x<1 ex 0, one bracket is +ve and one is -ve <Not Desired>; for 0, the inequality is not > 0 <not desired>

if we go still less it follows the first case

for x>-1 pretty much follows like the above. Hence nothing desired.--------(2)

Hence, x>1

A is sufficient.

First of all I wouldn't recommend to solve this question the way you are doing.

Next, when you consider both multiples to be negative and get $$x<-1$$ from the first one, then the second multiple automatically transformes to $$(-x-1)$$, since if $$x<-1<0$$ then $$|x|=-x$$. So, we have that $$-x-1<0$$ must also be true or $$x>-1$$, which contradicts the case for the first multiple ($$x<-1$$). So, both $$x+1$$ and $$|x|-1$$ can not be negative.

Also I think you got x<1 & x>-1 from |x|<1, and if yes, then it's not correct: $$|x|<1$$ means that $$-1<x<1$$. So, again $$x<-1$$ (for the first multiple to be negative) and $$-1<x<1$$ (for the second multiple to be negative) cannot simultaneously be true.

Hope it's clear.
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27 Jan 2013, 22:17
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fozzzy wrote:
Is x>1

1) (x+1)(lxl-1) > 0
2) lxl < 5

Statement 1:
For (x+1)(lxl-1) > 0, we should have either (x+1)>0 and (lxl-1) > 0 or (x+1)<0 and (lxl-1) < 0
when (x+1)>0 and (lxl-1) > 0
(x+1)>0 => x>-1
(lxl-1) > 0 => x>1 or x <-1
From above two, possible solution is x>1
when (x+1)>0 and (lxl-1) < 0
(x+1)<0 => x<-1
(lxl-1) < 0 => -1<x<1
Both of these can not be satisfied by any value of x.
Hence we get only 1 solution, x>1. which is what we wanted to ascertain. Sufficient.

Statement 2:
|x| <5
=> -5<x<5
Clearly not sufficient to tell whether x>1 or not.

Ans A it is.
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27 Jan 2013, 23:21
so you just have to find the common region if its done on a number line and we ignore one of the cases, since there isn't a common region?
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Last edited by fozzzy on 27 Jan 2013, 23:25, edited 1 time in total.
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06 Feb 2013, 02:21
andrew40 wrote:
Bunuel wrote:
Good question. +1.

Is x> 1?

(1) (x+1) (|x| - 1) > 0. Consider two cases:

If $$x>0$$ then $$|x|=x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (x - 1) > 0$$ --> $$x^2-1>0$$ --> $$x^2>1$$ --> $$x<-1$$ or $$x>1$$. Since we consider range when $$x>0$$ then we have $$x>1$$ for this case;

If $$x\leq{0}$$ then $$|x|=-x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (-x - 1) > 0$$ --> $$-(x+1) (x+1) > 0$$ --> $$-(x+1)^2>0$$ --> $$(x+1)^2<0$$. Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that $$(x+1) (|x| - 1) > 0$$ holds true only when $$x>1$$. Sufficient.

(2) |x| < 5 --> $$-5<x<5$$. Not sufficient.

Hope it's clear.

Sorry, I don't understand why we should consider the range where x>0. because of the absolute value?

We need to get rid of the modulus in the expression to solve it and this is the way to do that. Check here: is-x-134652.html#p1097668

Hope it helps.
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18 May 2013, 04:56
Bunuel wrote:
Good question. +1.

Is x> 1?

(1) (x+1) (|x| - 1) > 0. Consider two cases:

If $$x>0$$ then $$|x|=x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (x - 1) > 0$$ --> $$x^2-1>0$$ --> $$x^2>1$$ --> $$x<-1$$ or $$x>1$$. Since we consider range when $$x>0$$ then we have $$x>1$$ for this case;

If $$x\leq{0}$$ then $$|x|=-x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (-x - 1) > 0$$ --> $$-(x+1) (x+1) > 0$$ --> $$-(x+1)^2>0$$ --> $$(x+1)^2<0$$. Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that $$(x+1) (|x| - 1) > 0$$ holds true only when $$x>1$$. Sufficient.

(2) |x| < 5 --> $$-5<x<5$$. Not sufficient.

Hope it's clear.

Hi,

This is my first post so was little conscious to ask my doubt. In the above question, we took the roots as 1<x>1. However, in this question as mentioned below (unable to post the link as per new member rule)

is ((X-3)^2)^1/2 = 3-X ?

1) X does not = 3

2) -X|X| > 0

the roots are 3<x>3. Can you please explain the difference?
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Re: Is x > 1? [#permalink]

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22 May 2013, 15:45

When I try this question for statement 1:
(x+1) (|x| - 1) > 0
(x+1)|x|-(x+1)>0
(x+1)|x| >(x+1)
|x| > 1

x>1 ; x<-1
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Re: Is x > 1? [#permalink]

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23 May 2013, 01:14
smartyman wrote:

When I try this question for statement 1:
(x+1) (|x| - 1) > 0
(x+1)|x|-(x+1)>0
(x+1)|x| >(x+1)
|x| > 1

x>1 ; x<-1

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know its sign.

So you cannot reduce both parts of inequality (x+1)|x|>(x+1) by x+1 as you don't know the sign of x+1: if x+1>0 you should write |x|>1 BUT if x+1<0 you should write |x|<1 (flip the sign).

Hope it helps.
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Re: Is x > 1? [#permalink]

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02 Jun 2013, 09:59
Hi guys,

I just want to present an easier way to prove that I is sufficient.

For I to be positive what are the conditions?

1. x+1 > 0 this means x >-1
2.|x| -1 > 0 this means x > 1 or x < -1

then you must draw the number line ----------------[b]-1[b]-----0------1--------------> then draw this inequalities on this line and look for any unity. then you will fine x > 1

for the other side, I mean, x+1 <0 and |x| - 1 <0 you wont find any unity. so I is sufficient. Hope it clears. cheers
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Re: Is x > 1? [#permalink]

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23 Jun 2013, 16:59
So here is my question:

For #1 we have two cases: positive and negative

For x≥0
(x+1)(|x|-1)>0
(x+1)(x-1)>0
x^2-1>0
x^2>1
Of course, x could be 2 or negative two for all we know, so it seems like this is insufficient as we are testing for the range of x>1

For x≤0
(x+1)(|x|-1)>0
(x+1)(-x-1)>0
-x^2-1>0
-x^2>-1

I am a bit unsure where I went wrong here.
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Re: Is x > 1? [#permalink]

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24 Jun 2013, 01:11
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WholeLottaLove wrote:
So here is my question:

For #1 we have two cases: positive and negative

For x≥0
(x+1)(|x|-1)>0
(x+1)(x-1)>0
x^2-1>0
x^2>1

Of course, x could be 2 or negative two for all we know, so it seems like this is insufficient as we are testing for the range of x>1

x can never be -2, as because you have assumed that $$x\geq0$$. Also, from the inequality you have correctly arrived at, i.e. $$x^2>1$$ $$\to$$ x>1 OR x<-1. As assumption was $$x\geq0$$. thus only x>1 condition is valid. Also,as x>1 automatically makes $$x\geq0$$, thus the correct range is x>1. Sufficient.
Quote:
For x≤0 No need to include equality with zero twice.
(x+1)(|x|-1)>0
(x+1)(-x-1)>0 This leads to $$-(x+1)^2>0$$ and this is not possible for any real value of x. So,there is no solution for this.
-x^2-1>0
-x^2>-1

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28 Aug 2013, 12:08
I mean to say, we define |X| as +x when x is greater than equal to zero, and -x when x is less than zero. But I have noticed you at times have taken x as -x based on x less than equal to zero ? I mean equality cant be on both sides right ? Infact I saw you explained definition of mod function keeping equality on both +x and - x, I hope you are able to make out what doubt I have.
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29 Aug 2013, 03:34
ygdrasil24 wrote:
I mean to say, we define |X| as +x when x is greater than equal to zero, and -x when x is less than zero. But I have noticed you at times have taken x as -x based on x less than equal to zero ? I mean equality cant be on both sides right ? Infact I saw you explained definition of mod function keeping equality on both +x and - x, I hope you are able to make out what doubt I have.

The point is that |0|=-0=0. So, in that definition we can include = sign in both cases.

Hope it helps.
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Re: Is x > 1? [#permalink]

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04 Apr 2014, 21:15
enigma123 wrote:
Is x > 1?

(1) (x+1) (|x| - 1) > 0

(2) |x| < 5

Sol: (x+1) (|x| - 1) > 0

The RHS can only be positive if both are positive or both are negative
lets take -5 as value of x===> -ve *+ve ===>not possible
let's take 0 as value of x ==> 1* (-1) ===>not possible
let's take 1 as value (+ve) ===> 2* (0)===> not possible
let's take -1 as value (-ve) ====>0 *0 ====>not possible
so x != -ve, x != 0, x != 1 conditions apply. only remaining option is x>1 ....sufficient

2> ----(-5)@@@@@@(5)---------
-5<x<5 ...not sufficient since x is less than 1 here

final sol is A
Re: Is x > 1?   [#permalink] 04 Apr 2014, 21:15

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