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# If x is not equal to 0, is |x| less than 1? (1) x/|x|< x

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Intern
Joined: 26 Feb 2017
Posts: 5
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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17 Mar 2017, 01:48
Hi,
I'm new to this and my head is spining... so many details in all the posts, can't even begin to read them!

I thought in S2, x<0.
So we know all the signs in S1 and therefore we can solve it like a simple inequality...and we find |x|<1. S1 &S2 together suff.

But this is too simple!? Am i wrong in my approach even if the result is right?... I m a newbie with no GMAT confidence...help
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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13 May 2017, 08:25
Bunuel wrote:
Hussain15 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.

$$x\neq{0}$$, is $$|x|<1$$? Which means is $$-1<x<1$$? ($$x\neq{0}$$)

(1) $$\frac{x}{|x|}< x$$
Two cases:
A. $$x<0$$ --> $$\frac{x}{-x}<x$$ --> $$-1<x$$. But remember that $$x<0$$, so $$-1<x<0$$

B. $$x>0$$ --> $$\frac{x}{x}<x$$ --> $$1<x$$.

Two ranges $$-1<x<0$$ or $$x>1$$. Which says that $$x$$ either in the first range or in the second. Not sufficient to answer whether $$-1<x<1$$. (For instance $$x$$ can be $$-0.5$$ or $$3$$)

Second approach: look at the fraction $$\frac{x}{|x|}$$ it can take only two values:
1 for $$x>0$$ --> so we would have: $$1<x$$;
Or -1 for $$x<0$$ --> so we would have: $$-1<x$$ and as we considering the range for which $$x<0$$ then completer range would be: $$-1<x<0$$.

The same two ranges: $$-1<x<0$$ or $$x>1$$.

(2) $$|x| > x$$. Well this basically tells that $$x$$ is negative, as if x were positive or zero then $$|x|$$ would be equal to $$x$$. Only one range: $$x<0$$, but still insufficient to say whether $$-1<x<1$$. (For instance $$x$$ can be $$-0.5$$ or $$-10$$)

Or two cases again:
$$x<0$$--> $$-x>x$$--> $$x<0$$.
$$x>0$$ --> $$x>x$$: never correct.

(1)+(2) Intersection of the ranges from (1) and (2) is the range $$-1<x<0$$ ($$x<0$$ (from 2) and $$-1<x<0$$ or $$x>1$$ (from 1), hence $$-1<x<0$$). Every $$x$$ from this range is definitely in the range $$-1<x<1$$. Sufficient.

Should the answer not be A? because in option A, we can simply take the mod to the other side - x<|x|*x - and the sign will remain the same since mod can only be positive; further on. because x multiplied by mod x is is larger than x, it tells us that x must be positive, because were it negative, x multiplied by mod x would have been smaller than x.

Would really appreciate some light on this basic concept. Thanks.
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Joined: 02 Sep 2009
Posts: 44289
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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13 May 2017, 08:31
OreoShake wrote:
Bunuel wrote:
Hussain15 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.

$$x\neq{0}$$, is $$|x|<1$$? Which means is $$-1<x<1$$? ($$x\neq{0}$$)

(1) $$\frac{x}{|x|}< x$$
Two cases:
A. $$x<0$$ --> $$\frac{x}{-x}<x$$ --> $$-1<x$$. But remember that $$x<0$$, so $$-1<x<0$$

B. $$x>0$$ --> $$\frac{x}{x}<x$$ --> $$1<x$$.

Two ranges $$-1<x<0$$ or $$x>1$$. Which says that $$x$$ either in the first range or in the second. Not sufficient to answer whether $$-1<x<1$$. (For instance $$x$$ can be $$-0.5$$ or $$3$$)

Second approach: look at the fraction $$\frac{x}{|x|}$$ it can take only two values:
1 for $$x>0$$ --> so we would have: $$1<x$$;
Or -1 for $$x<0$$ --> so we would have: $$-1<x$$ and as we considering the range for which $$x<0$$ then completer range would be: $$-1<x<0$$.

The same two ranges: $$-1<x<0$$ or $$x>1$$.

(2) $$|x| > x$$. Well this basically tells that $$x$$ is negative, as if x were positive or zero then $$|x|$$ would be equal to $$x$$. Only one range: $$x<0$$, but still insufficient to say whether $$-1<x<1$$. (For instance $$x$$ can be $$-0.5$$ or $$-10$$)

Or two cases again:
$$x<0$$--> $$-x>x$$--> $$x<0$$.
$$x>0$$ --> $$x>x$$: never correct.

(1)+(2) Intersection of the ranges from (1) and (2) is the range $$-1<x<0$$ ($$x<0$$ (from 2) and $$-1<x<0$$ or $$x>1$$ (from 1), hence $$-1<x<0$$). Every $$x$$ from this range is definitely in the range $$-1<x<1$$. Sufficient.

Should the answer not be A? because in option A, we can simply take the mod to the other side - x<|x|*x - and the sign will remain the same since mod can only be positive; further on. because x multiplied by mod x is is larger than x, it tells us that x must be positive, because were it negative, x multiplied by mod x would have been smaller than x.

Would really appreciate some light on this basic concept. Thanks.

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Joined: 18 Mar 2017
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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15 Oct 2017, 14:02
For (1), wouldn't x<0 give you:

-x/x < -x
-1 < -x
1 > x?

When I plug in numbers for that, it doesn't work, but I don't see how we get x/-x < x when plugging in x<0?

Bunuel wrote:
Hussain15 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.

$$x\neq{0}$$, is $$|x|<1$$? Which means is $$-1<x<1$$? ($$x\neq{0}$$)

(1) $$\frac{x}{|x|}< x$$
Two cases:
A. $$x<0$$ --> $$\frac{x}{-x}<x$$ --> $$-1<x$$. But remember that $$x<0$$, so $$-1<x<0$$

B. $$x>0$$ --> $$\frac{x}{x}<x$$ --> $$1<x$$.

Two ranges $$-1<x<0$$ or $$x>1$$. Which says that $$x$$ either in the first range or in the second. Not sufficient to answer whether $$-1<x<1$$. (For instance $$x$$ can be $$-0.5$$ or $$3$$)

Second approach: look at the fraction $$\frac{x}{|x|}$$ it can take only two values:
1 for $$x>0$$ --> so we would have: $$1<x$$;
Or -1 for $$x<0$$ --> so we would have: $$-1<x$$ and as we considering the range for which $$x<0$$ then completer range would be: $$-1<x<0$$.

The same two ranges: $$-1<x<0$$ or $$x>1$$.

(2) $$|x| > x$$. Well this basically tells that $$x$$ is negative, as if x were positive or zero then $$|x|$$ would be equal to $$x$$. Only one range: $$x<0$$, but still insufficient to say whether $$-1<x<1$$. (For instance $$x$$ can be $$-0.5$$ or $$-10$$)

Or two cases again:
$$x<0$$--> $$-x>x$$--> $$x<0$$.
$$x>0$$ --> $$x>x$$: never correct.

(1)+(2) Intersection of the ranges from (1) and (2) is the range $$-1<x<0$$ ($$x<0$$ (from 2) and $$-1<x<0$$ or $$x>1$$ (from 1), hence $$-1<x<0$$). Every $$x$$ from this range is definitely in the range $$-1<x<1$$. Sufficient.

Math Expert
Joined: 02 Sep 2009
Posts: 44289
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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15 Oct 2017, 21:17
cgarmestani wrote:
For (1), wouldn't x<0 give you:

-x/x < -x
-1 < -x
1 > x?

When I plug in numbers for that, it doesn't work, but I don't see how we get x/-x < x when plugging in x<0?

Bunuel wrote:
Hussain15 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.

$$x\neq{0}$$, is $$|x|<1$$? Which means is $$-1<x<1$$? ($$x\neq{0}$$)

(1) $$\frac{x}{|x|}< x$$
Two cases:
A. $$x<0$$ --> $$\frac{x}{-x}<x$$ --> $$-1<x$$. But remember that $$x<0$$, so $$-1<x<0$$

B. $$x>0$$ --> $$\frac{x}{x}<x$$ --> $$1<x$$.

Two ranges $$-1<x<0$$ or $$x>1$$. Which says that $$x$$ either in the first range or in the second. Not sufficient to answer whether $$-1<x<1$$. (For instance $$x$$ can be $$-0.5$$ or $$3$$)

Second approach: look at the fraction $$\frac{x}{|x|}$$ it can take only two values:
1 for $$x>0$$ --> so we would have: $$1<x$$;
Or -1 for $$x<0$$ --> so we would have: $$-1<x$$ and as we considering the range for which $$x<0$$ then completer range would be: $$-1<x<0$$.

The same two ranges: $$-1<x<0$$ or $$x>1$$.

(2) $$|x| > x$$. Well this basically tells that $$x$$ is negative, as if x were positive or zero then $$|x|$$ would be equal to $$x$$. Only one range: $$x<0$$, but still insufficient to say whether $$-1<x<1$$. (For instance $$x$$ can be $$-0.5$$ or $$-10$$)

Or two cases again:
$$x<0$$--> $$-x>x$$--> $$x<0$$.
$$x>0$$ --> $$x>x$$: never correct.

(1)+(2) Intersection of the ranges from (1) and (2) is the range $$-1<x<0$$ ($$x<0$$ (from 2) and $$-1<x<0$$ or $$x>1$$ (from 1), hence $$-1<x<0$$). Every $$x$$ from this range is definitely in the range $$-1<x<1$$. Sufficient.

If x < 0, then |x| = -x.

Substitute |x| by -x in x/|x|< x to get x/(-x) < x and then to get -1 < x. Since we consider the range when x < 0, then -1 < x < 0.
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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29 Nov 2017, 07:09
Bunuel wrote:
udaymathapati wrote:
Hi Bunuel,
Thanks for detail explanation. I am finding it difficult only last intersection part. Can you explain it further. My doubut is...If we combine "-1<x<0" or "x>1" these two inequalities, how come range for x fall betweeen -1<x<1 since x>1 is area which will not fit into this equation. Can you explain?

Range from (1): -----(-1)----(0)----(1)---- $$-1<x<0$$ or $$x>1$$, green area;

Range from (2): -----(-1)----(0)----(1)---- $$x<0$$, blue area;

From (1) and (2): ----(-1)----(0)----(1)---- $$-1<x<0$$, common range of $$x$$ from (1) and (2) (intersection of ranges from (1) and (2)), red area.

Hope it's clear.

the red zone indicates that the range is -1<x<0 , how do we arrive at -1<x<1?
Math Expert
Joined: 02 Sep 2009
Posts: 44289
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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29 Nov 2017, 07:22
yousufkhan wrote:
Bunuel wrote:
udaymathapati wrote:
Hi Bunuel,
Thanks for detail explanation. I am finding it difficult only last intersection part. Can you explain it further. My doubut is...If we combine "-1<x<0" or "x>1" these two inequalities, how come range for x fall betweeen -1<x<1 since x>1 is area which will not fit into this equation. Can you explain?

Range from (1): -----(-1)----(0)----(1)---- $$-1<x<0$$ or $$x>1$$, green area;

Range from (2): -----(-1)----(0)----(1)---- $$x<0$$, blue area;

From (1) and (2): ----(-1)----(0)----(1)---- $$-1<x<0$$, common range of $$x$$ from (1) and (2) (intersection of ranges from (1) and (2)), red area.

Hope it's clear.

the red zone indicates that the range is -1<x<0 , how do we arrive at -1<x<1?

The question asks whether $$-1<x<1$$ is true. We got that $$-1<x<0$$. Any, x from $$-1<x<0$$ IS in the range from -1 to 1, so we have an YES answer to the question.
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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29 Nov 2017, 08:52
Hussain15 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Stmnt 2: |x| - x >0, which means x < 0... So, Insufficient
Stmnt 1: x (1-1/|x|)>0, which means x < 0 , 1-1/|x| < 0 or x > 0, 1-1/|x| >0.. Insufficeint

Combining 1 +2,

We know x <0, 1-1/|x| < 0.. which is |x| < 1.

Hope it helps.
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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27 Dec 2017, 03:15
Similar topic: https://gmatclub.com/forum/if-x-is-diff ... 26715.html
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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17 Feb 2018, 11:51
(1) We dont have to remove the abs(x)...we can treat it as a variable that is always positive.

So if we multiply both sides by abs(x),we dont have to change the sign of the inequality and we get

abs(x)*x> x

if x>0 ---> abs(x) >1
if x<0----> abs(x) <1
Not suf.

(2) abs(x) < x
x must be <0, but we cant say that abs (x)<1, because this will be true for any x<0.

(1)(2): We are restricted to the inference:
if x<0----> abs(x) <1
so we its suf.
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x   [#permalink] 17 Feb 2018, 11:51

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