If x is different from 0, is |x|

Question

If x is different from 0, is |x| < 1 ?

(1) x*|x| < x

(2) |x| > x

I think in this way.

From stat 1 : |x| < 1 and so we have 2 solution INSUFF

From stat 2: the same INSUFF.

At this point I'm lost how to evaluate together ????

Thanks

Bunuel · Accepted Answer

If x is different from 0, is |x| < 1 ?

(1) x*|x| < x

(2) |x| > x
_____________________

1) x*|x| < x : on first look it seems to be sufficient, Divide by X from both side and we will get the answer, But this is the catch here.
We only know that X != 0 . X can e negative as well and if X is negative then it will reverse the sign. So we get two conditions here. X +ve and x-ve

For X +ve = divide by X = |X| <1
For X -ve = divide by -X = |X| >1
So not sufficient.

2) |x| > x : It says that X is -ve. Only in this scenerio |X| can be greater than X. This is also not sufficient.

Now Lets take ! and 2 both
We know from 2 that X is -ve so from 1 we get |X| >1. Answer is no |X| is not less than 1.
So C is sufficient

Bunuel · Answer

OK. First of all: I marked "|x|<1" in red for (1), because you don't get that |x| < 1 from this statement. If it were so then then you'd have an YES answer right away (as the question asks exactly about the same thing "is |x|<1") Next, "is |x|<1 ?" means "is -1 x. Absolute value property: when x\geq{0} then |x|=x and when x<{0} (so when x is negative) then |x|=-x . For the second case |x|=positive>x=negative . That' why |x|>x means that x<0 (or as I wrote if x were positive we would have |x|=x not |x|>x ). You can also consider a simple example: |-3|>-3 --> -3<0 . Thus the range for (2) is: ------------- (0)-------- Also insufficient to answer the question. (1)+(2) Intersection of the ranges from (1) and (2) is: ---- (-1)----(0)----(1)---- --> x<-1 . Which makes the answer to the question "is -1math-absolute-value-modulus-86462.html Hope it's clear.

Carcass · Answer

I suppose that you have put in red my logic on stat 1 because not so clear.

Stat (1) : x > 1 or -x > 1.........x < -1

Stat (2) : x < x or -x > x.........x < -x

This is the logic that conduct me to say that the 2 stats are insufficient. I don't understand completely why in the 2 one |x| > x ........ x is < 0 ??'

I also read the topic that you suggested; your third reply is very helpful but the doubt remain on stat ( 2)

Thanks

Carcass · Answer

Thank you a lot.

Now right away I'll read absolute value chapter.

WholeLottaLove · Answer

If x is different from 0, is |x| < 1 ?
x≠0, is -1<x<1?

(1) x |x| < x 
For this to hold true: 0<x<1 or x<-1

0<x<1
1/2 * |1/2| < 1/2
1/4 < 1/2 Valid

x<-1
-2 * |-2| < -2
-4 < -2 Valid

x could be between 0 and 1 or it could be less than -1.
INSUFFICIENT

(2) |x| > x 
|x| is by definition ≥0 so for |x| to be greater than x, x must be negative. Depending on the value of -x, it may or may not satisfy the inequality.
INSUFFICIENT

1+2) 0<x<1 or x<-1 and x is negative. The only intersection is when x<-1 and "x is negative" meaning x<-1 and the inequality in the stem holds true. 
SUFFICIENT

(C)

pmsigns · Answer

i solved it in the following way.. what am i doing wrong ?

for x|x|<x

a. x>0: xx<x -> x<1 i.e. 0<x<1
b. x<0: -xx<x -> -x<1 -> x>-1 which is wrong.

ln another problem i solved x/|x| similarly, and got the correct answer .

for x/|x|< x

x>0: x/x<x -> 1<x
x<0: x/-x<x ->-1<x<0

i am confused ..

Bunuel · Answer

For (b) when x<0 you have: -x*x < x --> divie by negative x and flip the sign: -x>1 --> x<-1.

him1985 · Answer

If x is different from 0, is |x| < 1 ?

(1) x*|x| < x

(2) |x| > x
_____________________

1) x*|x| < x : on first look it seems to be sufficient, Divide by X from both side and we will get the answer, But this is the catch here.
We only know that X != 0 . X can e negative as well and if X is negative then it will reverse the sign. So we get two conditions here. X +ve and x-ve

For X +ve = divide by X = |X| <1
For X -ve = divide by -X = |X| >1
So not sufficient.

2) |x| > x : It says that X is -ve. Only in this scenerio |X| can be greater than X. This is also not sufficient.

Now Lets take ! and 2 both
We know from 2 that X is -ve so from 1 we get |X| >1. Answer is no |X| is not less than 1.
So C is sufficient

shiying · Answer

Hi Bunuel, can x also be x>-1 for Statement 1? Coz when I sub in the values of (-1/2) or (-1/4), x satisfies the equation. But on your number line it is in black, which indicates x>-1 does not hold true. Please kindly enlighten me.

Bunuel · Answer

Neither of these values satisfies x*|x| < x.

If x = -1/2, then (x*|x| = -1/4)  >  -1/2 NOT <.

Sabby · Answer

Hi,

I like to use smart numbers for this type of questions otherwise I feel completely lost.

Let x be : -2 or -1/2 or 1/2 or 2

1) cases where x*Ixl<x :

With x=-2
-2*l2l<-2 ==> -4<-2 and here l2l>1, is lxl<1? NO

With x= 1/2
1/2*1/2<1/2 ==> 1/4<1/2 and here l1/2l<1, is lxl<1? YES

we have a yes and no so it's insufficient

2) Cases where lxl>x

When x=-2
l2l>-2 ==> is lxl<1? NO

When x=-1/2
l1/2l>-1/2 ==> is lxl<1? YES

We have a yes and no so it's insufficient

1)+2) We have only one value that is consistent with the two statements and it's -2. Therefore, is lxl<1? the answer is NO.

Answer C)

GMATGuruNY · Answer

Statement 1: CRITICAL POINTS occur when the two sides of an inequality are EQUAL or when the inequality is UNDEFINED. x|x|=x when x=-1, x=0 or x=1 The critical points are x=-1, x=0 and x=1, implying the following number line: ..........-1..........0..........1.......... To determine which ranges for x are valid, test one value to the left and one value to the right of each critical point. If we test x=-2, x=-1/2, x=1/2 and x=2, only x=-2 and x=1/2 satisfy x|x|1, so the answer to the question stem is NO. Case 2: 0x for all negative values of x. If x=-1/2, then |x|<1, so the answer to the question stem is YES. If x=-2, then |x|>1, so the answer to the question stem is NO. INSUFFICIENT. Statements combined: Only Case 1 satisfies both statements. In Case 1, the answer to the question stem is NO. SUFFICIENT. C

TBT · Answer

Got to learn so much from this question! I was making mistake while defining the range. This brushes up your concept of inequality!

bumpbot · Answer

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If x is different from 0, is |x| < 1 ?

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