If x≠0, is |x| x

Question

If  x ≠ 0 , is  |x|<1 ?

(1)  \frac{x^2}{{|x|}} > x

(2)  \frac{x}{{|x|}} < x

I have a hard time to answer this question. Could someone please help to explain?

ganand · Answer

Hi, is |x|<1 ? => is -1 < x < 1 ? St 1: Case1: x>0 => |x| = x. \frac{x^2}{{|x|}} = \frac{x^{2}}{x} = x > x => No solution. Case2: x<0 => |x| = -x \frac{x^2}{{|x|}} = \frac{x^{2}}{-x} = -x > x => All negative values will satisfy this equation. Hence, not sufficient. St 2: Case 1: x>0 => |x| = x \frac{x}{{|x|}} = \frac{x}{x} = 1 < x => solution x>1. case 2: x<0 => |x| = -x \frac{x}{{|x|}} = \frac{x}{-x} = -1 < x => solution -1 < x < 0. From this statement, we have two sets of solutions. Hence, not sufficient. By St 1 and st 2, we have -1< x < 0. => Unique answer. Answer C. Hope this helps.

Bunuel · Answer

If x ≠ 0 , is |x|<1 ? The question asks whether -1 < x < 1 (x ≠ 0). (1) \frac{x^2}{{|x|}} > x --> reduce the LHS by |x|: |x| > x. This implies that x is negative. Not sufficient. (2) \frac{x}{{|x|}} < x : If x < 0, we'll have x/(-x) < x --> -1 < x. Since we consider x < 0, then we'll have -1 < x < 0. If x > 0, we'll have x/x < x --> 1 < x. So, we have that \frac{x}{{|x|}} < x is true for -1 < x < 0 and x > 1. Not sufficient. (1)+(2) Intersection of the ranges from (1) and (2) is -1 < x < 0. Thus the answer to the question is YES. Sufficient. Answer: C.

Alexey1989x · Answer

Hello Bunuel, could you please elaborate more on the (1) condition, I can't grasp how did you get that x is -ve.
Thx

Bunuel · Answer

We have |x| > x.

Now, if x were non-negative, then |x| would be equal to x, for example, if x=2, then |x| = |2| = 2 = x. Only for negative x we can have |x| > x. For example, consider x = -2: |x| = |-2| = 2 > x = 2.

MahmoudFawzy · Answer

Greetings Bunuel in analyzing statement 2, you mentioned : "If x < 0, we'll have x/(-x) < x --> -1 < x" my question: why the inequality is not flipped? isn't it right to flip the from '<' to '> ? I think I am confused in this point, please elaborate.

Bunuel · Answer

We are not dividing the entire inequality by x, we are simply reducing x/(-x) by x: x/(-x) = -(x/x) = -1.

GMATaspirant641 · Answer

Bunuel

Please let me know where I am going wrong in my approach with statement 1.

x^2 / |x| > x

we know that the LHS will always be positive.

So, x^2 / |x| > 0 (its not >= 0 since x cannot be 0)

x^2 / |x| > 0

x^2 > 0

|x| > 0

so, x>0 and x<0. Hence, x can be any non zero number.

With this approach, i get answer choice E.

I also understand the approaches mentioned by you and others in the thread. But, I am not able to understand where I am going wrong. Please help.

Bunuel · Answer

Yes,  \frac{x^2}{|x|} > 0  is true of all x's except 0. But we don't have  \frac{x^2}{|x|} > 0 , we have  \frac{x^2}{|x|} > x  and as shown above it won't hold true if x is positive. For example, if x = 2, then  \frac{x^2}{|x|} =2 , so  \frac{x^2}{|x|} = x .

rahul16singh28 · Answer

Hi  chetan2u   Bunuel

By adding two equations, we get -

\frac{x}{{|x|}}(x-1) > 0

Here, x can be -1,2.

Can you please advise where I am going wrong.

KanishkM · Answer

Why didn't i use the values used in A, to negate B , Nevertheless inline is my approach |x| < 1, will mean that -1 x when you use -0.5, the value will answer the statement and will answer the question as Yes but when you use -2, the value will answer the statement and will answer the question as No 4/2 > -2 (2) \frac{x}{{|x|}} < x when you use -0.5 , the value will answer the statement and will answer the question as Yes -0.5 /0.5 < -0.5 But when you use 2, the value will answer the statement and will answer the question as No Now when we combine both the statements for both the statements to be true together When you use x =-0.5, equality holds good C

Bunuel · Answer

We cannot add the inequalities the way you did. You can only apply subtraction when their signs are in the opposite directions : If a>b and c and < ) --> a-c>b-d (take the sign of the inequality you subtract from). Example: 3<4 and 5>1 --> 3-5<4-1 . You can only add inequalities when their signs are in the same direction: If a>b and c>d (signs in same direction: > and > ) --> a+c>b+d . Example: 3<4 and 2<5 --> 3+2<4+5 . For more check Manipulating Inequalities .

NinetyFour · Answer

Here's my reasoning:

Statement 1: x^2 > x{|x|}

X can be both a negative integer and a negative decimal. Thus we get a YES and a NO for the stem above.

Statement 2: this tells us  x < x{|x|} 
X can be a positive integer or a negative fraction. Once again a YES and a NO for the stem above.

Combining the two the overlapping set is a negative fraction. Hence we can conclude  |x|<1

ChairmanThe · Answer

I am having trouble understanding how you simplify the first condition to |x| > x. Can you please elaborate on that? 
Thanks

Bunuel · Answer

\frac{x^2}{|x|}>x ;

\frac{|x|*|x|}{|x|}>x ;

|x|>x .

ChairmanThe · Answer

But why are you using |x|∗|x| when you are factoring X^2. Couldn't it be -x *-x and give you the same value for X^2? 
Thanks

Bunuel · Answer

x^2 = |x|*|x| and if you write it that way you'd be able to simplify  \frac{x^2}{|x|}>x  and get  |x|>x .

If x≠0, is |x|<1? (1) x^2/|x| > x

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