If x <> 0, Is x^2/|x| <1 ?? 1.X<1 2. X>-1

[vinversa]

If x <> 0, Is x^2/|x| <1 ??
1. X<1
2. X>-1



|x| = +x if x>=0 (GMAT given rule)
|x| = -x if x<0 (ditto)

Here since x<>0 (given in Q)
|x| = -x
Then to prove X^2/-x < 1
i.e to prove -x<1

In option 1. X<1 (not enough to prove -x<1)
In option 2. X>-1 (not enough to prove -x<1)

Both 1 & 2 gives -1<x<1 - which is enough to prove -x<1. Hence OA = [C]

[dinesh19aug]

I would not approach the question in this manner. You concluded that we need to prove id -x<1, this is incorrect.

Here since x<>0 (given in Q)
|x| = -x
Then to prove X^2/-x < 1
i.e to prove -x<1

The guide says always reduce to a minimal.
So when we look at the question, it says if x^2/|x| <1

Case 1: x^2 / x <1 ===> x<1
case 2: x^2 / (-x) <1 ====> -x<1 ====> x> -1

So combining case 1 + 2 ====> -1 < x < 1
So we need to check if -1 < x <1 is true

Statement 1: x<1 , x could be 0, -1, -2. Hence we do not know the lower value. INSUFFICIENT

Statement 2: x> -1 , x could be 0, 1, 2 etc. Hence we do not know the upper value. INSUFFICIENT

Combining 1 + 2 ==> x> -1 and x<1. Align the signs
-1 < X
X < 1 ===> -1 < X < 1 SUFFICIENT

[vinversa]

Dinesh - your case 1 is wrong because

Case 1: x^2 / x <1 ===> x<1
Your case 1 assumes |x| = +x, which is wrong.
|x| can never be equal to +x. This is because -
x<>0 (I did'nt come up with this. It is given in the question)
Therefore
|x| = -x
|x| <> +x

Case 1: x^2 / x <1 ===> x<1
case 2: x^2 / (-x) <1 ====> -x<1 ====> x> -1


Here since x<>0 (given in Q)
|x| = -x

[vinversa]

|x| = +x if x>=0 (GMAT given rule)
|x| = -x if x<0 (GMAT given rule)

x<>0 (given in Q)
Therefore
|x| <> +x because (x<>0 does not completely satisfy |x| = +x if x>=0 (GMAT given rule))

Not paying attention to GMAT rule |x| = +x if x>=0 & just looking at x<>0, tells me two things
1. X is not equal to zero. Zero is neither +ve nor -ve
2. This could mean that |x| = +x or -x (like you suggested) May be I m wrong.... We will have to wait for experts to chime in...

[dinesh19aug]

Vinvers,
You are correct about
|x| = x, if x ≥ 0
|x| = -x, if x < 0.

and the question does say that x<> 0. This is given tell us that you cannot divide by 0 if x WAS able to acquirea 0 value.

However |x| > 0 - correct? that is the reason, my case 1 takes tha value of |x| = +x

[debmalya.dutta]

Either ways the question can be answered using both the statements right ?

Sorry guys..probably I am being naive to the discussions that have happened

[dinesh19aug]

Either ways the question can be answered using both the statements right ?

Sorry guys..probably I am being naive to the discussions that have happened

Deb,
"This" question might have been solved either ways. However the point is which is the correct way to solve it. There might be a question which will have similar pattern but it might not be solved
both ways that I and vinversa suggested. :-).

[debmalya.dutta]

Either ways the question can be answered using both the statements right ?

Sorry guys..probably I am being naive to the discussions that have happened

Thanks Dinesh.. Cool

[RonPurewal]

If x <> 0, Is x^2/|x| <1 ??
1. X<1
2. X>-1



|x| = +x if x>=0 (GMAT given rule)
|x| = -x if x<0 (ditto)

Here since x<>0 (given in Q)
|x| = -x
Then to prove X^2/-x < 1
i.e to prove -x<1

In option 1. X<1 (not enough to prove -x<1)
In option 2. X>-1 (not enough to prove -x<1)

Both 1 & 2 gives -1<x<1 - which is enough to prove -x<1. Hence OA = [C]

i'm not sure what "<>" means, but i'll assume from the context of the discussion that it means "≠".

as far as decoding the expression (x^2)/|x|, there are two ways to figure that out:

(1) just play around with the expression
just plug in a bunch of numbers:
plug x = 3 --> (x^2)/|x| = 3
plug x = -5 --> (x^2)/|x| = 5
plug x = 1/2 --> (x^2)/|x| = 1/2
plug x = -1 --> (x^2)/|x| = 1
etc.
you should notice that you're just getting the absolute value of the number every time.
therefore, (x^2)/|x| is just |x|.

(2) realize that √(x^2) = |x|
therefore, this works the same way as any other square root would work -- when you divide x^2 by its own square root (i.e., |x|), you get the square root (i.e., |x|) again.
therefore, (x^2)/|x| is just |x|.

--

once you realize that (x^2)/|x| is just |x|, you have the following question:
Is |x| < 1?
(1) x < 1
(2) x > -1

the question becomes "is -1 < x < 1?", so you need both of the statements. hence (c).