statement 1 seems to be INSUFF
consider x = 1/2, does satisfy x^2/|x| < 1
consider x = -2, does not satisfy x^2/|x| < 1

statement 2 seems to be INSUFF too
consider x = -1/2 and 2, the later does not satisfy x^2/|x| < 1

statement 1 and statement 2
-1 < x < 1 seems to satisfy x^2/|x| < 1

I think the answer should be C

Whats the OA?

OA - C.

Anyone with a different approach. Thanks!

beater wrote:If x &#8800; 0, is x^2/|x|< 1?
(1) x < 1
(2) x > &#8722;1

Let's start by simplifying the question.

Is x^2/|x|< 1?

Well, we know that |x| is positive (since x &#8800; 0), so it's safe to multiply both sides by |x| to get:

is x^2 < |x|?

When is the square of a number less than the number itself? If x is a positive fraction. Since in this question it's less than the abolute value of the number itself, a negative fraction will work as well.

So, the question really is: Is x a positive or negative fraction?

(1) x < 1

Could be a fraction, could be a negative non-fraction: insufficient.

(2) x > -1

Could be a fraction, could be a positive non-fraction: insufficient.

Combined:

-1 < x < 1

Well, the only numbers in that range are fractions and 0. Since x &#8800; 0, x MUST be a fraction: sufficient.

Together the statements are sufficient even though independently they're not: choose (C).

Stuart,
I picked numbers and determined the answer to be C.

I am really confused when mod functions appear. You said |x| is positive (since x &#8800; 0)

Is mod 0 undefined since we cant apply the mod rule

mod x = x if x is positive
mod x = -x if x is negative

I habe 2 other mod problems that I was confused on based on explanations given. Iknew I coudnt ask those questiosn here since I would mixing up this thread.

I may have to open 2 new threads for these.

Thanks agian for youer explanations (Always use the Pascals triangle approach of yours when it comes to problem where the prob is 1/2 )

cramya wrote:Stuart,
I picked numbers and determined the answer to be C.

I am really confused when mod functions appear. You said |x| is positive (since x &#8800; 0)

Is mod 0 undefined since we cant apply the mod rule

mod x = x if x is positive
mod x = -x if x is negative

I habe 2 other mod problems that I was confused on based on explanations given. Iknew I coudnt ask those questiosn here since I would mixing up this thread.

I may have to open 2 new threads for these.

Thanks agian for youer explanations (Always use the Pascals triangle approach of yours when it comes to problem where the prob is 1/2 )

|0| = 0

Absolute value measures the distance between what's inside the brackets and 0 on the number line. The distance from 0 to 0 is... 0!

Stuart Kovinsky wrote:

beater wrote:If x &#8800; 0, is x^2/|x|< 1?
(1) x < 1
(2) x > &#8722;1

Let's start by simplifying the question.

Is x^2/|x|< 1?

Well, we know that |x| is positive (since x &#8800; 0), so it's safe to multiply both sides by |x| to get:

is x^2 < |x|?

When is the square of a number less than the number itself? If x is a positive fraction. Since in this question it's less than the abolute value of the number itself, a negative fraction will work as well.

So, the question really is: Is x a positive or negative fraction?

(1) x < 1

Could be a fraction, could be a negative non-fraction: insufficient.

(2) x > -1

Could be a fraction, could be a positive non-fraction: insufficient.

Combined:

-1 < x < 1

Well, the only numbers in that range are fractions and 0. Since x &#8800; 0, x MUST be a fraction: sufficient.

Together the statements are sufficient even though independently they're not: choose (C).

hello stuart
Can you please clarify how |x| is +ve, if x!=0

Thanks in advance

hello stuart
Can you please clarify how |x| is +ve, if x!=0

By definition, the absolute value of a number is the distance between that number and 0 on the number line.

If x is not equal to 0, the distance from x to 0 will always be positive.

If x &#8800; 0, is x^2/|x|< 1?
(1) x < 1
(2) x > &#8722;1[/quote]

Absolute value is nothing but a distance. How far is something from 0 ?

So question asks whether square of an integer is smaller than its distance to 0.

Well we know that this can not be a positive number greater than 1

But we also know that numbers act FUNNY between the GMAT zone 0 and 1

Since the numbers get smaller between 0-1 we have to find this piece of information in the answers. Actually the same thing applies to numbers between -1 and 0 in this case because of the absolute value.

So actually the range we are looking for is:

-1<x<0
0>x>1

Remember 0 can not be a solution so

-1<x<1 where x &#8800; 0

(C)

Hi Ian,
I am looking for an algebric soln to these absolutes, is that possible.
Regards,
maihuna

Is it OK to cross multiply here as we know for sure |x| is positive?

i.e x^2 < |x|