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If x 0, is x^2/|x| < 1?

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If x ≠ 0, is \(\frac{x^2}{|x|} < 1?\)

(1) x < 1
(2) x > −1
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zaarathelab wrote:
If x ≠ 0, is x^2 / |x| < 1?
(1) x < 1
(2) x > −1


We can safely reduce LHS of inequality \(\frac{x^2}{|x|}<1\) by \(|x|\). We'll get: is \(|x|<1\)? Basically question asks whether \(x\) is in the range \(-1<x<1\).

Statements 1 and 2 are not sufficient, but together they are defining the range for \(x\) as \(-1<x<1\).

Answer: C.
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IMO c

the expression x^2/lxl<1 will always be +ve as the N is a sqaure and mod x is always +ve

the expression is true only for fractional value

s1) tells us that x can be a +ve or -ve fractional and also can be a -int hence insuff.
s2) tells us that x can be a +ve or -ve fractional and also can be a +int hence insuff....

combinig x is only a fraction and hence suff...
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Bunuel wrote:
zaarathelab wrote:
If x ≠ 0, is x^2 / |x| < 1?
(1) x < 1
(2) x > −1


We can safely reduce inequality \(\frac{x^2}{|x|}<1\) by \(|x|\). We'll get: is \(|x|<1\)? Basically question asks whether \(x\) is in the range \(-1<x<1\).

Statements 1 and 2 are not sufficient, but together they are defining the range for \(x\) as \(-1<x<1\).

Answer: C.


Bunuel - please clarify this divison -\(\frac{x^2}{|x|}<1\) by \(|x|\). We'll get: is \(|x|<1\)? can be done because |x| is always +ve or is there are any other reason ?
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ichha148 wrote:
Bunuel - please clarify this divison -\(\frac{x^2}{|x|}<1\) by \(|x|\). We'll get: is \(|x|<1\)? can be done because |x| is always +ve or is there are any other reason ?


We should never multiply (or reduce) inequality by variable (or expression with variable) if we don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.

But in this case we are not reducing the inequality we are reducing only one part of it. So, it's safe to do so.

For example if we had: \(\frac{x^4}{x^3}<0\) we can reduce LHS by \(x^3\) and write \(x<0\).
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If x ≠ 0, is x^2/|x|< 1?

1: x<1 - Insufficient. x can be -2 or 1/2
2:x>-1 - Insufficient. x can be 2 or 1/2

Together, sufficient. x must be a fraction, and thus the answer is C.

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If x ≠ 0, is x^2/|x| < 1?

(1) x < 1
(2) x > -1

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If x ≠ 0, is x^2/|x| < 1?

(1) x < 1
(2) x > -1

We can safely reduce the left hand side of the inequality \(\frac{x^2}{|x|}<1\) by \(|x|\). We'll get: is \(|x|<1\)? Hence, the question basically asks whether \(-1<x<1\)?

Neither statement is sufficient alone. When taken together they are defining the range for \(x\) as \(-1<x<1\). Sufficient.

Answer: C.
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If \(x \neq 0\), is \(\frac{x^2}{|x|}< 1\)?

(1) x < 1
(2) x > −1

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udaymathapati wrote:
If \(x \neq 0\), is \(\frac{x^2}{|x|}< 1\)?

(1) x < 1
(2) x > −1


If \(x\neq{0}\), is \(\frac{x^2}{|x|}<1\)? --> reduce by \(|x|\) --> is \(|x|<1\)? or is \(-1<x<1\)?

Two statements together give us the sufficient info.

Answer: C.
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Re: If x ≠ 0, is x^2/|x| < 1? [#permalink]

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udaymathapati wrote:
[img]
Attachment:
Maths1.JPG
[/img]

(1) x < 1
(2) x > −1


x^2/abs(x) <1 ?

Another way to look at it...

1. if you set x= positive decimal you get the original value which is <1
now you can try x = negative integer(-5) which results in the positive version which is >1 so INSUFF
2. this is INSUFF since x could be a huge positive number which makes it >1 OR it could be a small decimal number which makes it <1

combining you see -1< X <1 which means X is a +/- decimal which also means it will be <1 so C
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Re: If x ≠ 0, is x^2/|x| < 1? [#permalink]

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New post 09 Sep 2010, 04:38
Bunuel wrote:
udaymathapati wrote:
[img]
Attachment:
Maths1.JPG
[/img]

(1) x < 1
(2) x > −1


If \(x\neq{0}\), is \(\frac{x^2}{|x|}<1\)? --> reduce by \(|x|\) --> is \(|x|<1\)? or is \(-1<x<1\)?

Two statements together give us the sufficient info.

Answer: C.


Bunuel,
Can you explain how it reduce it to \(|x|\) from the expression?

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udaymathapati wrote:
Bunuel wrote:
udaymathapati wrote:
[img]
Attachment:
Maths1.JPG
[/img]

(1) x < 1
(2) x > −1


If \(x\neq{0}\), is \(\frac{x^2}{|x|}<1\)? --> reduce by \(|x|\) --> is \(|x|<1\)? or is \(-1<x<1\)?

Two statements together give us the sufficient info.

Answer: C.


Bunuel,
Can you explain how it reduce it to \(|x|\) from the expression?


Given: \(\frac{x^2}{|x|}<1\)

Consider this:
\(\frac{x^2}{|x|}=\frac{|x|*|x|}{|x|}=|x|\). It's basically the same as if it were \(\frac{x^2}{x}\) --> we could reduce this fraction by \(x\) and we would get \(x\), and when \(x\) is positive, result is positive and when \(x\) is negative, result is negative. Now, \(\frac{x^2}{|x|}\) is the ratio of two positive values and the result can not be negative, so we can not get \(x\), we should get \(|x|\) to guarantee that the result is positive.

OR:
\(x<0\)--> then \(|x|=-x\) --> \(\frac{x^2}{|x|}=\frac{x^2}{-x}=-x<1\) --> \(x>-1\);

\(x>0\)--> then \(|x|=x\) --> \(\frac{x^2}{|x|}=\frac{x^2}{x}=x<1\);

So \(-1<x<1\).
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Re: If x ≠ 0, is x^2/|x| < 1? [#permalink]

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New post 14 Oct 2010, 14:44
tatane90 wrote:
If x ≠ 0, is x^2/|x| < 1?
(1) x < 1
(2) x > −1


\(\frac{x^2}{|x|} \lt 1\)
If x>0, then this implies x<1
If x<0, then this implies x>-1
So it is only true if either 0<x<1 or -1<x<0

(1) Not sufficient clealry, as x is not bounded on lower side
(2) Not sufficient clealrly, as x is not bounded on upper side

(1+2) Exactly defined the range for which the inequality holds. Sufficient

Answer is (c)
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Re: If x ≠ 0, is x^2/|x| < 1? [#permalink]

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New post 16 Oct 2010, 04:49
Bunuel, did anybody tell you that you are a genius? Stay away from scientists, they might start researching on your brain :P

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Re: If x ≠ 0, is x^2/|x| < 1? [#permalink]

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\(x^2/|x|\) reduces to |x||x|/|x| which reduces the qn to is |x| <1 ? This again reduces to -1< x <1. Only combining (1) and two answers this qn. Hence answer is (C).

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x^2/|x| < 1
(x^2 = |x|*|x|)
SO
(|x|*|x|)/|x| < 1
Is |x|<1?
is x<1 or is x>-1
is -1<x<1?

(1) x < 1
The issue here is depending on what x is, x may not be in the range of -1<x<1.
INSUFFICIENT

(2) x > −1
The same problem that applied to a) applies to b).
INSUFFICIENT

a+b) this gives us a range of -1<x<1 which is what the question is looking for.
SUFFICIENT

(C)

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Re: If x ≠ 0, is x^2/|x| < 1? [#permalink]

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udaymathapati wrote:
If x#0, is |x|/x<1?

(1) x < 1
(2) x > −1



I think\(\frac{|x|}{x} <1\)

(1) x < 1
(2) x > −1

Here the answer should be E

x= 1/2 satisfies both the statements and answer to the stem is no, 1 is not less 1

X= - 1/2 satisfies both the statements and answer to the stem is yes , -1<1


but for question \(\frac{x^2}{x} <1\)

(1) x < 1
(2) x > −1


here the answer is C as shown above

Please do correct if I am missing something
thanks.
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Re: If x ≠ 0, is x^2/|x| < 1? [#permalink]

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New post 24 Sep 2013, 01:38
stne wrote:
udaymathapati wrote:
If x#0, is |x|/x<1?

(1) x < 1
(2) x > −1



I think\(\frac{|x|}{x} <1\)

(1) x < 1
(2) x > −1

Here the answer should be E

x= 1/2 satisfies both the statements and answer to the stem is no, 1 is not less 1

X= - 1/2 satisfies both the statements and answer to the stem is yes , -1<1


but for question \(\frac{x^2}{x} <1\)

(1) x < 1
(2) x > −1


here the answer is C as shown above

Please do correct if I am missing something
thanks.


If it were:
If x#0, is |x|/x<1?

(1) x < 1
(2) x > −1


Then the answer is E. The question basically asks whether x is negative and we cannot answer that even when we combine the statements given.

If it were:
If x#0, is x^2/x<1?

(1) x < 1
(2) x > −1


Then the answer is C. The question basically asks whether x<0 or 0<x<1. When we combine the statements, we get that -1<x<1 (x#0). So, the answer to the question is YES.
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Re: If x ≠ 0, is x^2/|x| < 1?   [#permalink] 24 Sep 2013, 01:38

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