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# If r is not equal to 0, is r^2/|r| < 1? (1) r > -1 (2)

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If r is not equal to 0, is r^2/|r| < 1? (1) r > -1 (2) [#permalink]

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27 Jun 2010, 12:43
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If r is not equal to 0, is r^2/|r| < 1?

(1) r > -1

(2) r < 1

AS far as i know the option B looks sufficient. Since, r<1, it can take values that are negative like -2 or fraction values like
1/2. in either case the value of r^2/ |R| is <1. The OA suggests other wise.
[Reveal] Spoiler: OA

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Re: Mod of R - DS [#permalink]

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27 Jun 2010, 13:10
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kylexy wrote:
If r is not equal to 0, is r^2/|r| < 1?

(1) r > -1

(2) r < 1

Hi pls help me out with a detailed explanation

AS far as i know the option B looks sufficient. Since, r<1, it can take values that are negative like -2 or fraction values like
1/2. in either case the value of r^2/ |R| is <1. The OA suggests other wise.

Is $$\frac{r^2}{|r|}<1$$? --> reduce by $$|r|$$ --> is $$|r|<1$$? or is $$-1<r<1$$?

Two statements together give us the sufficient info.

You made a mistake in calculation for statement (2). Given $$r<1$$: for $$-1<r<1$$, for example if $$r=-\frac{1}{2}$$, then $$\frac{(-\frac{1}{2})^2}{|-\frac{1}{2}|}=\frac{1}{2}<1$$ but if $$r\leq{-1}$$, for example if $$r=-2$$, then $$\frac{(-2)^2}{|-2|}=2>1$$.

Hope it's clear.
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Re: Mod of R - DS [#permalink]

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27 Jun 2010, 13:13
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The first thing to note is that the question isn't testing sign. They tell us that r is not 0, and by definition, both r^2 and |r| are positive. So neither of these statements would be more useful than the other alone.

Since pos/pos = pos, we are ok doing a little creative manipulation of r^2/|r| = |(r*r)/r| = |r|. This move (putting the absolute value sign around the whole thing) isn't a rule to memorize or anything. I'm just ignoring sign temporarily, cancelling, then just assuring the positive result I need with the bars.

This question is really asking "Is r a fraction, or is it larger than 1 (in absolute value)?"
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Re: Mod of R - DS [#permalink]

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27 Jun 2010, 23:21
C, value of 'r' shall fall in the range -1 and 1 for a single solution to exist.

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Re: Mod of R - DS [#permalink]

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28 Jun 2010, 02:37
Bunuel wrote:
kylexy wrote:
If r is not equal to 0, is r^2/|r| < 1?

(1) r > -1

(2) r < 1

Hi pls help me out with a detailed explanation

AS far as i know the option B looks sufficient. Since, r<1, it can take values that are negative like -2 or fraction values like
1/2. in either case the value of r^2/ |R| is <1. The OA suggests other wise.

Is $$\frac{r^2}{|r|}<1$$? --> reduce by $$|r|$$ --> is $$|r|<1$$? or is $$-1<r<1$$?

Two statements together give us the sufficient info.

You made a mistake in calculation for statement (2). Given $$r<1$$: for $$-1<r<1$$, for example if $$r=-\frac{1}{2}$$, then $$\frac{(-\frac{1}{2})^2}{|-\frac{1}{2}|}=\frac{1}{2}<1$$ but if $$r\leq{-1}$$, for example if $$r=-2$$, then $$\frac{(-2)^2}{|-2|}=2>1$$.

Hope it's clear.

I guess i did make a mistake in the calc....my bad!!! thanks for the info bunuel!!!
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Re: If r is not equal to 0, is r^2/|r| < 1? (1) r > -1 (2) [#permalink]

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18 Aug 2013, 20:48
I just did this question on MGMAT, and learning to use number lines as a tool to answer these questions.

And I got it going well so far. Here is how I did it.

First I simplified the statement, but this is how I did it.

Instead of using long drawn out algebra I made it into 2 conditions.

I made $$\frac{r^2}{|r|} < 1$$ into two conditions; first where, both r, and |r| is positive.

Creating the equation r<1, then I took the reverse and said that <-1 creating r>1

Then I took the absolute value into the picture and made it into and created r>1 and r>-1

Which makes it that the answer has to r is between -infinity and positive infinity.

And the only solution that satisfies those conditions is C.

I may have made an error in one of my rationales above but it works
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Re: Mod of R - DS [#permalink]

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06 Feb 2014, 10:54
Bunuel wrote:
kylexy wrote:
If r is not equal to 0, is r^2/|r| < 1?

(1) r > -1

(2) r < 1

Hi pls help me out with a detailed explanation

AS far as i know the option B looks sufficient. Since, r<1, it can take values that are negative like -2 or fraction values like
1/2. in either case the value of r^2/ |R| is <1. The OA suggests other wise.

Is $$\frac{r^2}{|r|}<1$$? -->reduce by $$|r|$$ --> is $$|r|<1$$? or is $$-1<r<1$$?

Two statements together give us the sufficient info.

You made a mistake in calculation for statement (2). Given $$r<1$$: for $$-1<r<1$$, for example if $$r=-\frac{1}{2}$$, then $$\frac{(-\frac{1}{2})^2}{|-\frac{1}{2}|}=\frac{1}{2}<1$$ but if $$r\leq{-1}$$, for example if $$r=-2$$, then $$\frac{(-2)^2}{|-2|}=2>1$$.

Hope it's clear.

How r^2/lrl reduce to lrl only ???
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Re: Mod of R - DS [#permalink]

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07 Feb 2014, 05:29
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sanjoo wrote:
Bunuel wrote:
kylexy wrote:
If r is not equal to 0, is r^2/|r| < 1?

(1) r > -1

(2) r < 1

Hi pls help me out with a detailed explanation

AS far as i know the option B looks sufficient. Since, r<1, it can take values that are negative like -2 or fraction values like
1/2. in either case the value of r^2/ |R| is <1. The OA suggests other wise.

Is $$\frac{r^2}{|r|}<1$$? -->reduce by $$|r|$$ --> is $$|r|<1$$? or is $$-1<r<1$$?

Two statements together give us the sufficient info.

You made a mistake in calculation for statement (2). Given $$r<1$$: for $$-1<r<1$$, for example if $$r=-\frac{1}{2}$$, then $$\frac{(-\frac{1}{2})^2}{|-\frac{1}{2}|}=\frac{1}{2}<1$$ but if $$r\leq{-1}$$, for example if $$r=-2$$, then $$\frac{(-2)^2}{|-2|}=2>1$$.

Hope it's clear.

How r^2/lrl reduce to lrl only ???

$$r^2=|r|*|r|$$ --> $$\frac{r^2}{|r|}$$ --> $$\frac{|r|*|r|}{|r|}$$ --> $$|r|$$.

Hope it's clear.
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Re: If r is not equal to 0, is r^2/|r| < 1? (1) r > -1 (2) [#permalink]

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06 Nov 2014, 01:36
kylexy wrote:
If r is not equal to 0, is r^2/|r| < 1?

(1) r > -1

(2) r < 1

AS far as i know the option B looks sufficient. Since, r<1, it can take values that are negative like -2 or fraction values like
1/2. in either case the value of r^2/ |R| is <1. The OA suggests other wise.

r^2/|r|<1 ---> r^2<|r|

Logically, the only way when any number squared is less than the same number not squared is when the number is between -1 and 1

S1. r>-1 only one part of interval, so INSUFFICIENT

S2. r<1 again, only one part of interval, INSUFFICIENT

S1+S2 gives full interval, SUFFICIENT

C

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If r is not equal to 0, is r^2/|r| < 1? (1) r > -1 (2) [#permalink]

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29 Jan 2017, 03:47
kylexy wrote:
If r is not equal to 0, is $$\frac{r^{2}}{|r|} < 1$$ ?

(1) $$r > -1$$

(2) $$r < 1$$

AS far as i know the option B looks sufficient. Since, $$r<1$$, it can take values that are negative like $$-2$$ or fraction values like
$$\frac{1}{2}$$ . in either case the value of $$\frac{r^2}{|r|}$$ is $$<1$$. The OA suggests other wise.

Since $$|r|$$ is always positive, we can multiply both sides of the inequality by $$|r|$$ and rephrase the question as: Is $$r^{2} < |r |$$ ? The only way for this to be the case is if $$r$$is a nonzero fraction between $$-1$$ and $$1$$.

(1) INSUFFICIENT: This does not tell us whether $$r$$ is between $$-1$$ and $$1$$. If $$r = - \frac{1}{2}$$ , $$|r| = \frac{1}{2}$$ and $$r^{2} = \frac{1}{4}$$ , and the answer to the rephrased question is YES.  However, if $$r = 4,$$ , $$|r| = 4$$and $$r^{2} = 16$$, and the answer to the question is NO.

(2) INSUFFICIENT: This does not tell us whether $$r$$ is between $$-1$$ and $$1$$. If $$r = \frac{1}{2}$$ , $$|r| = \frac{1}{2}$$ ans $$r^{2} = \frac{1}{4}$$ , and the answer to the rephrased question is YES.  However, if $$r = -4$$, $$|r| = 4$$ and $$r^{2}=16$$, and the answer to the question is NO.

(1) AND (2) SUFFICIENT: Together, the statements tell us that r is between $$-1$$ and $$1$$. The square of a proper fraction (positive or negative) will always be smaller than the absolute value of that proper fraction.

The correct answer is $$C$$.
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If r is not equal to 0, is r^2/|r| < 1? (1) r > -1 (2)   [#permalink] 29 Jan 2017, 03:47
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