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

C, value of 'r' shall fall in the range -1 and 1 for a single solution to exist.

I guess i did make a mistake in the calc....my bad!!! thanks for the info bunuel!!!

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

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.

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

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 \(C\).

If r is not equal to 0, is r^2/|r| < 1?

(1) r > -1
r = 2 NO
r = 1/2 YES
Insufficient

(2) r < 1
r = -2 NO
r = -1/2 YES
Insufficient

C: Sufficient
-1 < r < 1 and r =/ 0 implies that r is a fraction

fraction / fraction is always < 1 if the x/y and x < y

We know that: r^2 = |r|^2
Thus, the question essentially is:
Is |r|^2/|r| < 1?
i.e. Is |r| < 1?
i.e. Is -1 < r < 1

From (1): r > -1 is not sufficient since we need to also ensure that r < 1 => Insufficient
From (2): r < 1 is not sufficient since we need to also ensure that r > -1 => Insufficient

Combining (1) and (2): We have -1 < r < 1 => Sufficient
Answer C

Note that r^2 is the same as |r|^2.

\(\frac{r^2}{|r|} = \frac{|r|^2}{|r|}\)

Since we are given that r cannot be 0, we can easily cancel off |r| to get:
Is |r| < 1?

When will the distance of r from 0 be less than 1?
When r lies between -1 to 1. So we need both statements to say that |r| will be less than 1.

Answer (C)

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