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If r is not equal to 0, is r^2/|r| < 1? (1) r > -1 (2) [#permalink]
27 Jun 2010, 11:43

2

This post received KUDOS

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

68% (02:02) correct
32% (01:19) wrong based on 180 sessions

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.

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.

Answer: C.

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.

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)?" _________________

Emily Sledge | Manhattan GMAT Instructor | St. Louis

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.

Answer: C.

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!!! _________________

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.

Answer: C.

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 ??? _________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

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.

Answer: C.

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.