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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.
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Re: Mod of R  DS [#permalink]
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27 Jun 2010, 13:10
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. 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.
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Re: Mod of R  DS [#permalink]
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27 Jun 2010, 13:13
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. 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!!!
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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. 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 ???
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Re: Mod of R  DS [#permalink]
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07 Feb 2014, 05:29
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. 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 ??? \(r^2=r*r\) > \(\frac{r^2}{r}\) > \(\frac{r*r}{r}\) > \(r\). 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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