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(1) \(\frac{r}{3s}=\frac{1}{4}\) --> \(\frac{r}{s}=\frac{3}{4}\), so \(\frac{s}{r}=\frac{4}{3}\) --> \(\frac{r}{s}=\frac{3}{4}<\frac{4}{3}=\frac{s}{r}\) thus answer to the question is YES. Sufficient.
(2) \(s=r+4\) --> so \(s>r\) as given that \(r>0\) --> \(s>r>0\) --> \(\frac{s}{r}>1>\frac{r}{s}\), thus answer to the question is YES. Sufficient.
Re: If r > 0 and s > 0, is r/s < s/r?
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01 Sep 2013, 13:09
2
First I restated the problem since we are given r and s are greater than 0 --> therefore, the question can be solved by answering whether or not r^2<s^2.
(1) r = (3/4)s [r<s, so r^2 < s^2] Sufficient AD/BCE - elminate BCE (3) s = r + 4 [r<s, so r^2 < s^2] Sufficent - elminate A ... answered D
Re: If r > 0 and s > 0, is r/s < s/r?
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17 Nov 2015, 09:29
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
If r > 0 and s > 0, is r/s < s/r?
(1) r/(3s) = 1/4 (2) s = r + 4
In inequalities, the sign does not change when a positive integer is multiplied on both sides. If we modify the question, r/x<s/r, or is r^2<s^2, of is r^2-s^2<0?, or (r-s)(r+s)<0? and r>0 and s>0, so we want to know whether r-s>0? For condition 1, in r/s=3/4, r and s are positive, so s>r, which answers the question 'yes' and is sufficient. For condition 2, s-r=4>0. s>r, so this also answers the question 'yes' and is sufficient. The answer becomes (D).
Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
_________________
If r > 0 and s > 0, \(is \frac{r}{s} < \frac{s}{r}?\) The question stem tells us that both r and s are positive. What a relief, we can now do any operations on r and s without worrying about the polarity of the variable. Lets rephrase the statement Is \(\frac{r}{s} < \frac{s}{r}\)
THIS IS THE REAL QUESTION :- Is \(r^2 < s^2 ?\)
(1)\(\frac{r}{(3s)} = \frac{1}{4}\)
\(r=\frac{3}{4}*s\) Because r<s Therefore \(r^2 < s^2\) SUFFICIENT
(2) s = r + 4 Its obvious that s is bigger and r is smaller \(r<s\) and \(r^2<s^2\) SUFFICIENT
ANSWER IS D
_________________
Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly. FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
If r > 0 and s > 0, is r/s < s/r?
(1) r/(3s) = 1/4 (2) s = r + 4
In inequalities, the sign does not change when a positive integer is multiplied on both sides. If we modify the question, r/x<s/r, or is r^2<s^2, of is r^2-s^2<0?, or (r-s)(r+s)<0? and r>0 and s>0, so we want to know whether r-s>0? For condition 1, in r/s=3/4, r and s are positive, so s>r, which answers the question 'yes' and is sufficient. For condition 2, s-r=4>0. s>r, so this also answers the question 'yes' and is sufficient. The answer becomes (D).
Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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How do we arrive @ this.. r-s >0 ..I cant seem to figure it out Thanks
Re: If r > 0 and s > 0, is r/s < s/r?
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21 Aug 2018, 11:38
Bunuel wrote:
If r > 0 and S > 0, Is r/s < s/r?
Is \(\frac{r}{s}<\frac{s}{r}\)?
(1) \(\frac{r}{3s}=\frac{1}{4}\) --> \(\frac{r}{s}=\frac{3}{4}\), so \(\frac{s}{r}=\frac{4}{3}\) --> \(\frac{r}{s}=\frac{3}{4}<\frac{4}{3}=\frac{s}{r}\) thus answer to the question is YES. Sufficient.
(2) \(s=r+4\) --> so \(s>r\) as given that \(r>0\) --> \(s>r>0\) --> \(\frac{s}{r}>1>\frac{r}{s}\), thus answer to the question is YES. Sufficient.
Answer: D.
how from here \(\frac{r}{3s}=\frac{1}{4}\) we got this \(\frac{r}{s}=\frac{3}{4}\) ????
Re: If r > 0 and s > 0, is r/s < s/r?
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21 Aug 2018, 11:44
1
dave13 wrote:
Bunuel wrote:
If r > 0 and S > 0, Is r/s < s/r?
Is \(\frac{r}{s}<\frac{s}{r}\)?
(1) \(\frac{r}{3s}=\frac{1}{4}\) --> \(\frac{r}{s}=\frac{3}{4}\), so \(\frac{s}{r}=\frac{4}{3}\) --> \(\frac{r}{s}=\frac{3}{4}<\frac{4}{3}=\frac{s}{r}\) thus answer to the question is YES. Sufficient.
(2) \(s=r+4\) --> so \(s>r\) as given that \(r>0\) --> \(s>r>0\) --> \(\frac{s}{r}>1>\frac{r}{s}\), thus answer to the question is YES. Sufficient.
Answer: D.
how from here \(\frac{r}{3s}=\frac{1}{4}\) we got this \(\frac{r}{s}=\frac{3}{4}\) ????
Re: If r > 0 and s > 0, is r/s < s/r?
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30 Jul 2019, 17:28
I have a question here... so I simplified r/s <s/r and got r^2 - s^2 < 0 ( read equation1) and then for statement 2, I put the values s=r+4 on eq 1 and I got r<-2?
(1) \(\frac{r}{3s}=\frac{1}{4}\) --> \(\frac{r}{s}=\frac{3}{4}\), so \(\frac{s}{r}=\frac{4}{3}\) --> \(\frac{r}{s}=\frac{3}{4}<\frac{4}{3}=\frac{s}{r}\) thus answer to the question is YES. Sufficient.
(2) \(s=r+4\) --> so \(s>r\) as given that \(r>0\) --> \(s>r>0\) --> \(\frac{s}{r}>1>\frac{r}{s}\), thus answer to the question is YES. Sufficient.
Answer: D.
Could you tell me how did we get the highlighted part? it seems that we did not get the highlighted part dividing 0 by r, right?
Re: If r > 0 and s > 0, is r/s < s/r?
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23 Feb 2020, 03:07
Asad wrote:
Bunuel wrote:
If r > 0 and S > 0, Is r/s < s/r?
Is \(\frac{r}{s}<\frac{s}{r}\)?
(1) \(\frac{r}{3s}=\frac{1}{4}\) --> \(\frac{r}{s}=\frac{3}{4}\), so \(\frac{s}{r}=\frac{4}{3}\) --> \(\frac{r}{s}=\frac{3}{4}<\frac{4}{3}=\frac{s}{r}\) thus answer to the question is YES. Sufficient.
(2) \(s=r+4\) --> so \(s>r\) as given that \(r>0\) --> \(s>r>0\) --> \(\frac{s}{r}>1>\frac{r}{s}\), thus answer to the question is YES. Sufficient.
Answer: D.
Could you tell me how did we get the highlighted part? it seems that we did not get the highlighted part dividing 0 by r, right?
close
83% (01:26) correct 
\(\frac{r}{s}=\frac{3}{4}\) ???? 
