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If r > 0 and s > 0, is r/s < s/r?

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If r > 0 and s > 0, is r/s < s/r?  [#permalink]

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New post 17 Aug 2010, 09:02
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If r > 0 and s > 0, is r/s < s/r?

(1) r/(3s) = 1/4
(2) s = r + 4
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Re: GMAT - PAPER TEST QUESTION  [#permalink]

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New post 17 Aug 2010, 09:21
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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.
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Re: GMAT - PAPER TEST QUESTION  [#permalink]

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New post 01 Nov 2010, 08:44
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cmugeria wrote:
If R > 0 and S > 0, Is r/s < s/r?

1) r/3s =1/4
2) s = r + 4


Since r and s are both positive, we can simplify the inequality to \(r^2 < s^2\), and by taking the square root of both sides, \(r < s\).

(1) \(4r = 3s\)
\(r = \frac{3}{4}s\)
So r < s. Sufficient.

(2) Since \(r = s - 4\), \(r < s\). Sufficient.
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Re: If R > 0 and S > 0, Is r/s < s/r? 1) r/3s =1/4 2) s  [#permalink]

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New post 29 Nov 2011, 01:22
1. r/3s=1/4 so 3s = 4r , which is s>r - Sufficient
2. s=r+4 menas S>R - sufficient

Answer - D
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Re: If r > 0 and s > 0, is r/s < s/r?  [#permalink]

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New post 29 Jan 2013, 09:04
cmugeria wrote:
If r > 0 and s > 0, is r/s < s/r?

(1) r/(3s) = 1/4
(2) s = r + 4



For 1 --- \(\frac{r}{s}\)= 3/4we can deduce from this

From 2.... s = r+4

substituting this in the question---

\(\frac{r}{4+r}\)<\(\frac{4+r}{r}\)
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Re: If r > 0 and s > 0, is r/s < s/r?  [#permalink]

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New post 01 Sep 2013, 14:09
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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
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Re: If r > 0 and s > 0, is r/s < s/r?  [#permalink]

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New post 17 Nov 2015, 10:29
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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.
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If r > 0 and s > 0, is r/s < s/r?  [#permalink]

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New post 18 Jul 2016, 05:51
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cmugeria wrote:
If r > 0 and s > 0, is r/s < s/r?

(1) r/(3s) = 1/4
(2) s = r + 4


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
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If r > 0 and s > 0, is r/s < s/r?  [#permalink]

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New post 06 Nov 2017, 03:50
MathRevolution wrote:
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
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Re: If r > 0 and s > 0, is r/s < s/r?  [#permalink]

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New post 21 Aug 2018, 12:38
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 :? \(\frac{r}{s}=\frac{3}{4}\) ???? :-)
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Re: If r > 0 and s > 0, is r/s < s/r?  [#permalink]

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New post 21 Aug 2018, 12:44
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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 :? \(\frac{r}{s}=\frac{3}{4}\) ???? :-)


By multiplying both sides by 3.
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Re: If r > 0 and s > 0, is r/s < s/r?  [#permalink]

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New post 30 Jul 2019, 18:28
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?

Though the stem says r>0...

So where did I do wrong?

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Re: If r > 0 and s > 0, is r/s < s/r?  [#permalink]

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New post 30 Aug 2019, 08:37
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The statement can be rewritten as |r|<|s| i.e r/s<s/r => r^2<s^2 => |r|<|s|.
We need to prove magnitude of r is less than magnitude of s.

Statement 1: r/3s<1/4 => r/s=3/4=> |r|<|s| Sufficient.
Statement 2:s=r+4 Since, both r and s are >0 => |r|<|s|. Sufficient.

Answer D.

VeritasKarishma Bunuel Is the approach correct? Please reply.
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Re: If r > 0 and s > 0, is r/s < s/r?   [#permalink] 30 Aug 2019, 08:37
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