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Intern  Joined: 22 Dec 2009
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If r > 0 and s > 0, is r/s < s/r?  [#permalink]

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Question Stats: 83% (01:26) correct 17% (01:40) wrong based on 1682 sessions

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

(1) r/(3s) = 1/4
(2) s = r + 4
Math Expert V
Joined: 02 Sep 2009
Posts: 62357
Re: GMAT - PAPER TEST QUESTION  [#permalink]

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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.

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Re: GMAT - PAPER TEST QUESTION  [#permalink]

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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.
##### General Discussion
Intern  Joined: 09 Nov 2011
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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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1. r/3s=1/4 so 3s = 4r , which is s>r - Sufficient
2. s=r+4 menas S>R - sufficient

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

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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

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

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4
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.

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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1
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

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Intern  Joined: 07 Oct 2017
Posts: 1
If r > 0 and s > 0, is r/s < s/r?  [#permalink]

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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.

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.

Posted from my mobile device

Posted from my mobile device

How do we arrive @ this..
r-s >0 ..I cant seem to figure it out
Thanks
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Re: If r > 0 and s > 0, is r/s < s/r?  [#permalink]

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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.

how from here $$\frac{r}{3s}=\frac{1}{4}$$ we got this $$\frac{r}{s}=\frac{3}{4}$$ ???? Math Expert V
Joined: 02 Sep 2009
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Re: If r > 0 and s > 0, is r/s < s/r?  [#permalink]

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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.

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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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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1
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.

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

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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.

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

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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.

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?

Divide by r and then by s.
_________________ Re: If r > 0 and s > 0, is r/s < s/r?   [#permalink] 23 Feb 2020, 03:07
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