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If r and s are negative, is r/s less than 1 ?

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If r and s are negative, is r/s less than 1 ?  [#permalink]

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New post 07 Mar 2012, 16:24
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A
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D
E

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If r and s are negative, is r/s less than 1 ?

(1) r + 2s = s^2/r
(2) r/s is 2 less than s/r

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If r and s are negative, is r/s less than 1 ?  [#permalink]

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New post 07 Mar 2012, 16:37
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If r and s are negative, is \(\frac{r}{s}\) less than 1 ?

Since r and s are negative then either \(0<\frac{r}{s}<1\) or \(\frac{r}{s}\geq{1}\)

(1) \(r + 2s = \frac{s^2}{r}\). Divided both parts by \(s\): \(\frac{r}{s}+2= \frac{s}{r}\). We cannot have the second case because if \(\frac{r}{s}\geq{1}\) then its reciprocal \(\frac{s}{r}\leq{1}\) and it cannot equal to \(\frac{r}{s}+2=positive+2\), hence \(0<\frac{r}{s}<1\). Sufficient.

(2) \(\frac{r}{s}\) is 2 less than \(\frac{s}{r}\) --> \(\frac{r}{s}+2= \frac{s}{r}\). The same info as above. Sufficient.

Answer: D.
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Re: Integers R & S  [#permalink]

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New post 07 Mar 2012, 16:41
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Sorry Bunuel - didn't get this buddy.

How did you get first line which says Since r & s are negative ......
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Re: Integers R & S  [#permalink]

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Re: If r and s are negative, is r/s less than 1 ?  [#permalink]

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New post 18 Apr 2012, 08:42
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A gives us \(r + 2s = s^2/r\),
dividing by r we get ->\(1 + 2s/r = (s/r)^2\)
call s/r = a,
\(1 + 2a = a^2\)
\(a^2 - 2a = 1\)
adding 1 to both sides we get -> \((a -1)^2 = 2\)
or a -1 = \(+\sqrt{2}\) or a - 1 = \(-\sqrt{2}\)
since r and s are both -ve we know that s/r > 0 so a has to 1 + \sqrt{2}
=> \(s/r = 1+\sqrt{2}\).
Sufficient.

B give us same statement as A, just substitute s/r = a you will end up with (a -1)^2 = 2 and use the same logic as above. So, B is sufficient

Hence D.
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Re: If r and s are negative, is r/s less than 1 ?  [#permalink]

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New post 19 Apr 2012, 11:37
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enigma123 wrote:
If r and s are negative, is \(\frac{r}{s}\) less than 1 ?

(1) \(r + 2s = \frac{s^2}{r}\)

(2) \(\frac{r}{s}\) is 2 less than \(\frac{s}{r}\)


I am struggling badly on this one guys. Can you please help?


Even though it has already been answered by Bunuel, I would like to point out some takeaways.

1. Dont be in a hurry to bring the fractions in integer form. Look at the options and see how variables are used in the options.
2. If r and s are both negative, it means r/s is positive i.e. > 0
3. If r/s is a positive fraction (non 1), one of r/s and s/r will be between 0 and 1 (the smaller one) and the other will be to the right of 1. (greater than 1)
4. It is possible that the two statements give the same data. It they do and if one stmnt alone is enough, both together will be enough. If one is not enough, neither are they enough together, nor is the other one enough alone.
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If r and s are negative, is r/s less than 1 ?  [#permalink]

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New post 24 Mar 2016, 07:45
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If r and s are negative, is r/s less than 1 ?

(1) r + 2s = s^2/r
(2) r/s is 2 less than s/r

(1) r + 2s - (s^2/r)=0
r^2-s^2=-2rs taking RHS part since r & s are -ve then -2rs must be negative
so r^2-s^2<0
r^2<s^2
taking root both sides IrI<IsI
as both are -ve so IrI=-r & IsI=-s
So, now we can write as -r<-s------> r>s
so r/s is always <1
eg. say r=-2 & s=-3 (as r>s and both are -ve)
now -2/-3 < 1
So suff..

(2) r/s is 2 less than s/r
r/s=s/r-2
rearranging we again get r^2-s^2=-2rs
similar reasoning as (1).
Suff..

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Re: If r and s are negative, is r/s less than 1 ?  [#permalink]

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Re: If r and s are negative, is r/s less than 1 ?   [#permalink] 30 Mar 2019, 04:48
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