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M13-08

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M13-08  [#permalink]

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New post 16 Sep 2014, 00:48
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Question Stats:

49% (01:14) correct 51% (01:12) wrong based on 114 sessions

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M13-08  [#permalink]

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New post 30 Sep 2016, 19:45
2
ddb123 wrote:
mounirbr wrote:
Why you wrote down (r+s/rs)<2? I couldn't understand the logic used.



I am very certain that was a typo. He meant for 2 to be (1/2)


Hi,
It isn't a typo....
The original equation was rs/(r+s)>1/2....
Since both r and s are > than 1, they are positive..
So, you can cross multiply..
2*rs>1*(r+s)....
2/1>(r+s)/rs ....
(r+s)/rs<2...

Hope it is clear
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New post 16 Sep 2014, 00:48
Official Solution:


(1) \(r \gt 1\) and \(s \gt 1\). Since both variables are positive, we can rewrite the question as follows: is \(\frac{r+s}{rs} \lt 2\)? Or: is \(\frac{1}{s}+\frac{1}{r} \lt 2\)? Since it is given that \(r \gt 1\) and \(s \gt 1\), then both \(\frac{1}{s}\) and \(\frac{1}{r}\) are less than 1. Therefore, their sum is definitely less than 2. Sufficient.

(2) \(r \gt 2\) and \(s \gt 0\). The same here: is \(\frac{1}{s}+\frac{1}{r} \lt 2\)? If \(r=3\) and \(s=1\), then the answer is YES but if \(r=3\) and \(s=0.1\), then the answer is NO. Not sufficient.


Answer: A
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Re: M13-08  [#permalink]

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New post 11 Sep 2016, 17:26
Why you wrote down (r+s/rs)<2? I couldn't understand the logic used.
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Re: M13-08  [#permalink]

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New post 30 Sep 2016, 15:23
mounirbr wrote:
Why you wrote down (r+s/rs)<2? I couldn't understand the logic used.



I am very certain that was a typo. He meant for 2 to be (1/2)
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Re: M13-08  [#permalink]

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New post 30 Sep 2016, 20:17
chetan2u wrote:
ddb123 wrote:
mounirbr wrote:
Why you wrote down (r+s/rs)<2? I couldn't understand the logic used.



I am very certain that was a typo. He meant for 2 to be (1/2)


Hi,
It isn't a typo....
The original equation was rs/(r+s)>1/2....
Since both r and s are > than 1, they are positive..
So, you can cross multiply..
2*rs>1*(r+s)....
2/1>(r+s)/rs ....
(r+s)/rs<2...

Hope it is clear



Thanks for the clarification! I obviously wasn't looking at the information closely enough. Cheers.
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Re: M13-08  [#permalink]

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New post 03 Oct 2018, 17:26
let simplify the equation

2rs>r+s
rs-r>s-rs
r(s-1)>s(1-r)
r/s<(r-1)/(s-1)

stat-1

if r=3 and s=4 .... not true

but if r= 99 and s=2 ..... true

So can some one expain how the statement one itself sufficient.

Error detection will be appriciated on my expaination.......
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Re: M13-08  [#permalink]

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New post 03 Oct 2018, 21:17
rinkumaa4 wrote:
let simplify the equation

2rs>r+s
rs-r>s-rs
r(s-1)>s(1-r)
r/s<(r-1)/(s-1)


stat-1

if r=3 and s=4 .... not true

but if r= 99 and s=2 ..... true

So can some one expain how the statement one itself sufficient.

Error detection will be appriciated on my expaination.......


What is the point of algebraic manipulations you've done?

\(\frac{rs}{r + s} \gt \frac{1}{2}\) is true for r=3 and s=4 as well as for r= 99 and s=2.
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Re: M13-08  [#permalink]

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New post 14 Nov 2018, 05:35
Hi Bunuel,
Statement 1 is not sufficient
r > 1; s>1
Multiplying both we get rs >1;
Adding both we get r+s > 2;
if rs =2 and r+s =3, then rs/r+s > 1/2 - YES
If rs =1.5 and r+s =3, then rs/r+s =1/2 -- NO

Please clarify
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New post 14 Nov 2018, 07:18
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Re: M13-08  [#permalink]

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New post 15 Nov 2018, 01:55
rinkumaa4 wrote:
let simplify the equation

2rs>r+s
rs-r>s-rs
r(s-1)>s(1-r)
r/s<(r-1)/(s-1)


stat-1

if r=3 and s=4 .... not true

but if r= 99 and s=2 ..... true

So can some one expain how the statement one itself sufficient.

Error detection will be appriciated on my expaination.......


Hi

The problem that when you did as follows:

2rs>r+s
rs-r>s-rs
r(s-1)>s(1-r)
r(s-1)> -s(r-1)
Is r/s > -(r-1)/(s-1)??......When you changed the sign '<' you did not put the negative sign in next side
-r/s<(r-1)/(s-1)

Let take your examples again:
r=3 and s=4... -3/4 < 2/3....Yes
r= 99 and s=2... -99/2 < 98....Yes
In both cases left is negative side while right side is positive. So it is Sufficient

You could have stopped at the following step

r(s-1)>s(1-r)..........Based on Statement 1, the right will be always negative and right is positive..........Always Yes.

In same way, we can prove statement 2 is insufficient through following examples.

No need to duplicate work .Any example of the above will lead to yes.

Let s=1/2 & r =100

here if 0<S<1........the left side will always negative while the right side will be always positive...Answer is No.


I hope it helps.
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Re: M13-08  [#permalink]

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New post 03 Feb 2019, 02:42
Please correct me if my logic is wrong, but from statement (1), we can cross-multiply:

\(2rs>r+s?\)
Since r and s > 1, twice the product of r and s is always going to be larger than the sum.
E.g. if r = 1.1 and s = 1.2,
2rs = 2.64
r+s = 2.3
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Re: M13-08   [#permalink] 03 Feb 2019, 02:42
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