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# M13-08

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Math Expert
Joined: 02 Sep 2009
Posts: 59020
M13-08  [#permalink]

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16 Sep 2014, 00:48
00:00

Difficulty:

65% (hard)

Question Stats:

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

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Is $$\frac{rs}{r + s} \gt \frac{1}{2}$$?

(1) $$r \gt 1$$ and $$s \gt 1$$

(2) $$r \gt 2$$ and $$s \gt 0$$

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Posts: 8160
M13-08  [#permalink]

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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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Re M13-08  [#permalink]

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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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11 Sep 2016, 17:26
Why you wrote down (r+s/rs)<2? I couldn't understand the logic used.
Current Student
Joined: 01 Aug 2016
Posts: 12
Location: United States (AZ)
GMAT 1: 720 Q47 V41
GPA: 3.71
Re: M13-08  [#permalink]

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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)
Current Student
Joined: 01 Aug 2016
Posts: 12
Location: United States (AZ)
GMAT 1: 720 Q47 V41
GPA: 3.71
Re: M13-08  [#permalink]

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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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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.......
Math Expert
Joined: 02 Sep 2009
Posts: 59020
Re: M13-08  [#permalink]

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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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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
Math Expert
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Posts: 59020
Re: M13-08  [#permalink]

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14 Nov 2018, 07:18
sandeepg30 wrote:
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

rs =1.5 and r+s =3 does not give a solution for which both r and s are greater than 1, so this set is not possible.
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Re: M13-08  [#permalink]

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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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Joined: 18 Mar 2018
Posts: 27
Re: M13-08  [#permalink]

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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
Re: M13-08   [#permalink] 03 Feb 2019, 02:42
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# M13-08

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