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# M19-34

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:07
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25% (medium)

Question Stats:

76% (01:02) correct 24% (01:14) wrong based on 131 sessions

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Is $$\frac{x}{x + y} \gt 2$$ ?

(1) $$x \gt 1$$, $$y \gt 1$$

(2) $$x \gt 2$$, $$y \gt 0$$
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132696 [1], given: 12335

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16 Sep 2014, 01:07
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Expert's post
Official Solution:

Is $$\frac{x}{x + y} \gt 2$$ ?

(1) $$x \gt 1$$ and $$y \gt 1$$. Since both $$x$$ and $$y$$ are positive then $$0 \lt x \lt x+y$$, so $$\frac{x}{x+y} \lt 1$$. Sufficient.

(2) $$x \gt 2$$ and $$y \gt 0$$. The same here: since both $$x$$ and $$y$$ are positive then $$0 \lt x \lt x+y$$, so $$\frac{x}{x+y} \lt 1$$. Sufficient.

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07 Sep 2017, 05:56
Is $$\frac{x}{x + y} \gt 2$$ ?

(1) $$x \gt 1$$, $$y \gt 1$$ : since x and y are positive giving any minimum value of y and max value of x will turn the fraction less than 1:so sufficient ,never more than 2

(2) $$x \gt 2$$, $$y \gt 0$$[/quote] : since x and y are positive giving any minimum value of y and max value of x will turn the fraction less than 1:so sufficient ,never more than 2
therefore D

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07 Sep 2017, 05:59
Bunuel wrote:
Official Solution:

(1) $$x \gt 1$$ and $$y \gt 1$$. Since both $$x$$ and $$y$$ are positive then $$0 \lt x \lt x+y$$, so $$\frac{x}{x+y} \lt 1$$. Sufficient.

(2) $$x \gt 2$$ and $$y \gt 0$$. The same here: since both $$x$$ and $$y$$ are positive then $$0 \lt x \lt x+y$$, so $$\frac{x}{x+y} \lt 1$$. Sufficient.

I got the same answer through substitution. Not sure, if that's the right approach.

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07 Sep 2017, 06:57
Bunuel wrote:
Official Solution:

(1) $$x \gt 1$$ and $$y \gt 1$$. Since both $$x$$ and $$y$$ are positive then $$0 \lt x \lt x+y$$, so $$\frac{x}{x+y} \lt 1$$. Sufficient.

(2) $$x \gt 2$$ and $$y \gt 0$$. The same here: since both $$x$$ and $$y$$ are positive then $$0 \lt x \lt x+y$$, so $$\frac{x}{x+y} \lt 1$$. Sufficient.

Hi Bunuel,
This was my reasoning to solve the problem, but unfortunately I think is wrong...could you please tell me why?

x/x+y>2; x/x +x/y>2; 1+x/y>2;x/y>1;x>y So at the end is x>y?

1) insufficient
2)insufficient
1)+2) insufficient
Thus, E

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Math Expert
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07 Sep 2017, 07:11
juanito1985 wrote:
Bunuel wrote:
Official Solution:

(1) $$x \gt 1$$ and $$y \gt 1$$. Since both $$x$$ and $$y$$ are positive then $$0 \lt x \lt x+y$$, so $$\frac{x}{x+y} \lt 1$$. Sufficient.

(2) $$x \gt 2$$ and $$y \gt 0$$. The same here: since both $$x$$ and $$y$$ are positive then $$0 \lt x \lt x+y$$, so $$\frac{x}{x+y} \lt 1$$. Sufficient.

Hi Bunuel,
This was my reasoning to solve the problem, but unfortunately I think is wrong...could you please tell me why?

x/x+y>2; x/x +x/y>2; 1+x/y>2;x/y>1;x>y So at the end is x>y?

1) insufficient
2)insufficient
1)+2) insufficient
Thus, E

First of all, please format properly. x/(x + y) is NOT the same as x/x + y.

Next:
1. You cannot split x/(x + y) into x/x + x/y.
2. Even if you could, you cannot cross-multiply x/y > 1 to get x > y because you don't know the sign of y.

I think you should brush up fundamentals on inequities and algebra.

Check respective topics here: https://gmatclub.com/forum/ultimate-gma ... 44512.html
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07 Sep 2017, 07:13
Bunuel wrote:
juanito1985 wrote:
Bunuel wrote:
Official Solution:

(1) $$x \gt 1$$ and $$y \gt 1$$. Since both $$x$$ and $$y$$ are positive then $$0 \lt x \lt x+y$$, so $$\frac{x}{x+y} \lt 1$$. Sufficient.

(2) $$x \gt 2$$ and $$y \gt 0$$. The same here: since both $$x$$ and $$y$$ are positive then $$0 \lt x \lt x+y$$, so $$\frac{x}{x+y} \lt 1$$. Sufficient.

Hi Bunuel,
This was my reasoning to solve the problem, but unfortunately I think is wrong...could you please tell me why?

x/x+y>2; x/x +x/y>2; 1+x/y>2;x/y>1;x>y So at the end is x>y?

1) insufficient
2)insufficient
1)+2) insufficient
Thus, E

First of all, please format properly. x/(x + y) is NOT the same as x/x + y.

Next:
1. You cannot split x/(x + y) into x/x + x/y.
2. Even if you could, you cannot cross-multiply x/y > 1 to get x > y because you don't know the sign of y.

I think you should brush up fundamentals on inequities and algebra.

Check respective topics here: https://gmatclub.com/forum/ultimate-gma ... 44512.html

Thank you Bunuel

Sent from my iPhone using GMAT Club Forum mobile app

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07 Sep 2017, 10:05
I suck at this stuff but I thought the answer was B

For 1. I used x = 2 and y=4. Therefore bringing the first option to a fraction. Why would this way of thinking be incorrect? Is this a case that 1. is wrong?

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07 Sep 2017, 10:13
Ajbrooks wrote:
I suck at this stuff but I thought the answer was B

For 1. I used x = 2 and y=4. Therefore bringing the first option to a fraction. Why would this way of thinking be incorrect? Is this a case that 1. is wrong?

yes if you used x=2 and y=4 then the fraction you get is 2 /6 (x=2 /(x +y) =6) clearly this is less than 1 so it is always less than 2 so its sufficient
similarly second case says that x>2 and y>0 so assume a value of x suppose 3 and consider y to be 0.0001 then also x/(x+y) = 3/3.0001 clearly a value less than 1 so it is always less than 2 so its sufficient

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12 Sep 2017, 07:03
Bunuel wrote:
Is $$\frac{x}{x + y} \gt 2$$ ?

(1) $$x \gt 1$$, $$y \gt 1$$

(2) $$x \gt 2$$, $$y \gt 0$$

Bunuel Can we not modify this question as $$x > 2x+2y ==> -x>2y$$

Then Ans is D Am I correct?
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Math Expert
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Posts: 42249

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12 Sep 2017, 07:31
NandishSS wrote:
Bunuel wrote:
Is $$\frac{x}{x + y} \gt 2$$ ?

(1) $$x \gt 1$$, $$y \gt 1$$

(2) $$x \gt 2$$, $$y \gt 0$$

Bunuel Can we not modify this question as $$x > 2x+2y ==> -x>2y$$

Then Ans is D Am I correct?

No, you cannot cross-multiply this way because you don't know the sign of x+y.
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Re: M19-34   [#permalink] 12 Sep 2017, 07:31
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# M19-34

Moderators: Bunuel, chetan2u

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