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Re M1934 [#permalink]
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16 Sep 2014, 01:07



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Re: M1934 [#permalink]
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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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Re: M1934 [#permalink]
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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.
Answer: D I got the same answer through substitution. Not sure, if that's the right approach.



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Re: M1934 [#permalink]
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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.
Answer: D 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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Re: M1934 [#permalink]
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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.
Answer: D 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 crossmultiply 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/ultimategma ... 44512.html
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Re: M1934 [#permalink]
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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.
Answer: D 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 crossmultiply 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/ultimategma ... 44512.htmlThank you Bunuel Sent from my iPhone using GMAT Club Forum mobile app



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Re: M1934 [#permalink]
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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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Re: M1934 [#permalink]
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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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Re: M1934 [#permalink]
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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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Re: M1934 [#permalink]
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12 Sep 2017, 07:31



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Bunuel wrote: 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.
Answer: D Bunuel, do you have questions similar to this question, especially when the statements say not only positive, but negative option?
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Re: M1934 [#permalink]
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