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M19-34

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M19-34 [#permalink]

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New post 16 Sep 2014, 01:07
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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
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Re: M19-34 [#permalink]

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New post 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: M19-34 [#permalink]

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New post 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: M19-34 [#permalink]

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New post 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: M19-34 [#permalink]

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New post 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 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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M19-34 [#permalink]

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New post 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 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


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Re: M19-34 [#permalink]

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New post 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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New post 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: M19-34 [#permalink]

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New post 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: M19-34 [#permalink]

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New post 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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Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M19-34   [#permalink] 12 Sep 2017, 07:31
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