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(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.

(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

(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.

(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

(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.

(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.

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?

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