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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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06 Sep 2015, 00:52
Bunuel wrote: Are x and y both positive?(1) 2x2y=1 (2) x/y>1 (1) 2x2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x2y=1 > y=x1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient. (2) x/y>1 > x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient. (1)+(2) Again it can be done with different approaches. You should just find the one which is the less timeconsuming and comfortable for you personally. One of the approaches: \(2x2y=1\) > \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) > \(\frac{xy}{y}>0\) > substitute x > \(\frac{1}{y}>0\) > \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient. Answer: C. Discussed here: ds193964.html?hilit=number%20plugging%20consider%20approaches and also here along with other hard inequality problems: inequalityandabsolutevaluequestionsfrommycollection86939.htmlHope it helps. How do you form these terms in the underlined part? from x/y>1 to xy/y > 0 ? I do not understand this concept. Thanks.
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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06 Sep 2015, 04:20
reto wrote: Bunuel wrote: Are x and y both positive?(1) 2x2y=1 (2) x/y>1 (1) 2x2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x2y=1 > y=x1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient. (2) x/y>1 > x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient. (1)+(2) Again it can be done with different approaches. You should just find the one which is the less timeconsuming and comfortable for you personally. One of the approaches: \(2x2y=1\) > \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) > \(\frac{xy}{y}>0\) > substitute x > \(\frac{1}{y}>0\) > \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient. Answer: C. Discussed here: ds193964.html?hilit=number%20plugging%20consider%20approaches and also here along with other hard inequality problems: inequalityandabsolutevaluequestionsfrommycollection86939.htmlHope it helps. How do you form these terms in the underlined part? from x/y>1 to xy/y > 0 ? I do not understand this concept. Thanks. This is basic: \(\frac{x}{y}>1\); \(\frac{x}{y}1>0\); \(\frac{x}{y}\frac{y}{y}>0\); \(\frac{xy}{y}>0\). Was explained here: arexandybothpositive12x2x12xy63377.html#p1060399
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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06 Sep 2015, 05:13
reto wrote: Bunuel wrote: Are x and y both positive?(1) 2x2y=1 (2) x/y>1 (1) 2x2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x2y=1 > y=x1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient. (2) x/y>1 > x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient. (1)+(2) Again it can be done with different approaches. You should just find the one which is the less timeconsuming and comfortable for you personally. One of the approaches: \(2x2y=1\) > \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) > \(\frac{xy}{y}>0\) > substitute x > \(\frac{1}{y}>0\) > \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient. Answer: C. Discussed here: ds193964.html?hilit=number%20plugging%20consider%20approaches and also here along with other hard inequality problems: inequalityandabsolutevaluequestionsfrommycollection86939.htmlHope it helps. How do you form these terms in the underlined part? from x/y>1 to xy/y > 0 ? I do not understand this concept. Thanks. reto, look below: You get from statement 1, 2x2y=1 > \(x=\frac{1+2y}{2}\).....(1) From statement 2, \(\frac{x}{y} >1\)> \(\frac{x}{y}1>0\) >\(\frac{xy}{y} > 0\) , now substitute for x from (1) above, you get, \(\frac{\frac{1+2y}{2} y}{y} > 0\) > \(\frac{\frac{1+2y2y}{2}}{y} > 0\) > \(\frac{1}{2y} >0\) > \(\frac{1}{y} >0\) > \(y >0\)and as from (1), \(x=\frac{1+2y}{2}\), for y>0, x>0 as well. Alternately, you can view it as:\(\frac{x}{y} >1\) > x and y both have the same sign (i.e. both of them are either <0 or >0). x>0 > y >0 and when x<0 > y <0 and vice versa. This is true when \(\frac{x}{y} > 0\) \(\frac{x}{y} <1\) > x and y both have different sign i.e. x>0 > y <0 and when x<0 > y >0 and vice versa. This is true when \(\frac{x}{y} < 0\) Hope this helps.



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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06 Sep 2015, 20:31
Hi Bunuel I understoof your line of solving this question. But just wanted to say in GMAT two statements cannot contradict each other. Statement 1 says that slope of line is 1, which means x and y coordinates will always have same value, which means the ratio x/y = 1 always. However statement 2 says that x/y>1, how is this possible. This contradicts statement 1. Correct me if I am wrong.



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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06 Sep 2015, 20:48
Shreeya24 wrote: Hi Bunuel I understoof your line of solving this question. But just wanted to say in GMAT two statements cannot contradict each other. Statement 1 says that slope of line is 1, which means x and y coordinates will always have same value, which means the ratio x/y = 1 always. However statement 2 says that x/y>1, how is this possible. This contradicts statement 1. Correct me if I am wrong. You are correct in saying that 2x2y=1 has slope of 1 but it is not a must that this statement means that x/y = 1. Having slope of 1 does not mean that the x and y coordinates will be equal (to make x/y =1). You can confirm this by looking at the set of values satisfying the above equation: (1.5,1) (2.5,2) (3,2.5) So you see that none of these sets mean x/y > 1. But your understanding that an official DS question will never have contradictory statements is correct. Though this is not the case for this question. Hope this helps.



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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24 Sep 2015, 16:00
What happens when y = o?
X = 0.5 and y = 0, satisfies i and x/y is indeed > 0 since anything divided by 0 is infinity. But, despite this, y is not positive as it '0'.
Even in this case 
"One of the approaches: 2x−2y=1 > x=y+12 xy>1 > x−yy>0 > substitute x > 1y>0 > y is positive, and as x=y+12, x is positive too. Sufficient."
The moment you assume y = 0, you can easily see that 1/y is infinity that is greater than 0 and hence, satisfies the equation.
Am i missing something. My answer would be 'E'



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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24 Sep 2015, 16:16
Manbehindthecurtain wrote: Are x and y both positive?
(1) 2x2y = 1 (2) x/y > 1 Target question: Are x and y both positive? Statement 1: 2x  2y = 1 There are several pairs of numbers that satisfy this condition. Here are two: Case a: x = 1 and y = 0.5, in which case x and y are both positiveCase b: x = 0.5 and y = 1, in which case x and y are not both positiveSince we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT Statement 2: x/y > 1 This tells us that x/y is positive. This means that either x and y are both positive or x and y are both negative. Here are two possible cases: Case a: x = 4 and y = 2, in which case x and y are both positiveCase b: x = 4 and y = 2, in which case x and y are not both positiveSince we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT Statements 1 and 2Statement 1 tells us that 2x  2y = 1. Divide both sides by 2 to get: x  y = 1/2 Solve for x to get x = y + 1/2 Now take the statement 2 inequality (x/y > 1) and replace x with y + 1/2 to get: (y + 1/2)/y > 1 Rewrite as: y/y + (1/2)/y > 1 Simplify: 1 + 1/(2y) > 1 Subtract 1 from both sides: 1/(2y) > 0 If 1/(2y) is positive, then y must be positive. Statement 2 tells us that either x and y are both positive or x and y are both negative. Now that we know that y is positive, it must be the case that x and y are both positiveSince we can now answer the target question with certainty, the combined statements are SUFFICIENT Answer: C Cheers, Brent
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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24 Sep 2015, 16:21
jitendra31 wrote: What happens when y = o? X = 0.5 and y = 0, satisfies i and x/y is indeed > 0 since anything divided by 0 is infinity. But, despite this, y is not positive as it '0'. ' It may seem that 0.5/0 = infinity, but this is not the case. If we approach 0 from the positive side, then it looks like 0.5/0 is a REALLY BIG POSITIVE NUMBER 0.5/0.1 = 5 0.5/0.01 = 50 0.5/0.001 = 500 0.5/0.0001 = 5000 0.5/0.00001 = 50000 etc. But what if we approach 0 from the NEGATIVE side: 0.5/(0.1) = 5 0.5/(0.01) = 50 0.5/(0.001) = 500 0.5/(0.0001) = 5000 0.5/(0.00001) = 50000 Here it looks like 0.5/0 will be a REALLY BIG NEGATIVE NUMBER This is why we say that x/0 is undefined. Cheers, Brent
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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24 Sep 2015, 17:05
jitendra31 wrote: What happens when y = o?
X = 0.5 and y = 0, satisfies i and x/y is indeed > 0 since anything divided by 0 is infinity. But, despite this, y is not positive as it '0'.
Even in this case 
"One of the approaches: 2x−2y=1 > x=y+1/2 x/y>1 > x−y/y>0 > substitute x > 1/y>0 > y is positive, and as x=y+1/2, x is positive too. Sufficient."
The moment you assume y = 0, you can easily see that 1/y is infinity that is greater than 0 and hence, satisfies the equation.
Am i missing something. My answer would be 'E' As a rule for GMAT quant, a/0 = NOT DEFINED and as such you MUST NOT CONSIDER this case. BY definition, infinity can be both + or  . So your logic of using 'infinity' does not stand. Also , as mentioned before by me you should never consider any case with 0 in the denominator. If there is a case that is for example, a/(b2) = 3 then for GMAT, you must have b \(\neq\) 2 This is GMAT's rules and hence you have to follow them.



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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22 Jun 2016, 10:05
As per my solution i am getting E. Kindly help me understand where i am going wrong as the OA is 'C'
1. First statement: 2(x)2(y)=1 > xy=0.5
Now two cases:
(a) X=2.5 , y= 2 , difference is 0.5 > Both are positives (b) X= 2, y= 2.5 , difference is 0.5 > both are negatives
Hence insufficient
2. Second statement: (x/y) >1 > x>y
Both the numbers can be positive or negative hence insuff
1+2 : Both the statements combined
xy= 0.5 & x>y
Now two cases:
(a) X=2.5 , y= 2 , difference is 0.5 > Both are positives ( x>y) (b) X= 2, y= 2.5 > 2 (2.5) is equal to 2 + 2.5 which results into 0.5. Also 2 > 2.5
difference is 0.5 > both are negatives
This is insuff, hence 'E' should be the answer



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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22 Jun 2016, 10:15
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sgrover18 wrote: As per my solution i am getting E. Kindly help me understand where i am going wrong as the OA is 'C'
1. First statement: 2(x)2(y)=1 > xy=0.5
Now two cases:
(a) X=2.5 , y= 2 , difference is 0.5 > Both are positives (b) X= 2, y= 2.5 , difference is 0.5 > both are negatives
Hence insufficient
2. Second statement: (x/y) >1 > x>y
Both the numbers can be positive or negative hence insuff
1+2 : Both the statements combined
xy= 0.5 & x>y
Now two cases:
(a) X=2.5 , y= 2 , difference is 0.5 > Both are positives ( x>y) (b) X= 2, y= 2.5 > 2 (2.5) is equal to 2 + 2.5 which results into 0.5. Also 2 > 2.5
difference is 0.5 > both are negatives
This is insuff, hence 'E' should be the answer Couple of things: 1. x/y >1 DOES NOT mean that x>y (think x=3 and y=2). It only means that both x and y will be of same sign. 2. When x=2, y= 2.5, you are going against S2 that x/y >1 , your case makes it the opposite, x/y < 1. This is where you are making a mistake. By combining the 2 statements, you will get C as the correct answer.



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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22 Jun 2016, 11:20
Engr2012 wrote: sgrover18 wrote: As per my solution i am getting E. Kindly help me understand where i am going wrong as the OA is 'C'
1. First statement: 2(x)2(y)=1 > xy=0.5
Now two cases:
(a) X=2.5 , y= 2 , difference is 0.5 > Both are positives (b) X= 2, y= 2.5 , difference is 0.5 > both are negatives
Hence insufficient
2. Second statement: (x/y) >1 > x>y
Both the numbers can be positive or negative hence insuff
1+2 : Both the statements combined
xy= 0.5 & x>y
Now two cases:
(a) X=2.5 , y= 2 , difference is 0.5 > Both are positives ( x>y) (b) X= 2, y= 2.5 > 2 (2.5) is equal to 2 + 2.5 which results into 0.5. Also 2 > 2.5
difference is 0.5 > both are negatives
This is insuff, hence 'E' should be the answer Couple of things: 1. x/y >1 DOES NOT mean that x>y (think x=3 and y=2). It only means that both x and y will be of same sign. 2. When x=2, y= 2.5, you are going against S2 that x/y >1 , your case makes it the opposite, x/y < 1. This is where you are making a mistake. By combining the 2 statements, you will get C as the correct answer. Thank you so much ! That makes perfect sense. I guess I forgot the basic rule that if numbers are negative on the either side of the inequality then the sign of the inequality reverses (as you quoted 3<2)



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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22 Jun 2016, 12:49
From statement 1 : 2x2y=1 2*(xy)=1 xy=\(\frac{1}{2}\) x=\(\frac{1}{2}\)+y
So x>y Not sufficient
From statement 2 :
\(\frac{x}{y}\)>1>0
So, 2 cases :  x and y > 0 and x>y  x and y <0 and x<y Not Sufficient
From statement 1 & 2 combined :
We know that x>y from statement 1 and we know that x and y > 0 from statement 2
Sufficient, answer choice C



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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22 Jun 2016, 21:17
There are 2 variables (x and y) in the original condition. In order to match the number of variables and the number of equations, we need 2 equations. Since the condition 1) and 2) each has 1 equation, there is high chance that the correct answer is C. Using 1) and 2), from 2(xy)=1>0 we get 2(xy)>0, x>y. Using 2), if we multiply both sides by y^2, we get xy>y^2, xyy^2>0, y(xy)>0. Since xy>0, we get y>0. From x>y>0, x>0. The answer is always yes and the condition is sufficient. Hence, the correct answer is C. For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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22 Jun 2016, 22:03
Manbehindthecurtain wrote: Are x and y both positive?
(1) 2x2y = 1 (2) x/y > 1 Statement 1 2x2y = 1 x  y = (1/2) x = 3/2; y = 1 ; both x and y positive x = 1/4, y = 1/4; x is positive, y is negative so not sufficient Statement 2 (x/y) > 1 x = 3/2; y = 1 ; (x/y) is more than 1. both x and y positive x = 2; y = 1 ; (x/y) is more than 1. both x and y negative so not sufficient Combining Statement 1 and 2 x  y = 1/2 x = y+(1/2) x/y > 1 {y+(1/2)} / y is more than 1 1+(1/2y) > 1 (1/2y) > 0 y has to be positive. because x = y+1/2, so x also has to be positive. C is the answer.



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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22 Aug 2016, 08:45
Hey everyone,
Solved this question by simply picking numbers.
St:1 2x2y=1, xy = 1/2, Choice 1: x=3, y=2.5 (Satisfies the prompt), Choice 2: x=3, y=3.5 (Yields a diff result). NOT SUFFICIENT
St:2 x/y >1, Choice 1 satisfies the prompt, Assumes some negative values (such as x=5, y=10) (Yields a diff result). NOT SUFFICIENT.
Combining both: Choice 1 works. Hence, C.



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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27 Oct 2016, 00:26
Manbehindthecurtain wrote: Are x and y both positive?
(1) 2x2y = 1 (2) x/y > 1 clearly each aint suff both from 2 (xy)/y >0 and from 1 , xy = 1/2 , thus 1/2y>0 thus y is +ve and since x/y>1 and y is +ve therefore x is +ve too ... C



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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27 Oct 2016, 01:01
Manbehindthecurtain wrote: Are x and y both positive?
(1) 2x2y = 1 (2) x/y > 1 1. 2x2y = 1 ==> x= y +(1/2) ......[1] not suuficient 2. x/y>1 ==> x/y>1 .....[2] not sufficient combining, replace x from [1] into [2] (y+(1/2))/y > 1, ==> 1/y>2 ==> y is positive, as x>y [from [2]], x is also positive. Therefore, option C



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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27 Oct 2016, 17:16
Manbehindthecurtain wrote: Are x and y both positive?
(1) 2x2y = 1 (2) x/y > 1 Solution: We need to determine whether x and y are both positive. Statement One Alone: 2x – 2y = 1 Simplifying statement one we have: 2(x – y) = 1 x – y = ½ The information in statement one is not sufficient to determine whether x and y are both positive. For example, if x = 1 and y = ½, x and y are both positive; however, if x = 1/2 and y = 1, x and y are both negative. We can eliminate answer choices A and D. Statement Two Alone: x/y> 1 Using the information in statement two, we see that x and y can both be positive or both be negative. Statement two alone is not sufficient to answer the question. We can eliminate answer choice B. Statements One and Two Together: Using the information in statements one and two, we know that x – y = ½ and that x/y > 1. Isolating x in the equation we have: x = ½ + y. We can now substitute ½ + y for x in the inequality x/y > 1 and we have: (1/2 + y)/y > 1 (1/2)/y + y/y > 1 1/(2y) + 1 > 1 1/(2y) > 0 Thus, y must be greater than zero. Since x = ½ + y, x also must be greater than zero. Answer: C
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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11 Nov 2016, 21:56
I solved this using a combination of logic and testing general cases.
For (1), think of xy = 1/2 as the difference between two points on the number line. This difference can occur at if 1. x and y = + (AND x>y) 2. x and y =  (AND y>x) 3. x = +, y=  (AND x>y)
Clearly not sufficient.
(2) tells us that x and y have the same sign AND y>x. Also not sufficient on it's own.
Finally, using (2) we can rule out the first and third options for (1) and you end up with C as the answer.




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