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Re: xy>0 ? [#permalink]
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26 Jan 2011, 01:47
VeritasPrepKarishma wrote: 0987654312 wrote: Is xy>0? 1) xy> 2 2) x2y<6 Help!!?? ) As pointed out above, using graphs is extremely quick and efficient in such questions. The only region where xy> 2 and x2y<6 intersect is the first quadrant (as shown in the graph below). If you are uncomfortable with drawing accurate lines quickly, check out 'Bagging the graphs  Parts I, II and III' at http://www.veritasprep.com/blog/category/gmat/quarterwitquarterwisdom/Part III discusses a question exactly like this. Attachment: Ques1.jpg Dear Karishma, thank you very much for the link provided:very helpful! the approach with the graphs is very efficient! However, just one more comment as a clarification: I understand why the the lines are drawn in this way and i see that they intersect in Q1. However, how can i relate this to the question directily, i .e. what would the right logic be in order to answer the question (is xy>0?)?



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Re: xy>0 ? [#permalink]
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26 Jan 2011, 02:01
0987654312 wrote: VeritasPrepKarishma wrote: 0987654312 wrote: Is xy>0? 1) xy> 2 2) x2y<6 Help!!?? ) As pointed out above, using graphs is extremely quick and efficient in such questions. The only region where xy> 2 and x2y<6 intersect is the first quadrant (as shown in the graph below). If you are uncomfortable with drawing accurate lines quickly, check out 'Bagging the graphs  Parts I, II and III' at http://www.veritasprep.com/blog/category/gmat/quarterwitquarterwisdom/Part III discusses a question exactly like this. Attachment: Ques1.jpg Dear Karishma, thank you very much for the link provided:very helpful! the approach with the graphs is very efficient! However, just one more comment as a clarification: I understand why the the lines are drawn in this way and i see that they intersect in Q1. However, how can i relate this to the question directily, i .e. what would the right logic be in order to answer the question (is xy>0?)? Dear Karishma, one more question for you. I read the link you suggested and the explanations are brilliant ! I was wandering, can you provide me with a link to this: ( it is taken from the link posted below) "In part I of Graphs, I had also mentioned “Learn how to draw a line from its equation in under ten seconds and you shall solve the related question in under a minute" http://www.veritasprep.com/blog/2011/01 ... partiii/Thank you!!!!



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Re: xy>0 ? [#permalink]
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26 Jan 2011, 05:29
0987654312 wrote: VeritasPrepKarishma wrote: 0987654312 wrote: Is xy>0? 1) xy> 2 2) x2y<6 Help!!?? ) As pointed out above, using graphs is extremely quick and efficient in such questions. The only region where xy> 2 and x2y<6 intersect is the first quadrant (as shown in the graph below). If you are uncomfortable with drawing accurate lines quickly, check out 'Bagging the graphs  Parts I, II and III' at http://www.veritasprep.com/blog/category/gmat/quarterwitquarterwisdom/Part III discusses a question exactly like this. Attachment: Ques1.jpg Dear Karishma, thank you very much for the link provided:very helpful! the approach with the graphs is very efficient! However, just one more comment as a clarification: I understand why the the lines are drawn in this way and i see that they intersect in Q1. However, how can i relate this to the question directily, i .e. what would the right logic be in order to answer the question (is xy>0?)? Question: Is xy > 0 xy > 0 when either both x and y are positive or both are negative. If the intersection lies in only first quadrant, both x and y are positive (In I Quadrant, x > 0 and y > 0) If the intersection lies in only third quadrant then both x and y are negative because in III quadrant, x < 0 and y < 0. As long as your points lie in I and III quadrants only, xy will be > 0. If your points lie in II or IV quadrant too, xy can be negative too because either x or y (both not both) is negative in II and IV quadrant. Since in the graph above, the intersection lies only in I quadrant, it means x and y are both +ve and hence xy > 0.
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Re: xy>0 ? [#permalink]
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26 Jan 2011, 05:35



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Is xy > 0 ? 1) x  y > 2 2) x  2y < 6 [#permalink]
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Is xy > 0 ?
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Re: GMAT prep Inequality, positive negative [#permalink]
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01 Feb 2011, 02:58
tradinggenius wrote: Is xy > 0 ?
1) x  y > 2 2) x  2y < 6 Is xy>0?Note that question basically asks whether \(x\) and \(y\) have the same sign. (1) xy > 2 > we can have an YES answer, if for example \(x\) and \(y\) are both positive (\(x=10\) and \(y=1\)) as well as a NO answer, if for example \(x\) is positive and \(y\) is negative (\(x=10\) and \(y=10\)). Not sufficient. (2) x2y <6 > again it' easy to get an YES answer, if for example \(x\) and \(y\) are both positive (\(x=1\) and \(y=10\)) as well as a NO answer, if for example \(x\) is negative and \(y\) is positive (\(x=1\) and \(y=10\)). Not sufficient. You can get that the the two statement individually are not sufficient in another way too: we have (1) \(y<x+2\) and (2) \(y>\frac{x}{2}+3\). We are asked whether \(x\) and \(y\) have the same sign or whether the points (x,y) are in the I or III quadrant ONLY. But all (x,y) points below the line \(y=x+2\) (for 1) and all (x, y) points above the line \(y=\frac{x}{2}+3\) cannot lie only I or III quadrant: points above or below some line (not parallel to axis) lie at least in 3 quadrants. (1)+(2) Now, remember that we can subtract inequalities with the signs in opposite direction > subtract (2) from (1): \(xy(x2y)>2(6)\) > \(y>4\). As \(y>4\) and (from 1) \(x>y2\) then \(x>2\) (because we can add inequalities when their signs are in the same direction, so: \(y+x>4+(y2)\) > \(x>2\)) > we have that \(y>4\) and \(x>2\): both \(x\) and \(y\) are positive. Sufficient. Answer: C. Also discussed here: gprepdsisxyo87751.html
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Q: Is xy > 0 (1) XY > 2 (2) X2Y < 6 I received [#permalink]
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01 May 2011, 01:30
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Q: Is xy > 0
(1) XY > 2 (2) X2Y < 6
I received the following answer in a different post and cannot follow the logic  can anyone help?
together a and b , As the inequality signs are different, we have to subtract the equations. XY  (X2Y) > 2 + 6 Y > 4 meaning X > y 2 that is X>2 and Y >4 implies XY >0. Hence C.



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Re: Inequalities data sufficiency question [#permalink]
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01 May 2011, 01:59
chloeholding wrote: Q: Is xy > 0
(1) XY > 2 (2) X2Y < 6 Basically question is asking if x > 0 AND y >0 OR x < 0 AND y <0. Stmt1: x>y2. Now take y such that y>0. y=3, Check if x > 0 or not, x>1. Hence x >0. Hence xy >0. Take y =1, x >1. Hence x can be 0.5. Hence xy <0. Yes and No. Insufficient. Stmt2: x2y < 6. x<2(y3). Take y=2. x<2. Hence xy < 0. Take y=4, x<2. Hence xy>0 if x=1 and xy<0 if x=1. Yes and No. Insufficient. Combining, Multiplying (2) by 1 and change sign: x+2y > 6 Adding, xy+(x+2y) > 2+6 y > 4. and x>y2, hence x>2. On the number line, both x and y are to the right of 0. Hence both positive. x>0, y>0. xy>0 OA. C
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Re: Inequalities data sufficiency question [#permalink]
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01 May 2011, 02:07
Chloeholding: You are repeatedly making the same mistake. You must not post a PS problem in DS forum or viceversa. Please let me know if you have trouble following my instruction.
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Re: Inequalities data sufficiency question [#permalink]
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01 May 2011, 02:31
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chloeholding wrote: Q: Is xy > 0
(1) XY > 2 (2) X2Y < 6
I received the following answer in a different post and cannot follow the logic  can anyone help?
together a and b , As the inequality signs are different, we have to subtract the equations. XY  (X2Y) > 2 + 6 Y > 4 meaning X > y 2 that is X>2 and Y >4 implies XY >0. Hence C. doing same as jamifahad for both options as: Stmt1: x>y2. Now take y such that y>0. y=3, Check if x > 0 or not, x>1. Hence x >0. Hence xy >0. Take y =1, x >1. Hence x can be 0.5. Hence xy <0. Yes and No. Insufficient. Stmt2: x2y < 6. x<2(y3). Take y=2. x<2. Hence xy < 0. Take y=4, x<2. Hence xy>0 if x=1 and xy<0 if x=1. Yes and No. Insufficient. But for c xyx+2y>2(6) = y>8 so from 1 x > y2 x > 92 x > 7 so xy>0
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Re: Inequalities data sufficiency question [#permalink]
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02 May 2011, 10:16
chloeholding wrote: Q: Is xy > 0
(1) XY > 2 (2) X2Y < 6
I received the following answer in a different post and cannot follow the logic  can anyone help?
together a and b , As the inequality signs are different, we have to subtract the equations. XY  (X2Y) > 2 + 6 Y > 4 meaning X > y 2 that is X>2 and Y >4 implies XY >0. Hence C. 1. X+2>Y. Plug X=0 and Y=1. Answer is NO. Put X=1 and Y=2. Answer is YES. 2 answers. insufficient. 2. X+6<2Y. Plug different values and you can get 2 different answers. Insufficient 1 and 2 combined: Subtract eqn 1 from 2. Use the "GT/ LT"(GREATER THAN/ LESS THAN) terminology: 4<Y or Y= GT 4Put this value in eqn 1: X=GT42. This implies that X=GT2. Therefore both X and Y are positive. Hence XY>0. Answer C.
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Re: Inequalities data sufficiency question [#permalink]
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04 May 2011, 20:22
(1) x = 0, y = 0 xy = 0 > 2, but xy = 0 x = 1, y = 2, but xy > 0 Insufficient (2) x = 1, y = 5, xy > 0 (x  y = 1  10 = 9 < 6) x = 1, y = 3, xy < 0 (x  y = 7 < 6) (1) + (2) xy > 2 2y x > 6 (multiplying by 1) Adding both, y > 4 => x > y  2 => x is also positive xy > 0 Answer  C
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Re: Is xy > 0 ? 1) x  y > 2 2) x  2y < 6 [#permalink]
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12 Jul 2013, 11:44
tradinggenius wrote: Is xy > 0 ?
1) x  y > 2 2) x  2y < 6 Hi, statement 1: y<x+2===>this is straight line in coordinate system==>and all values below line(includes coordinates of all 4 coordinate ) will satisfy this condition==>hence xy can be positive or negative or zero==>not sufficient. statement 2: y>x/2+3===>again straight line==>and all values above line(includes coordinates of 1/2/3 coordinate ) will satisfy this condition==>hence xy can be positive or negative ==>not sufficient. combining 2 we can get the unique value for x/y==>hence we can say xy is positive/negative defenetely.....hence sufficient C
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Re: Is xy > 0 ? 1) x  y > 2 2) x  2y < 6 [#permalink]
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13 Jul 2013, 06:35
tradinggenius wrote: Is xy > 0 ?
1) x  y > 2 2) x  2y < 6 Is xy > 0 ? 1) x  y > 2 2) x  2y < 6 1) X = 2 and Y = 3 XY > 0 X =  0.5 and Y = 1 XY < 0 2) X = 2 and Y = 3 XY > 0 X = 0.5 and Y = 3 XY < 0 Now combine both X> Y  2 X < 6 + 2y The only values that suffice are values greater than 4 for y and greater than 2 for x. Hence C is the answer.
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Re: Is xy > 0? (1) x  y > 2 (2) x  2y < 6 [#permalink]
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Is xy>0?Note that question basically asks whether \(x\) and \(y\) have the same sign. (1) xy > 2 > we can have an YES answer, if for example \(x\) and \(y\) are both positive (\(x=10\) and \(y=1\)) as well as a NO answer, if for example \(x\) is positive and \(y\) is negative (\(x=10\) and \(y=10\)). Not sufficient. (2) x2y <6 > again it' easy to get an YES answer, if for example \(x\) and \(y\) are both positive (\(x=1\) and \(y=10\)) as well as a NO answer, if for example \(x\) is negative and \(y\) is positive (\(x=1\) and \(y=10\)). Not sufficient. You can get that the the two statement individually are not sufficient in another way too: we have (1) \(y<x+2\) and (2) \(y>\frac{x}{2}+3\). We are asked whether \(x\) and \(y\) have the same sign or whether the points (x,y) are in the I or III quadrant ONLY. But all (x,y) points below the line \(y=x+2\) (for 1) and all (x, y) points above the line \(y=\frac{x}{2}+3\) can not lie only I or III quadrant: points above or below some line (not parallel to axis) lie at least in 3 quadrants. (1)+(2) Now, remember that we can subtract inequalities with the signs in opposite direction > subtract (2) from (1): \(xy(x2y)>2(6)\) > \(y>4\). As \(y>4\) and (from 1) \(x>y2\) then \(x>2\) (because we can add inequalities when their signs are in the same direction, so: \(y+x>4+(y2)\) > \(x>2\)) > we have that \(y>4\) and \(x>2\): both \(x\) and \(y\) are positive. Sufficient. Answer: C. OPEN DISCUSSION OF THIS QUESTION IS HERE: isxy01xy22x2y114731.html
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Re: Is xy>0? (1) xy>2 (2) x2y<6 [#permalink]
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01 Nov 2017, 22:52
4test1 wrote: Is xy>0?
(1) xy>2 (2) x2y<6 Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Since we have 2 variables and 0 equation, C could be the answer most likely. 1) & 2) x  y > 2 x  2y < 6 <=> x + 2y > 6 When we add two inequalities, y > 4. And we have x > y  2 > 2. Both x and y are positive and xy > 0. 1) x = 3, y = 1 => xy > 0 : Yes x = 0, y = 1 => xy = 0 : No This is not sufficient. 2) x = 0, y = 4 => xy = 0 : No x = 1, y = 4 => xy > 0 : Yes Therefore, the answer is C. Normally 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 C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. 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 according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: Is xy>0? (1) xy>2 (2) x2y<6 [#permalink]
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