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Re: Gprep DS: Is xy>o? [#permalink]
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25 Jan 2011, 15:34
Dear Bunuel, cold you please explain this " Now, remember we can subtract inequalities with the signs in opposite direction." I seem to have forgotten basic arithmetic operations... (( Thank youuu



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Re: Gprep DS: Is xy>o? [#permalink]
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25 Jan 2011, 15:43
0987654312 wrote: Dear Bunuel, cold you please explain this " Now, remember we can subtract inequalities with the signs in opposite direction." I seem to have forgotten basic arithmetic operations... (( Thank youuu You can only add inequalities when their signs are in the same direction:If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) > \(a+c>b+d\). Example: \(3<4\) and \(2<5\) > \(3+2<4+5\). You can only apply subtraction when their signs are in the opposite directions:If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). Hope it's clear.
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Re: Gprep DS: Is xy>o? [#permalink]
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25 Jan 2011, 15:53
Bunuel wrote: 0987654312 wrote: Dear Bunuel, cold you please explain this " Now, remember we can subtract inequalities with the signs in opposite direction." I seem to have forgotten basic arithmetic operations... (( Thank youuu You can only add inequalities when their signs are in the same direction:If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) > \(a+c>b+d\). Example: \(3<4\) and \(2<5\) > \(3+2<4+5\). You can only apply subtraction when their signs are in the opposite directions:If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). Hope it's clear. Thank you so much!!!



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Re: xy>0 ? [#permalink]
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25 Jan 2011, 20:08
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 [ 15.35 KiB  Viewed 1022 times ]
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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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Re: Is xy>0? (1) xy>2 (2) x2y<6 [#permalink]
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Is xy>0? (1) xy>2 (2) x2y<6 [#permalink]
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06 Nov 2015, 22:47
hi bunuel, plz let me know where i m wrong at here, when i put x>2 in 1st stem , it pops ; y<x+2 , y could be ve , thus i zeroed to E, however i know oa is C




Is xy>0? (1) xy>2 (2) x2y<6
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