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Re: Gprep DS: Is xy>o? [#permalink]
25 Jan 2011, 14:34

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... ((

Re: Gprep DS: Is xy>o? [#permalink]
25 Jan 2011, 14:43

Expert's post

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 \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

Re: Gprep DS: Is xy>o? [#permalink]
25 Jan 2011, 14:53

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 \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

As pointed out above, using graphs is extremely quick and efficient in such questions. The only region where x-y> -2 and x-2y<-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/quarter-wit-quarter-wisdom/ Part III discusses a question exactly like this.

As pointed out above, using graphs is extremely quick and efficient in such questions. The only region where x-y> -2 and x-2y<-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/quarter-wit-quarter-wisdom/ 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?)?

As pointed out above, using graphs is extremely quick and efficient in such questions. The only region where x-y> -2 and x-2y<-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/quarter-wit-quarter-wisdom/ 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"

As pointed out above, using graphs is extremely quick and efficient in such questions. The only region where x-y> -2 and x-2y<-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/quarter-wit-quarter-wisdom/ 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. _________________

"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"

Re: Is xy>0? (1) x-y>-2 (2) x-2y<-6 [#permalink]
17 Jun 2014, 09:08

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