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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"

Note that question basically asks whether \(x\) and \(y\) have the same sign.

(1) x-y > -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) x-2y <-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): \(x-y-(x-2y)>-2-(-6)\) --> \(y>4\). As \(y>4\) and (from 1) \(x>y-2\) then \(x>2\) (because we can add inequalities when their signs are in the same direction, so: \(y+x>4+(y-2)\) --> \(x>2\)) --> we have that \(y>4\) and \(x>2\): both \(x\) and \(y\) are positive. Sufficient.

Q: Is xy > 0 (1) X-Y > -2 (2) X-2Y < -6 I received [#permalink]

Show Tags

01 May 2011, 00:30

1

This post received KUDOS

Q: Is xy > 0

(1) X-Y > -2 (2) X-2Y < -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. X-Y - (X-2Y) > -2 + 6 Y > 4 meaning X > y -2 that is X>2 and Y >4 implies XY >0. Hence C.

Re: Inequalities data sufficiency question [#permalink]

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01 May 2011, 00:59

chloeholding wrote:

Q: Is xy > 0

(1) X-Y > -2 (2) X-2Y < -6

Basically question is asking if x > 0 AND y >0 OR x < 0 AND y <0.

Stmt1: x>y-2. 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: x-2y < -6. x<2(y-3). 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, x-y+(-x+2y) > -2+6 y > 4. and x>y-2, 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
_________________

My dad once said to me: Son, nothing succeeds like success.

Re: Inequalities data sufficiency question [#permalink]

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01 May 2011, 01:07

Chloeholding: You are repeatedly making the same mistake. You must not post a PS problem in DS forum or vice-versa. Please let me know if you have trouble following my instruction.

Re: Inequalities data sufficiency question [#permalink]

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01 May 2011, 01:31

1

This post received KUDOS

chloeholding wrote:

Q: Is xy > 0

(1) X-Y > -2 (2) X-2Y < -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. X-Y - (X-2Y) > -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>y-2. 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: x-2y < -6. x<2(y-3). 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 x-y-x+2y>2-(-6) = y>8

so from 1 x > y-2 x > 9-2 x > 7 so xy>0
_________________

Re: Inequalities data sufficiency question [#permalink]

Show Tags

02 May 2011, 09:16

chloeholding wrote:

Q: Is xy > 0

(1) X-Y > -2 (2) X-2Y < -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. X-Y - (X-2Y) > -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 4

Put this value in eqn 1: X=GT4-2. This implies that X=GT2.

Therefore both X and Y are positive. Hence XY>0.

Answer C.
_________________

My GMAT debrief: http://gmatclub.com/forum/from-620-to-710-my-gmat-journey-114437.html

Re: Is xy > 0 ? 1) x - y > -2 2) x - 2y < -6 [#permalink]

Show Tags

12 Jul 2013, 10:44

tradinggenius wrote:

Is xy > 0 ?

1) x - y > -2 2) x - 2y < -6

Hi,

statement 1: y<x+2===>this is straight line in co-ordinate system==>and all values below line(includes co-ordinates of all 4 co-ordinate ) 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 co-ordinates of 1/2/3 co-ordinate ) 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
_________________

When you want to succeed as bad as you want to breathe ...then you will be successfull....

GIVE VALUE TO OFFICIAL QUESTIONS...

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Note that question basically asks whether \(x\) and \(y\) have the same sign.

(1) x-y > -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) x-2y <-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): \(x-y-(x-2y)>-2-(-6)\) --> \(y>4\). As \(y>4\) and (from 1) \(x>y-2\) then \(x>2\) (because we can add inequalities when their signs are in the same direction, so: \(y+x>4+(y-2)\) --> \(x>2\)) --> we have that \(y>4\) and \(x>2\): both \(x\) and \(y\) are positive. Sufficient.

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.
_________________