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Is xy > 0? (1) x  y > 2 (2) x  2y < 6 [#permalink]
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Is xy > 0? (1) x  y > 2 (2) x  2y < 6 OPEN DISCUSSION OF THIS QUESTION IS HERE: isxy01xy22x2y114731.html
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Originally posted by hasham222 on 03 Aug 2009, 04:11.
Last edited by Bunuel on 14 Sep 2014, 17:35, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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Re: Is xy > 0? (1) x  y > 2 (2) x  2y < 6 [#permalink]
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03 Aug 2009, 07:36
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If you try to rephrase this question, you basically have to find out if x and y have the same sign. So if x and y are negative ()() or positive (+)(+), then xy > 0. If they have opposite signs ()(+), then xy < 0.
In other words, what you are trying to find is if you can find combinations of x and y that satisfy the conditions. They will be SUFFICIENT if the condition only provides same sign combinations. They will be INSUFFICIENT if they allow both same sign combinations AND opposite sign combinations.
A. xy> 2 Insufficient x = 1, y = (1/2) 1  (1/2) = 1/2 > 2 x = 2, y = 1 2  (1) = 3 > 2
So you have both same sign combinations and opposite sign combinations that work. x*y can be both positive OR negative.
B. x2y<6 Insufficient x= 1 y = 7 1  14 = 8 < 6 x= 1 y = 8 1  16 = 15 < 6
Again, you have both same sign combinations and opposite sign combinations that work. x*y can be both positive OR negative.
C. So lets combine the two. This next part took some Googling because I did not know how to combine 2 separate inequalities. You can add inequalities like you can add two equations, however, you have to make sure that their signs are facing in the same direction.
Initially we have
x  y > 2 x  2y < 6
You flip the second one to
6 > x  2y
Now you can add the two equations
x  y > 2 6 > x  2y Add the two
x  y 6 > x  2y y > 4
So we know that Y > 8. Let's combine that with the inequality x  y > 2
x  y > 2 y > 8 Add the two
x  y + y > 2 + 8 x > 6
So x > 6 and Y > 4, both are positive. x*y MUST be positive. SUFFICIENT.



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Re: Is xy > 0? (1) x  y > 2 (2) x  2y < 6 [#permalink]
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03 Aug 2009, 08:30
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is xy>0 ie: do they have the same sign??
from 1: xy>2
can they have different sign in this inquality...yes....insuff
x2y<6 = 2y
can they have different signs in this inq...yes..insuff
both subtract
xyx+2y>2(6) = y>8, subst in any thus x is +ve too
C



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Re: Is xy > 0? (1) x  y > 2 (2) x  2y < 6 [#permalink]
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03 Aug 2009, 08:31
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hasham222 wrote: Guys, your 2 cents on this question please.
will post the OA soon. 1. x +2 > y 2. 2y  x > 6 Each statement alone is not suff. So, add 1 and 2: 2y + 2 > y + 6 y > 4 x > 2 C.
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Re: Is xy > 0? (1) x  y > 2 (2) x  2y < 6 [#permalink]
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1 and 2 alone can't be sufficient by themselves, since we can find counter examples to prove them wrong by setting either x or y to be negative and the other positive
when put 1 and 2 together, we get a solvable set: x  y > 2 2y  x > 6 (multiply statement 2 by 1 on both sides)
y > 4 x > 2



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Is xy>0? (1) xy>2 (2) x2y<6 [#permalink]
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Is xy>0?
(1) xy>2 (2) x2y<6
Originally posted by 4test1 on 08 Dec 2009, 09:07.
Last edited by Bunuel on 26 May 2012, 03:08, edited 1 time in total.
Edited the question and added the OA



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Re: Gprep DS: Is xy>o? [#permalink]
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4test1 wrote: Can't seem to find this in the forum. From the Gprep test:
Is xy>0? 1. xy > 2 2. x2y <6
Can't seem to get my head around this. Appreciate any help. answer c Statement 1: x = 5 and y = 4 answer is yes x = 1 and y = 0 answer is no Statement 2: x = 7 and y = 0 answer is no x = 10 and y = 1 answer is yes Combined: x  y > 2 x  2y < 6 or x + y < 2 x  2y <  6 y < 4 y > 4 so you know y is positive What about x..plug a value of y > 4 in the equation and you'll get x > 0



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Re: Gprep DS: Is xy>o? [#permalink]
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25 Dec 2009, 05:03
lagomez wrote: 4test1 wrote: Can't seem to find this in the forum. From the Gprep test:
Is xy>0? 1. xy > 2 2. x2y <6
Can't seem to get my head around this. Appreciate any help. answer c Statement 1: x = 5 and y = 4 answer is yes x = 1 and y = 0 answer is no Statement 2: x = 7 and y = 0 answer is no x = 10 and y = 1 answer is yes Combined: x  y > 2 x  2y < 6 or x + y < 2 x  2y <  6 y < 4 y > 4 so you know y is positive What about x..plug a value of y > 4 in the equation and you'll get x > 0 hi lagomez, nice method, but i m afraid there is some mistake in ur explanation. if y>4. say y=5 now plugging this value in eq2, i.e. option 2 x2y < 6 ===> x10<6 ===> x<4 ===> so x can also be ve... in this case xy<0 plugging the value of y=5 in eq1, i.e. option 1 xy > 2 ===> x>y+2 ===> x>7 ===> x=+ve.... in this case xy>0 two answers. hence data insuff. Answer:E here is my method.. as u know, 1 & 2 are individually insuff.. taking 1&2 together xy>2 ===> x>y+2 x2y<6 ===> x<2y6 so combining both the above equations (y+2)< x < (2y6) by plugging values, if y= 2 ==> 4<x<2 ==> x<2 & x>4 two cases arise. so, data insuff. Answer: E
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Re: Gprep DS: Is xy>o? [#permalink]
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25 Dec 2009, 06:22
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logan wrote: hi lagomez,
nice method, but i m afraid there is some mistake in ur explanation.
if y>4. say y=5 now plugging this value in eq2, i.e. option 2
x2y < 6 ===> x10<6 ===> x<4 ===> so x can also be ve... in this case xy<0
plugging the value of y=5 in eq1, i.e. option 1
xy > 2 ===> x>y+2 ===> x>7 ===> x=+ve.... in this case xy>0
two answers. hence data insuff. Answer:E
here is my method.. as u know, 1 & 2 are individually insuff.. taking 1&2 together
xy>2 ===> x>y+2 x2y<6 ===> x<2y6
so combining both the above equations (y+2)< x < (2y6)
by plugging values, if y= 2 ==> 4<x<2 ==> x<2 & x>4 two cases arise. so, data insuff. Answer: E Not so. Lagomez's answer is correct. Is xy>0?(1) xy > 2 (2) x2y <6 Note that question basically asks whether \(x\) and \(y\) have the same sign. Statements alone are not sufficient as was shown above. Now, remember we can subtract inequalities with the signs in opposite direction. \(xy(x2y)>2(6)\) > \(y>4\). As \(y>4\) and (from 1) \(x>y2\), hence \(x>2\) (we can add inequalities when their signs are in the same direction, so \(y+x>4+y2\) > \(x>2\)) > both \(x\) and \(y\) are positive. Sufficient. Answer: C. When you got that for y=5 > x<4, it's correct, but it doesn't mean that x can be negative, as this is not ALL information you have. If you plug the same value y=5 in (1) you'll get > x>3, which is also correct. Basically when y=5 > 3<x<4. Hope it's clear.
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Re: Gprep DS: Is xy>o? [#permalink]
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25 Dec 2009, 06:51
This question is best seen graphically. It is a very quick method. See the graphs below. They only meeting in 1st Quadrant. So both equations together tell us that x>0 y>0. Attachment:
xygr2.GIF [ 3.63 KiB  Viewed 6289 times ]
Attachment:
eqn2.GIF [ 2.88 KiB  Viewed 6285 times ]
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Re: Gprep DS: Is xy>o? [#permalink]
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01 Jan 2010, 17:04
lagomez wrote: 4test1 wrote: Can't seem to find this in the forum. From the Gprep test:
Is xy>0? 1. xy > 2 2. x2y <6
Can't seem to get my head around this. Appreciate any help. answer c Statement 1: x = 5 and y = 4 answer is yes x = 1 and y = 0 answer is no Statement 2: x = 7 and y = 0 answer is no x = 10 and y = 1 answer is yes Combined: x  y > 2 x  2y < 6 or x + y < 2 x  2y <  6 y < 4 y > 4 so you know y is positive What about x..plug a value of y > 4 in the equation and you'll get x > 0 Correct answer is C... As pointed by Bunuel, when u plug in y > 4 in both the above equations, you get x range as 3<x<4.... Therefore xy > 0! You shouldn't consider only x  2y < 6 for determining the value of x. Both the equations should be considered! Cheers! JT
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Re: Gprep DS: Is xy>o? [#permalink]
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01 Jan 2010, 17:08
burnttwinky wrote: Bunuel wrote: chetan2u wrote: i think ans should be E... agreed y>4.. substitute in eq 1...x>2... substitute in eq 2... x<2y6...x<86...x<2..... combining the two , can we get a value of x, which is both < 2 and>2,, i dont think When you substitute in equation 2, you take the value of y=4 (which can not be as y>4). C is correct. If I substituted x=2 into each of the equations to solve for Y, i got two different answers. Y>4 and Y<4, depending on which equation you substitute it into. I don't quite understand what's your point. Answer to the original question is C. xy>0 means that the point (x,y) is either in I quadrant or in III. (1) xy > 2 the points satisfying this inequality can be in ALL four quadrants, refer to the first graph msunny provided (blue area). Hence insufficient. (2) x2y <6 the points satisfying this inequality can be in I, II or in III quadrants, refer to the second graph msunny provided (blue area). Hence insufficient. (1)+(2) Points satisfying BOTH inequalities are only in I quadrant: intersection of blue areas. Hence sufficient. Hope it's clear.
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Re: Gprep DS: Is xy>o? [#permalink]
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01 Jan 2010, 17:49
I apologize if my point is unclear, but my question is that, after I calculated X>2 (I solved for X first), I plugged in X=2 into the first equation, and I got Y<4. What I am confused about is that if Y<4, that means Y can be negative; however, x>2, so if Y is negative and X is always positive, then you would be in quadrant IV.
If plugged in X=2 into the second equation, then you get y>4, which is sufficient since X>2 and Y>4, so you will always be in quadrant I.



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Re: Gprep DS: Is xy>o? [#permalink]
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burnttwinky wrote: I apologize if my point is unclear, but my question is that, after I calculated X>2 (I solved for X first), I plugged in X=2 into the first equation, and I got Y<4. What I am confused about is that if Y<4, that means Y can be negative; however, x>2, so if Y is negative and X is always positive, then you would be in quadrant IV.
If plugged in X=2 into the second equation, then you get y>4, which is sufficient since X>2 and Y>4, so you will always be in quadrant I. Your X value is > 2, do not substitute 2 as X in the equation. Substituting X = 3 would be a right approach. On doing that for first eq, you get: 3y>2 i.e. y < 5  [1] Substituting X = 3 in equation 2, you get: 32y<6 i.e. y > 4.5.... Therefore if X = 3, then y range is from 4.5 to 5 which means both X and Y are positive. Hence XY > 0. Hope this is clear.. Thanks, JT
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Re: Gprep DS: Is xy>o? [#permalink]
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01 Jan 2010, 18:16
burnttwinky wrote: I apologize if my point is unclear, but my question is that, after I calculated X>2 (I solved for X first), I plugged in X=2 into the first equation, and I got Y<4. What I am confused about is that if Y<4, that means Y can be negative; however, x>2, so if Y is negative and X is always positive, then you would be in quadrant IV.
If plugged in X=2 into the second equation, then you get y>4, which is sufficient since X>2 and Y>4, so you will always be in quadrant I. First of all when you solved system of inequalities and got x>2 and you want to continue by number plugging you should plug correct numbers. You got x>2, hence you should plug x more than 2. Any value more than 2 for x will give you the RANGE for y which is positive. x=3: xy > 2 > 3y>2 > y<5 and x2y <6 > 32y<6 > y>4.5 So, for x=3, y is in the range {4.5, 5}. You'll get only positive range for y, when you test ANY value of x>2, as the solution of this system of inequalities is x>2 and y>4 (refer to the solution in my first post in this topic). And viseversa: if you try values of y>4 you'll get only positive range for x. Hope it's clear.
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Re: Gprep DS: Is xy>o? [#permalink]
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20 Dec 2010, 02:54
you are wrong. we CAN NOT ADD INEQUALITY, though we can add equality. a>b and c>d we can not have a+c>b+d . because there is some redundant we use this method y2<x<2y6 y2<2y6 4<y that means x>2, C is correct.
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Re: Gprep DS: Is xy>o? [#permalink]
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20 Dec 2010, 03:01
thangvietnam wrote: you are wrong.
we CAN NOT ADD INEQUALITY, though we can add equality.
a>b and c>d
we can not have a+c>b+d . because there is some redundant
we use this method
y2<x<2y6 y2<2y6 4<y that means x>2, C is correct. The red part is not correct. 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, 15:34
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: 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 1794 times ]
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