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Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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07 Nov 2015, 09:54
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Jamboree and GMAT Club Contest Starts QUESTION #2:Is xy < 0? (1) x + y > 5 (2) 3y  x < 9 Check conditions below: For the following two weekends we'll be publishing 4 FRESH math questions and 4 FRESH verbal questions per weekend. To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific. Then a week later, respective forum moderators will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6months access to GMAT Club Tests. PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize. He/She can opt for one of the following as a Grand Prize. It will be a choice for the winner:  GMAT Online Comprehensive ( If the student wants an online GMAT preparation course)  GMAT Classroom Program ( Only if he/she has a Jamboree center nearby and is willing to join the classroom program) Bookmark this post to come back to this discussion for the question links  there will be 2 on Saturday and 2 on Sunday! There is only one Grand prize and student can choose out of the above mentioned too options as per the conditions mentioned in blue font.All announcements and winnings are final and no whining GMAT Club reserves the rights to modify the terms of this offer at any time. NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.
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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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07 Nov 2015, 10:42
tough and tricky one!
Is xy < 0?
this is a Yes/No DS question. In order to answer this question, we need to know the signs of X and Y. In order to satisfy this condition one must be positive and the other negative. Any other cases will yield a positive number
(1) x + y > 5 from this, we can rearrange y>x+5. it can be the case that x=3 and y>2 in this case, the answer is yes. in the same time, x can be 1, and y must be greater than 6. in this case, we'll have a positive number, and the answer is no.
From the above said, statement 1 is insufficient. (2) 3y  x < 9 rearrange: 3y < x9 x can be 81, which means that 3y must be less than 72. Y must be less than 24. Since Y can be both positive and negative, statement 2 is insufficient.
We can now cross out answer choices A, B, and D, and if we don't know how to solve further, we can take a smart guess, with a chances of answering correctly 1/2.
let's take now 1+2: y>x+5 3y<x9 what can we do with this system of equations? let's multiply first with 1. we get y<x5. Note that when multiplying with a negative number, the inequality sign switches. Since the inequality sign is the same, we can add both inequalities and still get the same result: 2y < 14, and Y < 7. Ok, now we know for sure Y is negative. But don't get excited. To answer the question, we need to know the sign for X as well. TO find the sign for X, multiply the first equation with 3, and get 3y<3x15 again, add the 2 equations: 3y<3x15 3y<x9 we get: 0<2x24 or 24<2x or 12<x. Multiply this by 1, switch the inequality sign: 12>x. X is thus negative. Knowing the signs for X and Y, we can give a definite answer to the question. the answer is thus C.



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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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07 Nov 2015, 11:01
Solved through the graph. IMO, the answer is E



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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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07 Nov 2015, 11:25
1. x+y>5 => y> 5+x x= 0 , y=6 => xy=0 . Is xy<0 . No x= 10 , y= 1 =>xy = 10 . Is xy<0 .Yes Not sufficient 2. 3y x < 9 => x > 3y + 9 x= 10 , y= 0 => xy=0 .Is xy<0 . No x= 7 ,y =1 => xy=7 .Is xy<0 . Yes Not sufficient Combining, we get Let's switch the signs of the second inequality. x 3y > 9  2 x+y>5 Multiply the first inequality by 3 , 3x + 3y > 15  1 Adding 1 and 2 , we get 2x> 24 => x>12 => x< 12 This means that x is negative . Now in the first inequality, x+y>5 => y> 5+x if x=13 , then y> 8 .Depending on value of y , xy can be negative or positive. Answer E
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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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07 Nov 2015, 12:23
For option 1, y can only take positive values more than 7 while x can take both negative n positive values hence no unique answer can be found out through equation while in equation 2 y has to b negative n x has to be positive to meet the given condition so unique answer can b found through 2nd equation alone, hence b is the correct option.



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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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07 Nov 2015, 12:38
IMO E is correct
I) x+y>5
If we start plugging some values x=11 y=2 11+2>5 Here xy<0
Also id x=11 y=2 112>5 here xy>0 So Statement I is insufficient
II) 3yx<9
again after plugging again we find x= 15 y = 1 315<9 here xy>0
IF x= 1y=5 151<9 Here xy<0
Now lets check both statements together We need to identify value of x and y which satisfy both equations After plugging certain values you get a knack they both cannot be satisfied together
Further to check and confirm we can draw graph of x+y>5 which contains area most in II quandrant and for 3yx<9 area is in IV quadrant so there doesn't exist any common values
So answer is E



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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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07 Nov 2015, 14:38
Ans is E
Neither statements conclusively prove product xy >,=or < 0



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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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07 Nov 2015, 15:23
Is xy < 0? We are basically asked whether x and y are of different signs (also please note that none of them shall be equal to 0) (1)x + y > 5 . Here I would rather proceed with numberpicking. We can have that y is extremely large, let’s say 100 while x can be extremely small in absolute value (e.g. +/1). In case y=100,x=1 We have that xy>0, while in case y=100,x=1 we have xy<0, so the answer is (1) is not sufficient. (2) 3y  x < 9. Again, let’s pick some numbers. In this case y can be extremely small (e.g. 100), while x can be extremely small in absolute value (e.g. +/1). In case y=100,x=1 we will have xy<0, while in the case y=100 and x=1, then xy>0. Statement (2) also seems insufficient. Let’s take (1) and (2) together. Let’s sum both conditions, but before that, we should assure that they have similar signs, so let’s multiply the first condition by 1. We will have xy<5 . After summing the first and the second condition, we get.2y<14 or y<7. Let’s depict x in terms of y using first argument. We will have y5>x, this suggests that x is more negative than y. Considering that we found that both variables are negative, then we will have that their product is positive. Since, the sign of the product of these variables can be identified, then using both, statement (1) and statement (2), we can answer this question, which means the correct answer is C.



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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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07 Nov 2015, 22:29
Answer is E. QUESTION #2:
Is xy < 0?
(1) x + y > 5 (2) 3y  x < 9
i) Considering statement 1, Assume x=6 and y=16, both x and y are positive. Means, xy>0. So the answer to the question is No. Now assume, x=1 and y=6 , as x and y have different signs, so xy<0. So the answer to the question is Yes. This statement is not sufficient.
ii) Statement ii says 3y  x < 9, it follows same concept above, the answer to Is xy < 0? can be "Yes" or "No" so this statement is also not sufficient.
Combining both the statements we get y<7 and x<y5 means x can be zero, +ve or ve so answer to the question Is xy < 0? differs.
So the answer is "E".



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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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07 Nov 2015, 23:00
xy<0 only when \(xy\neq{0}\) and exactly one of the two (x and y) is less than 0.
1. x+y>5>Not Sufficient y5>x...................................................(1)
y and x could take any values. y=10 x=4; xy>0 y=10 x=1; xy<0 y=2 x=8; xy>0
2. 3yx<9>Not Sufficent
3y+9<x.................................................(2) y=1 x=13; xy>0 y=2 x=4; xy<0 y=4 x=2 xy>0
1 and 2: Sufficient
Combining (1) and (2)
3y+9<x<y5 2y<14 y<7
y5<75 y5<12 but (2) says x<y5 So, x<12
Since both y and x are negatives, we know for sure that xy>0. So the answer to the question is "No".
Answer: C



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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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08 Nov 2015, 00:35
Is xy negative?
1) y  x > 5 6  0 > 5 > xy = 0 6  (1) > 5 > xy = 6 So 1 alone is not sufficient
2) 3y  x < 9 3(4)  0 < 9 > xy = 0 3(4)  1 < 9 > xy = 4 So 2 alone is not sufficient
Combining 1 and 2,
Multiply St.1 by 3 on both sides, 3y  3x > 15 3y  x  2x > 15 From St. 2 we have, 3y  x < 9.. Assume 3y  x = 9 9  2x > 15 2x > 24 x < 12
If x < 12, 3y  x < 9.. Assume the value of x to be 12 3y  (12) < 9 3y < 21 > y < 7.
Combining St1 and St2 we get that both x and y are negative. Product of 2 negative numbers is positive. So xy is nonnegative.
Answer: C



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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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08 Nov 2015, 00:53
Basically the question asks about the sign of x and y .If any of x & y is negative, the answer is YES, otherwise NO. Statement 1 says (1) x + y > 5 ==> y > x+5Clearly x & y can take either positive OR negative sign and still satisfy the equation. So we will get dual answer of YES & NO. Insufficient. Example : x= 10 & y = 2 => xy > 0. NOx= 10 & y = .2 => xy < 0. YESStatement 2 says (1) 3y  x < 9 ==> 3y < x9Here too x & y can take any sign (+ve / ve) and still satisfy the equation. Again we will get dual answer of YES & NO. Insufficient. Example :x=  .2 & y = 10 => xy > 0. NOx= 10 & y =  .2 . => xy < 0. YESAnother quick & efficient way is finding the quadrants the points of (x,y) lie. The question asks whether xy<0 ; that means whether points of x & y lie in quadrant II and IV, where x & y will have different sign. After having a close look in equation [I] y > x+5 and [ii] 3y < x9; ==> y = \(\frac{x}{3}3\), we realize that points of given 2 lines will lie in at least 3 quadrants. Thus the sign of point (x , y) may have the same sign OR different sign. So, Statement 1 & 2 alone are insufficient.Combining 1 & 2 3y  y < x9x5; ==>2y < 14; ==> y<7 Plugging y in Stmt 1, we get 7 > x+5 ; x <12Plugging y in Stmt 2, we get 3(7) < x9 ; ==> x >12 i.e. value of x could be both positive and negative. thus combination of 1+ 2 is also insufficient. So, Option (E) wins.
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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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08 Nov 2015, 04:25
Is xy < 0? (1) x + y > 5 (2) 3y  x < 9 This can be easily done using graphs. Look at the attached graph: (1) x + y > 5 It can be seen that the graph of this equation that this equation has solutions in I, II and III quadrants. For I and III quadrants, xy>0 while for II quadrant, xy<0. Hence insufficient. (2) 3y  x < 9 It can be seen that the graph of this equation that this equation has solutions in I, IV and III quadrants. For I and III quadrants, xy>0 while for IV quadrant, xy<0. Hence insufficient. (1)+(2) It can be seen that the graphs for 1 and 2 overlaps in III quadrant only and for III quadrants, xy>0, hence it can concluded that xy is not less than 0. Hence sufficient.
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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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08 Nov 2015, 07:05
Ans E The question asks if \(XY <0\) 1.\(x + y>5 => y>5+x\) clearly x,y may be +ve Nos e.g \(For X=1 Y > 6 therefore XY >0\) or ve Nos say x= 6 therefore \(Y>1 XY >0\) Not Sufficient. 2. \(3Yx<9 => 3y+9<x =>x >3y+9\) again not sufficient. Since for any value of X,Y the solution will hold. 1+2 \(=> x > 3(5+x) +9 => x> 3x + 24 => x < 12\) But we know from 2 that y > 5+x therefore y is also < 0 but this condition is also satisfied if\(y =0\) therefore Ans E
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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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08 Nov 2015, 10:17
C is my answer
St 2 shows that x> y bt is not sufficient alone St 1 tells that either x or y is +ve, or both x and y can be +ve So alone not sufficient
By combining both statement we can get the answer, both are +ve



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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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08 Nov 2015, 11:25
QUESTION #2:
Is xy < 0?
(1) x + y > 5 (2) 3y  x < 9
Option 1: Not sufficient. As X can be 2 & Y can be 4 or X can be 2 & Y can be 8, resulting in XY<0 or XY>0. Option 2: Not Sufficient; This can be arranged to form X3Y > 9. Here X can be 20 & Y can be 2, resulting in XY>0. or X can be 7 & Y can be 1, resulting in XY<0. 1+2 : Sufficient. Combining & adding two equation : X + Y > 5 X  3Y > 9  2Y > 14 or, Y < 7. Suppose Y = 8, then X has be 14; means X & Y both has be negative and in that case XY>0, which itself is sufficient to proved that.



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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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08 Nov 2015, 13:03
Is xy < 0? (1) x + y > 5 (2) 3y  x < 9 Ans:C statement wants to know is xy<0 That means either x and y both have positive sign or both have negative signs. Taking Only 1) x+y>5 we can see by putting numbers that XY could be positive or negative. example if you take x as 2 and y as 8 => 2+8=6>5 so xy is positive Take x as 1 and y as 7 (1)+7=6>5 but xy=negativeSo we can not come to ans yes or no cos we have both yes and noA is ruled out Taking only 2) 3yx<9 we can see again by putting numbers that XY could be positive as well as negative. example if you take y=3 and x=3=> 3(3)3=12<9 and xy is negative but if you take y=4 and x=1 => 3(4)(1)=12+1=11<9 but XY is positive.Again we get yes and no both B ruled out. Only options left are C and E. If we combine both statements by multiplying 1 in statement 2..we get x+y>5 3y+x>9 .......We can add both as equality sign is in the same direction. So 2y> 14 and y<7 so with this we get y<7 for x putting the value of y in both equations we get x<13 and x >15 so we get value in between and x is negative too..So finally x and y both negative giving XY>0
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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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Updated on: 12 Nov 2015, 10:48
we can solve this using algebra
(1) X+Y >5
Here X can be negative or positive. y also can be positive or negative so it is in sufficient
(2) 3YX<9
here also X can be positive or negative. same thing with y
we can't define whether xy<0 or xy>0
it is also insufficient
combining two equations.(1)+(2)
X+Y>5 9>3YX _______________
X+Y9>3YX+5 95>3YY 14 >2Y 7>Y so Y is negative
substitute in equation 1 Y5>X
some negative number > X
so X and Y are negative XY must be greater than zero..
XY>0.
so answer is option (C)
Originally posted by vinod265 on 11 Nov 2015, 08:30.
Last edited by vinod265 on 12 Nov 2015, 10:48, edited 1 time in total.



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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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11 Nov 2015, 13:07
QUESTION #2: Is xy < 0? (1) x + y > 5 or y>x+5 If x is +ve y is +ve if x is ve y can be either positive or negative I is not sufficient (2) 3y  x < 9 3y+9<x or x>3y+9 if y is ve (1) x is +ve(6) if y=5 x>6 i.e., can be negative or +ve(5 or 4) y i s+ve then x is +ve II is not sufficient 1&2 y>x+5 and x>3y+9 y>3y+9+5 2y>14 y>7 y<7 y is ve always x> 3(8)+9 x>15 x can be either +ve or negative. So E is the correct answer
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Re: Jamboree and GMAT Club Contest: Is xy < 0? [#permalink]
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12 Nov 2015, 14:20
Please follow the uploaded file to see the solution. Graphical Method is used to solve the problem.
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Re: Jamboree and GMAT Club Contest: Is xy < 0?
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