Jamboree and GMAT Club Contest: Is xy

Question

Jamboree and GMAT Club Contest Starts

QUESTION #2:

Is xy < 0?

(1) -x + y > 5
(2) 3y - x < -9

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Official Answer

C

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tarunk31 · Accepted Answer

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.

Attachments

IMG_20151108_164322.jpg [ 2.39 MiB | Viewed 6166 times ]

mvictor · Answer

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 < x-9 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 3yx. 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.

karant · Answer

Solved through the graph. IMO, the answer is E

Skywalker18 · Answer

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

nehagupta0123 · Answer

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.

kanigmat011 · Answer

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
11-2>5
here xy>0
So Statement I is insufficient

II) 3y-x<-9

again after plugging again
we find x= 15 y = 1
3-15<-9
here xy>0

IF x= 1y=-5
-15-1<-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 3y-x<-9
area is in IV quadrant so
there doesn't exist any common values

So answer is E

ronny123 · Answer

Ans is E

Neither statements conclusively prove product xy >,=or < 0

justdan · Answer

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 number-picking. 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 x-y<-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 y-5>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.

Celerma · Answer

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<y-5 means x can be zero, +ve or -ve so answer to the question Is xy < 0? differs.

So the answer is "E".

AmoyV · Answer

xy<0 only when xy eq{0} and exactly one of the two (x and y) is less than 0. 1. -x+y>5--->Not Sufficient y-5>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. 3y-x<-9--->Not Sufficent 3y+90 y=-2 x=4; xy<0 y=-4 x=-2 xy>0 1 and 2: Sufficient Combining (1) and (2) 3y+90. So the answer to the question is "No". Answer: C

Kurtosis · Answer

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 non-negative.

Answer: C

musunna · Answer

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+5
Clearly 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. NO
x= -10 & y = .2 => xy < 0. YES

Statement 2 says (1) 3y - x < -9 ==> 3y < x-9
Here 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. NO
x= 10 & y = -.2. => xy < 0. YES

Another 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 < x-9; ==> 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 < x-9-x-5; ==>2y < -14; ==> y<-7
Plugging y in Stmt 1, we get -7 > x+5 ; x <-12
Plugging y in Stmt 2, we get 3(-7) < x-9 ; ==> 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.

Dutta · Answer

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. 3Y-x<9 => 3y+9x >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

anceer · Answer

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

JackH · Answer

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 X-3Y > 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.

sahil7389 · Answer

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=negative--------So we can not come to ans yes or no cos we have both yes and no--A is ruled out
Taking only 2) 3y-x<-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

vinod265 · Answer

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) 3Y-X<-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>3Y-X
_______________

-X+Y-9>3Y-X+5
-9-5>3Y-Y
-14 >2Y
-7>Y
so Y is negative

substitute in equation 1

Y-5>X

some negative number > X

so X and Y are negative XY must be greater than zero..

XY>0.

so answer is option (C)

Nevernevergiveup · Answer

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

saunakdey · Answer

Please follow the uploaded file to see the solution.

Graphical Method is used to solve the problem.

Jamboree and GMAT Club Contest: Is xy < 0?

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