SOLUTIONS:

1. If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y – x = 3
(2) x^3< 0

First let's simplify given expression \(6*x*y = x^2*y + 9*y\):

\(y*(x^2-6x+9)=0\) --> \(y*(x-3)^2=0\). Note here that we CAN NOT reduce this expression by \(y\), as some of you did. Remember we are asked to determine the value of \(xy\), and when reducing by \(y\) you are assuming that \(y\) doesn't equal to \(0\). We don't know that.

Next: we can conclude that either \(x=3\) or/and \(y=0\). Which means that \(xy\) equals to 0, when y=0 and x any value (including 3), OR \(xy=3*y\) when y is not equal to zero, and x=3.

(1) \(y-x=3\). If y is not 0, x must be 3 and y-x to be 3, y must be 6. In this case \(xy=18\). But if y=0 then x=-3 and \(xy=0\). Two possible scenarios. Not sufficient.

OR:

\(y-x=3\) --> \(x=y-3\) --> \(y*(x-3)^2=y*(y-3-3)^2=y(y-6)^2=0\) --> either \(y=0\) or \(y=6\) --> if \(y=0\), then \(x=-3\) and \(xy=0\) \(or\) if \(y=6\), then \(x=3\) and \(xy=18\). Two different answers. Not sufficient.

(2) \(x^3<0\). x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.

Answer: B.

This one was quite tricky and was solved incorrectly by all of you.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

Never multiply (or reduce) inequality by variable (or expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.
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Question Stem : Is the distance between X and -Y greater than the distance between X and Y?
Note : Using number line approach.

St. (1) : |x| > |y|
OR, The distance between X and the origin is greater than the distance between Y and the origin.

Now we can have two cases :

(a) When X is positive : In this case, X > Y for the above condition to be true.

_________|_________|_________|_________|_________|__________
_________________-Y______Origin______Y........ Region of X.........

Thus we can see that the distance between X and - Y will always be greater than the distance between X and Y. Hence question stem is true.

(b) When X is negative : In this case, X < -Y for the statement to be true.

_________|_________|_________|_________|_________|__________
.... Region of X.....-Y_____Origin_______Y___________

Thus we can see that the distance between X and -Y will always be less than the distance between X and Y. Hence, question stem becomes false.

Due to conflicting statements, St (1) becomes Insufficient.

St. (2) : |x-y| < |x|
OR, the distance between X and Y is less than the distance between X and the origin.

Now again let us consider the different cases :

(a) When X is positive : For the statement to be true and for X to be positive, X must be greater than Y/2. For any value of X less than Y/2 the statement will become false. The statement will be true for any value greater than Y/2.

Thus we see that only one case comes into the picture. Now let us see how it relates to the question stem.

_________|_________|____;____|_________|________
________-Y____Origin__y/2...Y...... Region of X....

Thus we can see that the distance between X and - Y will be greater than the distance between X and Y for all values of X > Y/2. Thus question stem is true.
St. (2) is sufficient.

Answer : B
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ahh..yes...fresh meat

1. x < 0
you will always get x minus itself so always 0

2. y < 1
y is an integer so y<=0
y can't be negative because x minus itself is always zero

answer d

I'm getting B for this one

1. (x-1)^2 <= 1
x can be 0 which would make the question no
or x can be 1/2 which would make the answer yes
so 1 is insufficient

2. x^2 - 1 > 0
means x^2>1
so x<-1 or x>1
both of which make the question no
so sufficient

I'm getting c

1. s can be 3 and r can be 3 which makes question yes
s can be 3 and r can be 2 which makes question no
insufficient

2. r can be 3 and s can be 3 makes question yes
r can be 3 s can be 2 makes question no
insufficient

combining:
|r|>=s means
r>=s or r<=-s

and -s<=r<=s means
-s<=r and r<=s

now we have -s<=r and -s>=r so -s = r or s = r
r>=s and r<=s so s = r

10. If n is not equal to 0, is |n| < 4 ?
(1) n^2 > 16
(2) 1/|n| > n

answer A
because in number 2 n can be negative or a fraction

Bunuel, thanks for the questions. Please provide the OA's too. It would be great if you can provide them soon. I am having my GMAT this week, so kinda tensed and impatient. Also, I am yet to give my MGMAT CAT's, so tell me whether should I solve the questions on the forum because if the questions are from the MGMAT CAT's or Gmat Prep then it may overestimate my result. I would appreciate your response. Thanks once again.

Quality questions as always... Thanks Bunuel! +1

(1) (x-1)^2 <= 1
x is 0 to 2.
If x = 2, yes.
If x < 2, No.

(2) x^2 - 1 > 0
x cannot be -1 to 1 i.e. x<-1 or x>1. NSF.

From 1 and 2: x is >1 but <=2. NSF..

E.

These questions are from various sources. Couple of questions might be from MGMAT CAT or Gmat Prep, but not more than that.

I'll provide OA in a day or two, after discussions. Tell me if you want the answers for the specific questions earlier than that and I'll mail you.
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Not sure about this one...

First I reduced the given equation (divided out the y) and solved for x:
6*x*y = x^2*y + 9*y
6*x = x^2 + 9
0 = x^2 - 6*x + 9
0 = (x-3)^2
x = 3

Statement 1:

y-x=3
y-3=3
y=6
xy=3*6=18

SUFFICIENT

Statement 2:

x^3<0

We have no idea what the value of y is from this statement. The only thing that made me look twice was the face that if x^3 is true, then x should be a negative value... did I calculate the value of x incorrectly above?

INSUFFICIENT

ANSWER: A.

Another way of looking at the problem is to ask, is x<0? Because if it is, then we know that y is zero. The only case in which y will not be zero is if x is positive.

Statement 1:

x<0... answers my question above.

SUFFICIENT

Statement 2:

y<1

Because y is an integer, it must be one of the following values: 0, -1, -2, -3...

BUT |x| + x can never be a negative value. The lowest value that it can be is 0.

Hence, y can never be negative and the only possible value it can be then is 0.

SUFFICIENT

ANSWER: D.

3)
I) (x+y)^2=9a x^2+y^2=9a-2xy NS
II) (x-y)^2=a x^2+y^2=a+2xy NS
Together 2(x^2+y^2)=10a x^2+y^2=5a
If either x or y are larger than 0, the stem would be true, but if they’re both zero the stem is false, hence E

4)
I don’t get the two clues; they seem to be mutually exclusive

5)
I) 3|x^2-4|=y-2 either y=3x^2-10 or y=14-3x^2 NS
II) |3-y|=11 either y=-8 or y=14 NS
Together -8=3x^2-10 so 3x^2=2 ok 14=3x^2-10 so 3x^2=28 ok, hence E

6)
I) x+1>0 so x={0, 1, 2, …} NS
II) xy>0 so x and y have the same sign and none of them is zero NS
Together, x={1, 2, 3, ..} and y has the same sign, hence C

7) |x+2|=|y+2| either x+2=y+2 or x+2=-y-2 (the other two combinations can be transformed into these by multiplying by -1)
Reordering: x-y=0 or x+y=-4
I)xy<0, hence x and y have different signs and none of them is zero. The only possibility is x+y=-4 S
II) x>2, y<2 hence x#y. The only possibility is x+y=-4 S, therefore D

8)a*b#0, hence a and b are both non-zero
I) |a*b|=a*b a and b have the same sign and the stem is always true S
II) |a|/|b|=|a/b| this is true regardless of the values of a and b, and nothing can be said about the stem NS, therefore A

9)
I) –n=|-n| n<=0 NS
II) n^2=16 n=+/-4 NS
Together n=-4 therefore C

10)n#0
I) n^2>16, so |n|>4 S
II) 1/|n|>n true for n<-1 NS, therefore A

11) Plugging in numbers I get B, but there’s no rime or reason to my solution

12)
I) –s<=r<=s obviously NS. Since s>=-s, s is either positive or zero
II)|r|>=s obviously NS
Together: I) tells us that s>=0; II) tells us that r>=s or r<=-s. The only case in which I and II are simultaneously satisfied is r=s, therefore C

13) x=(0:2) with 0 and 2 excluded
I) (x-1)^2<=1, hence x=[0:2] with 0 and 2 included, hence NS
II) x^2-1>0 x<-1 or x>1. For x=1.5 the stem is true, for x=3 it is false, hence NS
Together, for x=1.5 the stem is true, for x=2 it is false, hence E

Yes there was a typo in 4. Edited. Great job Marco83. Even though not every answer is correct, you definitely know how to deal with this kind of problems.
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4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

Statement 1:

2(1)-2(1/2)=1 , x,y are both positve

2(1/2)-2(-1/2)=1 x is positive, y is negative

INSUFFICIENT

Statement 2:

Either (x,y) are both positive or both negative

INSUFFICENT

Statement 1 and 2:

With both requirements x must be greater than y and satisfy this equation: 2x-2y=1

2(1)-2(1/2)=1 , x,y are both positve and x>y

2(1/2)-2(-1/2)=1 x is positive, y is negative and x>y

Answer: E

12. Is r=s?

(1) -s<=r<=s

(2) |r|>=s


E – for this - both can be true or false when 0< r < 1
For example , take r as 0.8
S = 0.86 i.e. -0.86 < = 0.8 < = 0.86
|0.8|>= 0.86 i.e. 1 >= 0.86
Combining , any values can be taken , on values > =1 , both r and s
will be same

3. Is x^2 + y^2 > 4a?

(1) (x + y)^2 = 9a

(2) (x – y)^2 = a
C is the answer

Combined both and the equation will give x^2 + y^2 = 5a
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Statement 1:

Two equations, two unknowns... INSUFFICIENT

Statement 2:

|3 - y| = 11
(3-y)=11 or (3-y)=-11
y=-8, 14

INSUFFICIENT

Statements 1 and 2:

y must be 14 because 3|x^2 -4| can never be a negative value (no matter what you plug in for x, you will get a positve value because of the absolute value signs).

SUFFICIENT

ANSWER: C.

Statement 1:

Nothing about y... INSUFFICIENT

Statement 2:

two equations, two unknowns... INSUFFICIENT

Statements 1 and 2:

From x +1 > 0 and the fact that x must be an integer, we know that x must be [0,1,2,3...]

Because we know that xy > 0, we know that x cannot be 0... therefore y must be a positive integer!

SUFFICIENT

ANSWER: C.