Inequality and absolute value questions from my collection

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.

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.

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

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.

4)
I) 2x-2y=1 so y=x-1/2 NS
II)x/y>0 so x and y have the same sign and the modulus of x has to be larger than the modulus of y NS
Together, to satisfy both clues needs to be larger than 1/2 and x becomes larger than 0; the stem is true, therefore C

Your last choice of numbers: x=1/2, y=-1/2 does not satisfy clue I, because 2*(1/2)-2*(-1/2)=2, not 1

Taking the modulus does not mean rounding up to the nearest integer; it means removing the negative sign if present. |0.8|<0.86



Nowhere it is said that x and y are non-zero. If x and y are zero, 5a=0, therefore a=0, and the stem is false (x^2+y^2=0)

\(6*x*y = x^2*y + 9*y\)
\(6*x = x^2 + 9 = 0\)
\(x^2 - 6*x + 9 = 0\)
\((x-3)^2 = 0\)
\(x = 3\)

St. (1) : y - x = 3
y = 6
Sufficient.

St. (2) : x^3 < 0
Invalid statement. Does not give us value of y.
Insufficient.

Answer : A

Question Stem gives us :

(a) If x > 0 ; y = 2x
(b) If x < 0 ; y = 0

St. (1) : x < 0
Sufficient.

St. (2) : y < 1
Since y is an integer and y cannot be less than 0 (question stem part b) therefore y must be 0.
Sufficient.

Answer : D

St. (1) : (x + y)^2 = 9a
x^2 + y^2 + 2xy = 9a
Insufficient.

St. (2) : (x - y)^2 = a
x^2 + y^2 - 2xy = a
Insufficient.

St. (1) and (2) together : x^2 + y^2 = 5a
When either x or y is not 0, question stem holds true.
When x and y are both 0, question stem is false.

Hence insufficient.

Answer : E

Question Stem : x > 0 ; y > 0 ?

St. (1) : 2x -2y = 1
x = y + 0.5
Equation can be satisfied for both positive and negative values of x and y.
Hence Insufficient.

St. (2) : x/y > 1
Equation can be satisfied when both x and y are either positive or negative.
Hence Insufficient.

St. (1) and (2) together : (y + 0.5)/y > 1
1 + 0.5/y > 1
0.5/y > 1
For this to be true, y must be positive.
If y is positive then x will also be positive.
Hence Sufficient.

Answer : C

Question stem : What is the exact value of y?

St. (1) : 3*|x^2 -4| = y - 2
y = 3*|x^2 -4| + 2
From this we can infer that y will be a positive value. That is, y > 0. However, we want to know the exact value of y.
Therefore, Insufficient.

St. (2) : |3 - y| = 11
(a) When (3 - y) > 0 ; 3 - y = 11 ; y = -8.
(b) When (3 - y) < 0 ; - (3 - y) = 11 ; y = 14.
Thus we can see that there are two possible values for y.
Hence Insufficient.

St. (1) and (2) together : y > 0 ; y = 14 or -8.
Obviously since y has to be greater than 0, it cannot be -8. Therefore value of y = 14.
Hence Sufficient.

Answer : C

St. (1) : x + 1 > 0
Tells us nothing about y.
Insufficient.

St. (2) : xy > 0
Both x and y can either be positive or negative. Neither x nor y can be 0.
Insufficient.

St. (1) and (2) together :
Since x is an integer and cannot hold the value 0, it has to be greater than 1 in order to satisfy St. (1).
Since we know that x will be positive, y will also have to be a positive integer in order to satisfy St. (2).
Hence Sufficient.

Answer : C