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
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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
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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
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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

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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
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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
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Question stem :
Note: Since the equations is symmetrical, there will only be two distinct cases. However, for the sake of explanation, I have illustrated all 4.
(a) When both x and y are greater than - 2 ; x + 2 = y + 2 ; x = y
(b) When both x and y are less than - 2 ; - x - 2 = - y - 2 ; x = y
(c) When x is less than -2 and y is greater than -2 ; - x - 2 = y + 2 ; x + y = - 4
(d) When x is greater than -2 and y is less than -2 ; x + 2 = - y - 2 ; x + y = - 4

St. (1) : xy < 0
This implies that one is negative and the other is positive. Therefore, in order for xy to be less than 0, x cannot be equal to y. Thus in order to satisfy the question stem, it can only be cases (c) and (d).
Thus Sufficient.

St. (2) : x > 2 ; y < 2
Again, this implies that x and y cannot be equal. Thus, in order to satisfy the question stem it can only be cases (c) and (d).
Thus Sufficient.

Answer : D
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Question stem : Neither a nor b can hold the value 0 ; |a|/|b|=a/b
For condition to be true, both a and b must hold the same sign.

St. (1) : |a*b| = a*b
This condition will be satisfied only when both a and b are either both positive or both negative.
Hence Sufficient.

St. (2) : |a|/|b| = |a/b|
This condition can be satisfied when a and b are same sign as well as opposite sign.
Hence Insufficient.

Answer : A
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Question Stem : Is n negative?

St. (1) : -n = |-n|
Let -n = A ; therefore the statement becomes : A = |A|.
This can only be valid when A is positive (or equal to 0). This in turn means that n must be negative (or equal to 0).
-n=|-n| also for n=0, hence not sufficient.

St. (2) : n^2 = 16
n = ±4.
Thus Insufficient.



Answer : C
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Question Stem : n # 0 ; -4 > n > 4 (excluding 0)

St. (1) : n^2 > 16
(n - 4)*(n +4) = 0 Therefore boundary conditions are 4 and - 4. Thus we can write it as : n < -4 and n > 4.
Sufficient.

St. (2) : 1/|n| > n
This condition will be valid for all n < 1 excluding 0.
Thus it will be impossible to tell whether |n| < 4.
Hence Insufficient.

Answer : A

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St. (1) : -s <= r < = s
Clearly Insufficient.

St. (2) : |r| >= s
When r > 0 ; r >= s.
When r < 0 ; -r >= s ; r <= -s
Therefore, this statement can be rewritten as : -s >= r >= s
Insufficient.

St. (1) and (2) : -s <= r < = s ; -s >= r >= s
For both statements to be simultaneously valid, r must be equal to s.
Hence Sufficient.

Answer : C
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Question Stem : Is |x-1| < 1 ?
When x > 1 ; x - 1 < 1 ; x < 2.
When x < 1 ; -x + 1 < 1 ; x > 0.
Thus it can be written as : 0 < x < 2.

St. (1) : (x-1)^2 <= 1
x^2 + 1 - 2x <= 1
x^2 - 2x <= 0
x(x - 2) <= 0 ; Thus boundary values are 0 and 2.
Therefore statement can be written as : 0 <= x <= 2.
Since the values are inclusive of 0 and 2, it cannot give us the answer.
Insufficient.

St. (2) : x^2 - 1 > 0
(x + 1)*(x - 1) > 0
Statement can be written as x > 1 and x < -1.
Thus it is possible for x to hold values which make the question stem true as well as false.
Insufficient.

St. (1) and (2) : 0 <= x <= 2 ; x > 1 and x < -1
Thus combined, the statements become : 1 < x <= 2.
Since it is inclusive of 2, it will give us conflicting solutions for the question stem.
Hence Insufficient.

Answer : E

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-n=|-n| also for n=0, hence not sufficient.

Everything else is as you suggested, therefore C

Yes, you are right. I overlooked that.
Thanks.
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2. If y is an integer and y = |x| + x, is y = 0?

Notice that from \(y=|x|+x\) it follows that y cannot be negative:
If \(x>0\), then \(y=x+x=2x=2*positive=positive\);
If \(x\leq{0}\) (when x is negative or zero) then \(y=-x+x=0\).

(1) \(x<0\) --> \(y=|x|+x=-x+x=0\). Sufficient.

(2) \(y<1\). We found out above that y cannot be negative and we are given that y is an integer, hence \(y=0\). Sufficient.


Answer: D.
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3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

(1) (x + y)^2 = 9a --> x^2+2xy+y^2=9a. Clearly insufficient.

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.
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4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches:
\(2x-2y=1\) --> \(x=y+\frac{1}{2}\)
\(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.
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