liarish wrote:
Hi Bunuel,
I have read all the responses to Q4. But I am still confused why C is the answer. Here is how I solved it.

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

1) Insufficient . First reduced equation to x-y=0.5 . Plugged in 2 positive and 2 negative values. I chose x=3, y=2.5 => 3-2.5=0.5 works.. then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5. works again. So x and y can be both +ve and -ve. So 1) is Insufficient.
2) x/y>1. Just tells us that both x and y have same sign. both are -ve or both are +ve. So Insufficient.

Now Combining,
Picking the same values used in 1) x=3, y=2.5. both signs positive and x-y=0.5. works
then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5 both signes negative. works as well. So we still don't know if both signs are +ve or -ve. So my answer is E.

Could you please take a look at my solution and tell me where I am going wrong? That would be a big help. Thanks a ton!


If x= -1 and y=-1.5, then x/y=2/3<1, so these values don't satisfy the second statement.

This question is also discussed here: are-x-and-y-both-positive-1-2x-2y-1-2-x-y-93964.html

Hope it helps.

2. If y is an integer and y = |x| + x, is y = 0?
(1) x < 0
(2) y < 1

Note: as \(y=|x|+x\) then \(y\) is never negative. For \(x>{0}\) then \(y=x+x=2x\) and for \(x\leq{0}\) then (when x is negative or zero) then \(y=-x+x=0\).

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

(2) \(y<1\), as we concluded y is never negative, and we are given that \(y\) is an integer, hence \(y=0\). Sufficient.

Answer: D.

Hi Bunuel,

Thanks for the explanation to the above Q.

Regarding st 1 i.e X less than zero then y=|x|+x = -x+x=0,

1. we know any value in modulus is positive then ideally the above should be interpreted as y=|x|+x--> y=x-x=0.
2.Also if from St 1 if we x<0 then y=|x|+x= -x-x=-2x

3. Where as we also know that |x|= -x for X<0 and |x|= x for X>/ 0

So can you please tell me where am I going wrong with the concept.

Thanks
Mridul

5. What is the value of y?
(1) 3|x^2 -4| = y - 2
(2) |3 - y| = 11

(1) As we are asked to find the value of y, from this statement we can conclude only that y>=2, as LHS is absolute value which is never negative, hence RHS als can not be negative. Not sufficient.

(2) |3 - y| = 11:

y<3 --> 3-y=11 --> y=-8
y>=3 --> -3+y=11 --> y=14

Two values for y. Not sufficient.

(1)+(2) y>=2, hence y=14. Sufficient.

Answer: C.

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.

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.

Bunuel , Can you please show how we can reach to C using graphical approach ?