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!

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.

Great.. I get it now.. Thanks Bunuel.

I am also stuck at Q10 :

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

We need to see if |n|<4 (this means -4<n<4)

1) n^2>16 => n<-4 and n>4
So from n<-4, |n|<|-4| = |n|<4 (works)
But n>4 does not work.
Doesn't that make 1) Insufficient?

Could you please tell me what I am doing wrong here ??

Bunuel I have very clear the 1 statement but not so much the secon one : 1/|n|< n this is true only for negative value. So we could have i. e. : -1 or -6 so insuff but how we you arrive to this conclusion:

Correct me if I'm wrong because the concepts are always the same but the gmat blonds them and as consequence blow your mind.

in other words, you are saying

1/|n| > n 2 cases

1/-n > n ----> n^2 > -1 this implies that any squared number is positive and therefore greater than -1, all negative n values work as solutions. your n < 0

1/n > n -----> 1 > n^2 ----> n^ 2 < 1 --------> -1 < n < 1 . your second range. so in the end we have all this information and we are not sure of course of - 4 < n < 4.

Correct ???

Thanks

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.

9. Is n<0?
(1) -n=|-n|
(2) n^2=16

(1) -n=|-n|, means that either n is negative OR n equals to zero. We are asked whether n is negative so we can not be sure. Not sufficient.

(2) n^2=16 --> n=4 or n=-4. Not sufficient.

(1)+(2) n is negative OR n equals to zero from (1), n is 4 or -4 from (2). --> n=-4, hence it's negative, sufficient.

Answer: C.

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

Hello Bunuel,

I got A as the answer to the Q.

From St1, we have -n=|-n|---> -n=n (As Mod value is +ve)---> we have 2n=0 or -2n=0. In both case we can say that n=0 and hence Ans should be A.

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.

Bunuel, I think I need some conceptual help. Why should we not solve statement 1 by rewriting the two statements and then adding them together? (Besides the fact that it's time consuming....) I rewrote them and found 3x^2 -10 = y for the positive absolute vlaue, and -3x^2+14=y for the negative abs value. From this, I added them together and got y=4..

Can you please explain what I'm getting wrong conceptually? Thanks so much!!!! I appreciate your kindness.

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.

hi Bunuel,

Thank you very much for all the explanations. I have a query on this one

Combining both we get x^2+y^2=5a or x,y,a = 0

aren't those sufficient to answer the question is x^2+y^2>4a

Is the first case where x^2+y^2=5a, the answer is yes

Second case where x,y,a=0, the answer is no

Kindly do elaborate. Thanks

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.

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

Question basically asks is -4<n<4 true.

(1) n^2>16 --> n>4 or n<-4, the answer to the question is NO. Sufficient.

(2) 1/|n| > n, this is true for all negative values of n, hence we can not answer the question. Not sufficient.

Answer: A.

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

Hi Bunuel ,

I know saying (1/|n|) < n will be true for all n<0 is quite clear logically. Still I want to reach this conclusion mathematically.

I got swayed solving for n|n| < 1 .