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.

Also insufficient as x,y, and a could be 0
Why did you assume that ALL COULD be zero.

11. Is |x+y|>|x-y|?
(1) |x| > |y|
(2) |x-y| < |x|

To answer this question you should visualize it. We have comparison of two absolute values. Ask yourself when |x+y| is more then than |x-y|? If and only when x and y have the same sign absolute value of x+y will always be more than absolute value of x-y. As x+y when they have the same sign will contribute to each other and x-y will not.

5+3=8 and 5-3=2
OR -5-3=-8 and -5-(-3)=-2.

So if we could somehow conclude that x and y have the same sign or not we would be able to answer the question.

(1) |x| > |y|, this tell us nothing about the signs of x and y. Not sufficient.

(2) |x-y| < |x|, says that the distance between x and y is less than distance between x and origin. This can only happen when x and y have the same sign, when they are both positive or both negative, when they are at the same side from the origin. Sufficient. (Note that vise-versa is not right, meaning that x and y can have the same sign but |x| can be less than |x-y|, but if |x|>|x-y| the only possibility is x and y to have the same sign.)

Answer: B.

Hi,

I didnt quite understand this part "(2) |x-y| < |x|, says that the distance between x and y is less than distance between x and origin. This can only happen when x and y have the same sign, when they are both positive or both negative, when they are at the same side from the origin."

what in the second equation tells us that they both have to be the same sign?

Thanks

I get super confused in solving the below kind of inequalities.
Can some expert help?

Are the below working's correct?

1/| n | >n
• Case 1: n > 0… n 2 < 1… - 1 < n < 1
• Case 2: n < 0… - n 2 < 1… n 2 > -1…

x/| x | < x
• Case 1: x > 0… x^2 > x… as x > 0, divide by x without changing signs… x > 1
• Case 2: x < 0… - x^2 > x… as x < 0, divide and only flip sign… - x < 1… x > -1

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.

Is this also a right way to solve this:
x^2+2xy+y^2=9a --> If we have to find x^2 + y^2 > 4a we should know whether 2xy>5a or 2xy<5a...No info on this.Hence Insuff
x^2-2xy+y^2=a---> If we have to find x^2 + y^2 > 4a we should know whether 2xy>3a or 2xy<3a...No info..Hence Insuff

Combining both..we still dont know as both the cases are possible and hence E..

Bunuel Please Confirm..

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.

Hi Bunnel,

I don't quite get the part on "substitute x". Does it mean \(\frac{1}{y}(x-y)>0\) and so (x-y) is zero?

Thanks in advance!

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.

hi
m a little confused here. How is A sufficient

isn't n^2>16 ----> n^2-16>0 ---> (N^2-4^2) >0 ----> n+4>0 or n-4>0 ----> n> -4 or n>4.....how did u get n<-4....

kindly correct me if m wrong

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

Answer: C.

Great question.
But -0=|-0| did not make sense to me because I thought "0 is neither +ve nor -ve so why negate it and then find absolute value". I did not know that was a possible direction. Is there some logic i'm missing or is it just a 'possible calculation'?
Thank you

13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

Is |x-1| < 1? Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.

Answer: E.

About S1: can i say |x-1|<=1 after rooting the given and putting it in modulus? I recall reading about squaring being used to remove modulus, so I thought to reverse that. Please correct me.

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.

Bunuel

Cant get thru stmt 1. Opening the mod we get either -n=-n or -n=n. First case gives null so am left with n=0 and hence sufficient. Where am I wrong in opening up the mod.?

First of all |-n| = |n|, so we are given that -n = |n|, which implies that n <= 0.

How did we arrive at |-n| = |n|?

would you think this is a 700+ question?