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Re: Inequality and absolute value questions from my collection
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03 Oct 2012, 07:13
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) 2x2y=1 (2) x/y>1
1) Insufficient . First reduced equation to xy=0.5 . Plugged in 2 positive and 2 negative values. I chose x=3, y=2.5 => 32.5=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 xy=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: arexandybothpositive12x2y12xy93964.htmlHope it helps.
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Re: Inequality and absolute value questions from my collection
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03 Oct 2012, 12:26
Quote: 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) 2x2y=1 (2) x/y>1
1) Insufficient . First reduced equation to xy=0.5 . Plugged in 2 positive and 2 negative values. I chose x=3, y=2.5 => 32.5=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 xy=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: arexandybothpositive12x2y12xy93964.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 ??



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Re: Inequality and absolute value questions from my collection
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04 Oct 2012, 03:39
liarish wrote: Quote: 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) 2x2y=1 (2) x/y>1
1) Insufficient . First reduced equation to xy=0.5 . Plugged in 2 positive and 2 negative values. I chose x=3, y=2.5 => 32.5=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 xy=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: arexandybothpositive12x2y12xy93964.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 ?? If n<4, then n, for example can be 4.5 > 4.5=4.5>4, so n<4 doesn't hold true. If n is not equal to 0, is n < 4 ?Question basically asks whether \(4<n<4\), so whether \(n\) is some number from this range. (1) n^2>16. This implies that either \(n>4\) or \(n<4\). No number from these ranges is between 4 and 4, thus the answer to the question whether \(4<n<4\) is NO. Since we have a definite answer then this statement is sufficient. Hope it's clear.
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Re: Inequality and absolute value questions from my collection
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Updated on: 04 Oct 2012, 04:03
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:
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Re: Inequality and absolute value questions from my collection
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04 Oct 2012, 04:00
carcass wrote: 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: Given: 1/n > n. Now, 1/n is always positive (because of the absolute value in the denominator), so if n is negative then this inequality will always hold true: 1/n=positive>n=negative. Hope it's clear. P.S. Complete solution for 1/n > n is n<0 or 0<n<1.
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Re: Inequality and absolute value questions from my collection
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04 Oct 2012, 04:13
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 those information and we are not sure of course of  4 < n < 4. Correct ??? Thanks
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Re: Inequality and absolute value questions from my collection
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04 Oct 2012, 04:20
carcass wrote: 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 1/n > n > 2 cases: If n<0, then n=n, so we'll have that 1/n>n > multiply by n and flip the sign (since we consider negative n): 1<n^2 > which holds true for any n from this range, so for any negative n. If n>0, then n=n, so we'll have that 1/n>n > multiply by positive n, this time: 1>n^2 > 1<n<1, since we consider n>0, then finally we'll get 0<n<1. So, 1/n > n holds true for n<0 and 0<n<1. Hope it's clear.
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Re: Inequality and absolute value questions from my collection
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23 Dec 2012, 04:40
Bunuel wrote: 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 [m]y=x+x = x+x=0, 1. we know any value in modulus is positive then ideally the above should be interpreted as [m]y=x+x> [m]y=xx=0. 2.Also if from St 1 if we x<0 then [m]y=x+x= xx=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
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Re: Inequality and absolute value questions from my collection
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23 Dec 2012, 05:29
Bunuel wrote: 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. 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. From your explanation, it is very clear that either n<0 or n=0. Could you tell me what was your approach to this Question. I mean did you assume values of 1. n as less than zero, 2. ngreater than zero and 3. n equal to zero and check under which condition the St1 holds true. If so, would this be a standard way of doing a modulus Question because clearly I just considered only 1 of the above conditions here. Your inputs please Thanks Mridul
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Re: Inequality and absolute value questions from my collection
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23 Dec 2012, 05:37
mridulparashar1 wrote: Bunuel wrote: 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=xx=0. 2.Also if from St 1 if we x<0 then y=x+x= xx=2x
3. Where as we also know that x= x for X<0 and x= x for X>/ 0So can you please tell me where am I going wrong with the concept. Thanks Mridul Absolute value properties:When \(x\leq{0}\) then \(x=x\), or more generally when \(some \ expression\leq{0}\) then \(some \ expression={(some \ expression)}\). For example: \(5=5=(5)\); When \(x\geq{0}\) then \(x=x\), or more generally when \(some \ expression\geq{0}\) then \(some \ expression={some \ expression}\). For example: \(5=5\); So, if \(x<0\), then \(x=x\) and \(y=x+x=x+x=0\). For more check here: mathabsolutevaluemodulus86462.html
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Re: Inequality and absolute value questions from my collection
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23 Dec 2012, 05:40
mridulparashar1 wrote: Bunuel wrote: 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. 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. First of all: \(n=n\), so \(n=n\) is the same as \(n=n\), which means that \(n\leq{0}\).
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Re: Inequality and absolute value questions from my collection
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21 Feb 2013, 21:23
Bunuel wrote: 5. What is the value of y? (1) 3x^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 > 3y=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.
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Re: Inequality and absolute value questions from my collection
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22 Feb 2013, 01:27
JJ2014 wrote: Bunuel wrote: 5. What is the value of y? (1) 3x^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 > 3y=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. x^24=x^24 when x^24>0; x^24=(x^24) when x^24<=0. So, the two equations you'll get from the original are relevant for different ranges of x. Hence, you cannot consider them as two separate equations and solve. To put it simply: we cannot get the single value of y from 3x^2 4 = y  2. Consider y=2 and x=2 OR y=11 and x=1. Hope it's clear.
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24 Apr 2013, 05:04
Bunuel wrote: 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^22xy+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
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Re: Inequality and absolute value questions from my collection
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24 Apr 2013, 05:26
Transcendentalist wrote: Bunuel wrote: 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^22xy+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 First of all when we combine we get that x^2+y^2=5a. If \(xya\neq{0}\), then the answer is YES but if \(xya={0}\), then the answer is NO. Next, it's a YES/NO DS question. In a Yes/No Data Sufficiency question, statement(s) is sufficient if the answer is “always yes” or “always no” while a statement(s) is insufficient if the answer is "sometimes yes" and "sometimes no".Hope it's clear.
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Re: Inequality and absolute value questions from my collection
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21 Sep 2013, 22:21
Bunuel wrote: 4. Are x and y both positive? (1) 2x2y=1 (2) x/y>1
(1) 2x2y=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. 2x2y=1 > y=x1/2, 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 timeconsuming and comfortable for you personally.
One of the approaches: \(2x2y=1\) > \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) > \(\frac{xy}{y}>0\) > 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 ?
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21 Sep 2013, 23:30
Bunuel wrote: 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 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 nn < 1 .
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22 Sep 2013, 05:27
StormedBrain wrote: Bunuel wrote: 4. Are x and y both positive? (1) 2x2y=1 (2) x/y>1
(1) 2x2y=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. 2x2y=1 > y=x1/2, 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 timeconsuming and comfortable for you personally.
One of the approaches: \(2x2y=1\) > \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) > \(\frac{xy}{y}>0\) > 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 ? 4. Are x and y both positive? The question asks whether point (x, y) is in the first quadrant. (1) 2x2y=1 > draw line y=x1/2: Attachment:
graph.png [ 6.13 KiB  Viewed 2291 times ]
Not sufficient. (2) x/y>1 > Draf line x/y=1. The solutions is the green region: Attachment:
graph (2).png [ 5.95 KiB  Viewed 2269 times ]
Not sufficient. (1)+(2) Intersection is the portion of the blue line which lies in the first quadrant. Sufficient. Answer: C. Hope it helps.
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Re: Inequality and absolute value questions from my collection
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22 Sep 2013, 06:00
StormedBrain wrote: Bunuel wrote: 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 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 nn < 1 . n*n < 1. If n<0, then we'll have n^2<1 > n^2>1. Which is true. So, n*n < 1 holds true for any negative value of n. If n>0, then we'll have n^2<1 > 1<n<1. So, n*n < 1 also holds true for 0<n<1. Thus 1/n > n holds true if n<0 and 0<n<1. Does this make sense?
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Re: Inequality and absolute value questions from my collection
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13 Oct 2013, 23:13
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^22xy+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.




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