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Inequality and absolute value questions from my collection [#permalink]
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Guys I didn't forget your request, just was collecting good questions to post. So here are some inequality and absolute value questions from my collection. Not every problem below is hard, but there are a few, which are quite tricky. Please provide your explanations along with the answers. 1. If \(6*x*y = x^2*y + 9*y\), what is the value of xy? (1) \(y – x = 3\) (2) \(x^3< 0\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p6536902. If y is an integer and \(y = x + x\), is \(y = 0\)? (1) \(x < 0\) (2) \(y < 1\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p6536953. Is \(x^2 + y^2 > 4a\)?(1) \((x + y)^2 = 9a\) (2) \((x – y)^2 = a\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p6536974. Are x and y both positive?(1) \(2x2y=1\) (2) \(\frac{x}{y}>1\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p6537095. What is the value of y?(1) \(3x^2 4 = y  2\) (2) \(3  y = 11\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p6537316. If x and y are integer, is y > 0? (1) \(x +1 > 0\) (2) \(xy > 0\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p6537407. \(x+2=y+2\) what is the value of x+y?(1) \(xy<0\) (2) \(x>2\), \(y<2\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p653783 AND http://gmatclub.com/forum/inequalityan ... l#p11117478. \(a*b \neq 0\). Is \(\frac{a}{b}=\frac{a}{b}\)?(1) \(a*b=a*b\) (2) \(\frac{a}{b}=\frac{a}{b}\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p6537899. Is n<0?(1) \(n=n\) (2) \(n^2=16\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p65379210. If n is not equal to 0, is n < 4 ?(1) \(n^2 > 16\) (2) \(\frac{1}{n} > n\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p65379611. Is \(x+y>xy\)?(1) \(x > y\) (2) \(xy < x\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p65385312. Is r=s?(1) \(s \leq r \leq s\) (2) \(r \geq s\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p65387013. Is \(x1 < 1\)?(1) \((x1)^2 \leq 1\) (2) \(x^2  1 > 0\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p653886Official answers (OA's) and detailed solutions are in my posts on pages 2 and 3.PLEASE READ THE WHOLE DISCUSSION BEFORE POSTING A QUESTION.
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Re: Inequality and absolute value questions from my collection [#permalink]
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21 Feb 2013, 10:57
Thanks Bunuel. These questions are of great value indeed!



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Re: Inequality and absolute value questions from my collection [#permalink]
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21 Feb 2013, 19:55
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. I got C. Can you please explain why this is incorrect? Statement 1 gives x^2+2xy+y^2 = 9a, and I rewrote it to 9a2xy = x^2+y^2 Statement 2 gives x^22xy+y^2=a, and i rewrote this to x^2+y^2= a +2xy Together, I have 9a2xy = a+2xy, which leads to 8a = 4xy, but 8a is also equal to x^2+y^2? ... this last part must be wrong?
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Re: Inequality and absolute value questions from my collection [#permalink]
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21 Feb 2013, 20:23
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 [#permalink]
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22 Feb 2013, 00:18
JJ2014 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. I got C. Can you please explain why this is incorrect? Statement 1 gives x^2+2xy+y^2 = 9a, and I rewrote it to 9a2xy = x^2+y^2 Statement 2 gives x^22xy+y^2=a, and i rewrote this to x^2+y^2= a +2xy Together, I have 9a2xy = a+2xy, which leads to 8a = 4xy, but 8a is also equal to x^2+y^2? ... this last part must be wrong? The red part is not right. If you sum the two equations you'll get 2(x^2+y^2)=10a > x^2+y^2=5a. Below posts might help with this question: inequalityandabsolutevaluequestionsfrommycollection8693980.html#p687991inequalityandabsolutevaluequestionsfrommycollection86939100.html#p746278Hope it helps.
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Re: Inequality and absolute value questions from my collection [#permalink]
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22 Feb 2013, 00:27
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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Re: Inequality and absolute value questions from my collection [#permalink]
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22 Feb 2013, 06:14
Bunuel wrote: JJ2014 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. I got C. Can you please explain why this is incorrect? Statement 1 gives x^2+2xy+y^2 = 9a, and I rewrote it to 9a2xy = x^2+y^2 Statement 2 gives x^22xy+y^2=a, and i rewrote this to x^2+y^2= a +2xy Together, I have 9a2xy = a+2xy, which leads to 8a = 4xy, but 8a is also equal to x^2+y^2? ... this last part must be wrong? The red part is not right. If you sum the two equations you'll get 2(x^2+y^2)=10a > x^2+y^2=5a. Below posts might help with this question: inequalityandabsolutevaluequestionsfrommycollection8693980.html#p687991inequalityandabsolutevaluequestionsfrommycollection86939100.html#p746278Hope it helps. So I understand that you summed the equations and got that answer. But why is setting them equal to each other wrong in this case? I'm trying to figure out what concept I'm missing so that I don't end up doing it again.
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22 Feb 2013, 06:26
Bunuel wrote: 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. This is clear. Thank you!!
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Re: Inequality and absolute value questions from my collection [#permalink]
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22 Feb 2013, 07:09
JJ2014 wrote: Bunuel wrote: JJ2014 wrote: I got C. Can you please explain why this is incorrect?
Statement 1 gives x^2+2xy+y^2 = 9a, and I rewrote it to 9a2xy = x^2+y^2 Statement 2 gives x^22xy+y^2=a, and i rewrote this to x^2+y^2= a +2xy
Together, I have 9a2xy = a+2xy, which leads to 8a = 4xy, but 8a is also equal to x^2+y^2? ... this last part must be wrong?
The red part is not right. If you sum the two equations you'll get 2(x^2+y^2)=10a > x^2+y^2=5a. Below posts might help with this question: inequalityandabsolutevaluequestionsfrommycollection8693980.html#p687991inequalityandabsolutevaluequestionsfrommycollection86939100.html#p746278Hope it helps. So I understand that you summed the equations and got that answer. But why is setting them equal to each other wrong in this case? I'm trying to figure out what concept I'm missing so that I don't end up doing it again. What are you trying to get when setting "them" equal? Anyway, you won't be able to solve two equations with three unknowns.
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Re: Inequality and absolute value questions from my collection [#permalink]
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28 Feb 2013, 05:35
7. x+2=y+2 what is the value of x+y? (1) xy<0 (2) x>2 y<2 The solution seem confusing to me as I see four cases: a] x<2, y<2 b]x>2, y>2 c] x<2, y>2 d]x>2, y<2
case [a] and [b] support x=y while case [c] and [d] support x+y=4
when xy<0, the case [c]or[d] always do not apply, for example: x=3 and y=3 would come under case[c] and x=1 and y=3 would come under case [b] , so it is insufficient.
when x>2 , y<2, we have a case [b] with x=3, y=1 and a case [d] with x=3,y=3. So insufficient
when we combine(1)+(2) , we have a case as shown above , it is also insufficient.
So my answer choice would be E.
Can somebody help if I am wrong.



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Re: Inequality and absolute value questions from my collection [#permalink]
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28 Feb 2013, 05:42
piealpha wrote: 7. x+2=y+2 what is the value of x+y? (1) xy<0 (2) x>2 y<2 The solution seem confusing to me as I see four cases: a] x<2, y<2 b]x>2, y>2 c] x<2, y>2 d]x>2, y<2
case [a] and [b] support x=y while case [c] and [d] support x+y=4
when xy<0, the case [c]or[d] always do not apply, for example: x=3 and y=3 would come under case[c] and x=1 and y=3 would come under case [b] , so it is insufficient.
when x>2 , y<2, we have a case [b] with x=3, y=1 and a case [d] with x=3,y=3. So insufficient
when we combine(1)+(2) , we have a case as shown above , it is also insufficient.
So my answer choice would be E.
Can somebody help if I am wrong. Please read the thread: 11 pages of good discussion. Links to OA's and solutions are given in the original post: inequalityandabsolutevaluequestionsfrommycollection86939200.html#p652806OA for this question is D, not E. Discussed here: inequalityandabsolutevaluequestionsfrommycollection8693940.html#p653783 and here: inequalityandabsolutevaluequestionsfrommycollection86939160.html#p1111747Hope it helps.
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Re: Inequality and absolute value questions from my collection [#permalink]
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28 Feb 2013, 07:36
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Question 1:
\(6xy = x^2 y + 9y\)
\(y(x^2 6x +9) = 0\)
\(y(x3)^2 = 0\)
either y =0, or x=3
statement 1: yx =3 If y= 0, xy =0, irrespective of x If x=3, y =6, xy= 18
So, A & D are not correct
statement 2:
\(x^3 < 0 => x <0\)
=> x is not equal to 3 so y=0, and xy = 0
Correct Answer B



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Re: Inequality and absolute value questions from my collection [#permalink]
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17 Apr 2013, 03:18
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 Bunnel Solved the 1st statement like this  \((x + y)^2 = 9a\) Since \(x^2 + y^2 >= 2xy\) \(x^2 + y^2 + x^2 + y^2 >= 9a\) \(2(x^2 + y^2) >= 9a\) \(x^2 + y^2 >= 4.5a\) Now this would have been sufficient if a is not = 0 had been given in the stem Is this approach to the problem alright?? Is this st sufficient if it is given that a is not equal to 0 Thanks



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Re: Inequality and absolute value questions from my collection [#permalink]
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17 Apr 2013, 05:34
Dipankar6435 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 Bunnel Solved the 1st statement like this  \((x + y)^2 = 9a\) Since \(x^2 + y^2 >= 2xy\) \(x^2 + y^2 + x^2 + y^2 >= 9a\) \(2(x^2 + y^2) >= 9a\) \(x^2 + y^2 >= 4.5a\) Now this would have been sufficient if a is not = 0 had been given in the stem Is this approach to the problem alright?? Is this st sufficient if it is given that a is not equal to 0 Thanks Yes, but we don't know whether x is 0.
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Re: Inequality and absolute value questions from my collection [#permalink]
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24 Apr 2013, 04:04
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 [#permalink]
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24 Apr 2013, 04:26
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 [#permalink]
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27 May 2013, 22:57
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. Hi Bunnel, A small doubt: On GMAT (x2)^1/2 = mod X so in 2  is mod x= mod 4? ....= +/x= +/4 ? still this will give x = +/4, but i just wanted this clarification



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Re: Inequality and absolute value questions from my collection [#permalink]
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27 May 2013, 23:50
cumulonimbus 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. Hi Bunnel, A small doubt: On GMAT (x2)^1/2 = mod X so in 2  is mod x= mod 4? ....= +/x= +/4 ? still this will give x = +/4, but i just wanted this clarification Yes, you can do this way: \(n^2=16\) > \(\sqrt{n^2}=\sqrt{16}=4\) > \(n=4\) > \(n=4\) or \(n=4\). Hope it's clear.
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Re: Inequality and absolute value questions from my collection [#permalink]
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22 Jun 2013, 10:48
hariunplugs wrote: Bunuel wrote: 7. x+2=y+2 what is the value of x+y? (1) xy<0 (2) x>2 y<2 This one is quite interesting.
First note that x+2=y+2 can take only two possible forms:
A. x+2=y+2 > x=y. This will occur [b]if and only x and y are both >= than 2 OR both <= than 2. In that case x=y. Which means that their product will always be positive or zero when x=y=0. B. x+2=y2 > x+y=4. This will occur when either x or y is less then 2 and the other is more than 2.[/b]
When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=4. Also note that viseversa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.
(1) xy<0 > We have scenario B, hence x+y=4. Sufficient.
(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=4. Sufficient.
Hi,Can u explain the logic behind bolded part?...i tried my best..but not able comprehend much..thanks in advance.
Answer: D. Check the following posts: inequalityandabsolutevaluequestionsfrommycollection8693960.html#p666223inequalityandabsolutevaluequestionsfrommycollection8693980.html#p677045inequalityandabsolutevaluequestionsfrommycollection86939140.html#p1033611inequalityandabsolutevaluequestionsfrommycollection86939140.html#p1049828Alternative solution: inequalityandabsolutevaluequestionsfrommycollection86939160.html#p1111747Hope it helps.
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Re: Inequality and absolute value questions from my collection [#permalink]
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27 Jun 2013, 17:12
1. If 6*x*y = x^2*y + 9*y, what is the value of xy?
6xy = yx^2 + 9y
6xy  yx^2 9y = 0 y(6x  x^2  9) = 0 y(x^2  6x +9) = 0 y = 0 OR (x  3)^2 = 0 x  3 = 0 x = 3
(CANNOT reduce by y as that would be assuming y is equal to zero and we are looking for what xy is)
(1) y – x = 3 x = y  3
IF x = 3 3 = y  3 6 = y (x = 3, y = 6) xy = 18
IF y = 0 x = 0  3 x = 3 xy = 0
INSUFFICIENT
(2) x^3< 0
x < 0
(FROM STEM: x = 3 OR y = 0)
X = 3 6xy = yx^2 + 9y 6(3)y = y(3^2) + 9y 18y = 9y + 9y 18y = 18y y=y
Y = 0 6xy = yx^2 + 9y 6x(0) = (0)x^2 + 9(0) 0 = 0
We are told that x is negative, meaning x cannot = 3. If x doesn't = 3 then y = 0 in which case xy = 0 SUFFICIENT.



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Re: Inequality and absolute value questions from my collection [#permalink]
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27 Jun 2013, 18:36
2. If y is an integer and y = x + x, is y = 0?
y ≥ 0
(1) x < 0 If x is negative and y = x + x then the positive value of x will be canceled out by the addition of  x. y = 0 SUFFICIENT
(2) y < 1 If y ≥ 0 and it's less than 1, y must be zero as y is an integer. SUFFICIENT
(D)




Re: Inequality and absolute value questions from my collection
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