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Inequality and absolute value questions from my collection [#permalink]
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16 Nov 2009, 11:33
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#p653709 Graphic approach: https://gmatclub.com/forum/inequalitya ... l#p12698025. 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 https://gmatclub.com/forum/inequalitya ... 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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08 Oct 2015, 13:34
Bunuel wrote: jayaddula 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 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.
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.
Answer: D. Hi Bunuel, I am getting E and just cannot understand D. Please see my solution below  I used number picking. A. xy<0, x=+ and y= For this condition choosing different values of x and y (x=2,y=6: x=3, y=7)satisfies the given condition in modulus. Hence x=y can be different value or x= and y=+  This condition doesn't satisfy the modulus condiotion B x>2 and y<2  As per the above stmt 1  condition 1, there can be various values for x and y, hence x+y is different. Hence E. I know I am going wrong some where, please help. thanks jay In your example, both pairs give the same value for x+y: 26=4 and 37=4. We can solve this question in another way: 7. x+2=y+2 what is the value of x+y?Square both sides: \(x^2+4x+4=y^2+4y+4\) > \(x^2y^2+4x4y=0\) > \((x+y)(xy)+4(xy)=0\) > \((xy)(x+y+4)=0\) > either \(x=y\) or \(x+y=4\). (1) xy<0 > the first case is not possible, since if \(x=y\), then \(xy=x^2\geq{0}\), not \(<0\) as given in this statement, hence we have the second case: \(x+y=4\). Sufficient. (2) x>2 and y<2. This statement implies that \(x\neq{y}\), therefore \(x+y=4\). Sufficient. Answer: D. Hope it's clear. Hi @Buenel, i'm having a really hard time understanding this question. First, I don't understand why x=y should imply a unique answer for x+y. Same for the second stantement, I don't fully understand why having the equation x+y=4 ensures a unique answer. Maybe I am missing some steps. Would greatly appreciate your help, or any1 else's help (maybe different approaches will help me understand better). Thanks in advance!
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Re: Inequality and absolute value questions from my collection [#permalink]
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09 Oct 2015, 21:22
petocities wrote: 7. x+2=y+2 what is the value of x+y? (1) xy<0 (2) x>2 y<2
Hi @Buenel, i'm having a really hard time understanding this question. First, I don't understand why x=y should imply a unique answer for x+y. Same for the second stantement, I don't fully understand why having the equation x+y=4 ensures a unique answer. Maybe I am missing some steps. Would greatly appreciate your help, or any1 else's help (maybe different approaches will help me understand better).
Thanks in advance! Hi petocities, I think this question requires more of observation about given information x+2=y+2 will be true for two possible cases of x and y Case 1: when x = y Case 2: For Values like (x=1 and y=5) or (x=2 and y=6) or (x=3 and y=7) ...etc. Case 1 gives inconsistent answers because for each different value of x and y, x+ywill be different but Case 2 always gives a consistent value of x+y=4 (Check all set of values mentioned above in Case 2) Statement 1 suggests that x and y are not equal (for x and y equal, their product must be NonNegative) i.e. Case 2 prevails which always gives us x+y=4 i.e. SUFFICIENT Statement 2 also rules out the scenario in which x and y may be equal i.e. Case 2 prevails again leading to a consistent value of x+y=4 i.e. SUFFICIENT Answer: Option D
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Re: Inequality and absolute value questions from my collection [#permalink]
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18 Oct 2015, 13:18
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've solved this one correctly, but have one question. A is ok  no questions. Can we manipulate Statement 2 and say n*n<1 as n is always positive we must be able to do this  but n*n can be also positive as it's not stated that n must be an integer, let's say 1/2*1/2<1 and it can be also any negative value as stated above.
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Re: Inequality and absolute value questions from my collection [#permalink]
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18 Oct 2015, 13:25
BrainLab 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've solved this one correctly, but have one question. A is ok  no questions. Can we manipulate Statement 2 and say n*n<1 as n is always positive we must be able to do this  but n*n can be also positive as it's not stated that n must be an integer, let's say 1/2*1/2<1 and it can be also any negative value as stated above. \(\frac{1}{ n }>n\) > multiply by \(n\) (we can safely do that since n>0): \(n*n < 1\). If \(n>0\), then we'll have \(n^2<1\) > \(1<n<1\). Since we consider the range when \(n>0\), then for this range we'll have \(0<n<1\). If \(n<0\), then we'll have \(n^2<1\) > \(n^2>1\). Which is true for any n from the range we consider. So, \(n*n < 1\) holds true for any negative value of n. Thus \(\frac{1}{ n }>n\) holds true if \(n<0\) and \(0<n<1\).
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Re: Inequality and absolute value questions from my collection [#permalink]
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04 Nov 2015, 19:46
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. How did you figure out that \(\frac{x}{y}>1\) > \(\frac{xy}{y}>0\) ? Any help is appreciated. Thank you.



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Re: Inequality and absolute value questions from my collection [#permalink]
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05 Nov 2015, 01:28
Johnbreeden85 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. How did you figure out that \(\frac{x}{y}>1\) > \(\frac{xy}{y}>0\) ? Any help is appreciated. Thank you. This is explained couple of times on the previous pages: \(\frac{x}{y}>1\) \(\frac{x}{y}1>0\) \(\frac{x}{y}\frac{y}{y}>0\) \(\frac{xy}{y}>0\). Hope it's clear.
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Re: Inequality and absolute value questions from my collection [#permalink]
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23 Feb 2016, 00:25
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. hey bunuel can you please clear my doubt? in statement 1 you've written either n is negative OR n equals to zero but as per my knowledge shouldn't n be negative only because I've read it in many post that are on absolute value, here's a link: mathabsolutevaluemodulus86462.htmlcorrect me if i'm wrong!



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Re: Inequality and absolute value questions from my collection [#permalink]
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23 Feb 2016, 00:43
nishantdoshi 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. hey bunuel can you please clear my doubt? in statement 1 you've written either n is negative OR n equals to zero but as per my knowledge shouldn't n be negative only because I've read it in many post that are on absolute value, here's a link: mathabsolutevaluemodulus86462.htmlcorrect me if i'm wrong! An absolute value cannot be negative but it CAN be 0. For this particular case 0 fits: n=n > 0 = 0 > 0 = 0.
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Re: Inequality and absolute value questions from my collection [#permalink]
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23 Feb 2016, 00:53
Bunuel wrote: nishantdoshi 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. hey bunuel can you please clear my doubt? in statement 1 you've written either n is negative OR n equals to zero but as per my knowledge shouldn't n be negative only because I've read it in many post that are on absolute value, here's a link: mathabsolutevaluemodulus86462.htmlcorrect me if i'm wrong! An absolute value cannot be negative but it CAN be 0. For this particular case 0 fits: n=n > 0 = 0 > 0 = 0. thanks for the reply my understanding about this topic is that... if x>=0 then x=xand if x<0 then x=x am i wrong? please reply!!!



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Re: Inequality and absolute value questions from my collection [#permalink]
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14 May 2016, 00:34
In your step you have taken it as (x3)^2 but isn't (x^26x+9) = (x3)^2 or (3x)^2? Also, what does 'reduce the expression by y' mean? Bunuel wrote: SOLUTIONS:
1. If 6*x*y = x^2*y + 9*y, what is the value of xy? (1) y – x = 3 (2) x^3< 0
First let's simplify given expression \(6*x*y = x^2*y + 9*y\):
\(y*(x^26x+9)=0\) > \(y*(x3)^2=0\). Note here that we CAN NOT reduce this expression by \(y\), as some of you did. Remember we are asked to determine the value of \(xy\), and when reducing by \(y\) you are assuming that \(y\) doesn't equal to \(0\). We don't know that.
Next: we can conclude that either \(x=3\) or/and \(y=0\). Which means that \(xy\) equals to 0, when y=0 and x any value (including 3), OR \(xy=3*y\) when y is not equal to zero, and x=3.
(1) \(yx=3\). If y is not 0, x must be 3 and yx to be 3, y must be 6. In this case \(xy=18\). But if y=0 then x=3 and \(xy=0\). Two possible scenarios. Not sufficient.
OR:
\(yx=3\) > \(x=y3\) > \(y*(x3)^2=y*(y33)^2=y(y6)^2=0\) > either \(y=0\) or \(y=6\) > if \(y=0\), then \(x=3\) and \(xy=0\) \(or\) if \(y=6\), then \(x=3\) and \(xy=18\). Two different answers. Not sufficient.
(2) \(x^3<0\). x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.
Answer: B.
This one was quite tricky and was solved incorrectly by all of you.
Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.
Never multiply (or reduce) inequality by variable (or expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.



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Re: Inequality and absolute value questions from my collection [#permalink]
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14 May 2016, 01:06
ameyaprabhu wrote: In your step you have taken it as (x3)^2 but isn't (x^26x+9) = (x3)^2 or (3x)^2? Also, what does 'reduce the expression by y' mean? Bunuel wrote: SOLUTIONS:
1. If 6*x*y = x^2*y + 9*y, what is the value of xy? (1) y – x = 3 (2) x^3< 0
First let's simplify given expression \(6*x*y = x^2*y + 9*y\):
\(y*(x^26x+9)=0\) > \(y*(x3)^2=0\). Note here that we CAN NOT reduce this expression by \(y\), as some of you did. Remember we are asked to determine the value of \(xy\), and when reducing by \(y\) you are assuming that \(y\) doesn't equal to \(0\). We don't know that.
Next: we can conclude that either \(x=3\) or/and \(y=0\). Which means that \(xy\) equals to 0, when y=0 and x any value (including 3), OR \(xy=3*y\) when y is not equal to zero, and x=3.
(1) \(yx=3\). If y is not 0, x must be 3 and yx to be 3, y must be 6. In this case \(xy=18\). But if y=0 then x=3 and \(xy=0\). Two possible scenarios. Not sufficient.
OR:
\(yx=3\) > \(x=y3\) > \(y*(x3)^2=y*(y33)^2=y(y6)^2=0\) > either \(y=0\) or \(y=6\) > if \(y=0\), then \(x=3\) and \(xy=0\) \(or\) if \(y=6\), then \(x=3\) and \(xy=18\). Two different answers. Not sufficient.
(2) \(x^3<0\). x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.
Answer: B.
This one was quite tricky and was solved incorrectly by all of you.
Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.
Never multiply (or reduce) inequality by variable (or expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. Both (x3)^2 and (3x)^2 are the same.
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02 Sep 2016, 23:52
Bunuel wrote: 13. Is x1 < 1? (1) (x1)^2 <= 1 (2) x^2  1 > 0
Last one.
Is x1 < 1? Basically the question asks is 0<x<2 true?
(1) (x1)^2 <= 1 > x^22x<=0 > x(x2)<=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. Hi Bunuel, Can you please explain why have you not considered (for option1) the other case. I mean, x(x2)<=0 can lead to two possiblities, one that you have mentioned, and the other one could be just the opposite x<=0 and x>=2. I need to understand this, please reply. Thanks.



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04 Sep 2016, 04:50
rpradhan25 wrote: Bunuel wrote: 13. Is x1 < 1? (1) (x1)^2 <= 1 (2) x^2  1 > 0
Last one.
Is x1 < 1? Basically the question asks is 0<x<2 true?
(1) (x1)^2 <= 1 > x^22x<=0 > x(x2)<=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. Hi Bunuel, Can you please explain why have you not considered (for option1) the other case. I mean, x(x2)<=0 can lead to two possiblities, one that you have mentioned, and the other one could be just the opposite x<=0 and x>=2. I need to understand this, please reply. Thanks. x(x2)<=0 is true for 0<=x<=2 and not true for any other range. Check the links below: Inequalities Made Easy!Solving Quadratic Inequalities  Graphic ApproachInequality tipsWavy Line Method Application  Complex Algebraic InequalitiesDS Inequalities Problems PS Inequalities Problems 700+ Inequalities problemsinequalitiestrick91482.htmldatasuffinequalities109078.htmlrangeforvariablexinagiveninequality109468.htmleverythingislessthanzero108884.htmlgraphicapproachtoproblemswithinequalities68037.htmlHope it helps.
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Re: Inequality and absolute value questions from my collection [#permalink]
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27 Dec 2016, 05:11
[quote="Bunuel"]SOLUTIONS:
1. If 6*x*y = x^2*y + 9*y, what is the value of xy? (1) y – x = 3 (2) x^3< 0
First let's simplify given expression \(6*x*y = x^2*y + 9*y\):
\(y*(x^26x+9)=0\) > \(y*(x3)^2=0\). Note here that we CAN NOT reduce this expression by \(y\), as some of you did. Remember we are asked to determine the value of \(xy\), and when reducing by \(y\) you are assuming that \(y\) doesn't equal to \(0\). We don't know that.
Next: we can conclude that either \(x=3\) or/and \(y=0\). Which means that \(xy\) equals to 0, when y=0 and x any value (including 3), OR \(xy=3*y\) when y is not equal to zero, and x=3.
(1) \(yx=3\). If y is not 0, x must be 3 and yx to be 3, y must be 6. In this case \(xy=18\). But if y=0 then x=3 and \(xy=0\). Two possible scenarios. Not sufficient.
OR:
\(yx=3\) > \(x=y3\) > \(y*(x3)^2=y*(y33)^2=y(y6)^2=0\) > either \(y=0\) or \(y=6\) > if \(y=0\), then \(x=3\) and \(xy=0\) \(or\) if \(y=6\), then \(x=3\) and \(xy=18\). Two different answers. Not sufficient.
(2) \(x^3<0\). x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.
Answer: B. This one was quite tricky and was solved incorrectly by all of you.
Hi Bunuel,
I am a big fan of your posts and all your easy explanations, just want to point out a possible correction in your explanation of question 1 here. From the statement (ii) x^3<0, it is clear that x<0. But how did you arrive at the value of x=3 without combining statement (i) which gives two values of x=3,3? That is the reason I think the answer is (C) when we get x=3 using both statement (i) and statement (ii) and hence the value of xy.
Let me know if I am correct.
Thanks!



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27 Dec 2016, 06:52
naren01 wrote: Bunuel wrote: SOLUTIONS:
1. If 6*x*y = x^2*y + 9*y, what is the value of xy? (1) y – x = 3 (2) x^3< 0
First let's simplify given expression \(6*x*y = x^2*y + 9*y\):
\(y*(x^26x+9)=0\) > \(y*(x3)^2=0\). Note here that we CAN NOT reduce this expression by \(y\), as some of you did. Remember we are asked to determine the value of \(xy\), and when reducing by \(y\) you are assuming that \(y\) doesn't equal to \(0\). We don't know that.
Next: we can conclude that either \(x=3\) or/and \(y=0\). Which means that \(xy\) equals to 0, when y=0 and x any value (including 3), OR \(xy=3*y\) when y is not equal to zero, and x=3.
(1) \(yx=3\). If y is not 0, x must be 3 and yx to be 3, y must be 6. In this case \(xy=18\). But if y=0 then x=3 and \(xy=0\). Two possible scenarios. Not sufficient.
OR:
\(yx=3\) > \(x=y3\) > \(y*(x3)^2=y*(y33)^2=y(y6)^2=0\) > either \(y=0\) or \(y=6\) > if \(y=0\), then \(x=3\) and \(xy=0\) \(or\) if \(y=6\), then \(x=3\) and \(xy=18\). Two different answers. Not sufficient.
(2) \(x^3<0\). x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.
Answer: B. This one was quite tricky and was solved incorrectly by all of you.
Hi Bunuel,
I am a big fan of your posts and all your easy explanations, just want to point out a possible correction in your explanation of question 1 here. From the statement (ii) x^3<0, it is clear that x<0. But how did you arrive at the value of x=3 without combining statement (i) which gives two values of x=3,3? That is the reason I think the answer is (C) when we get x=3 using both statement (i) and statement (ii) and hence the value of xy.
Let me know if I am correct.
Thanks! No, you are not correct. From \(y*(x3)^2=0\) it follows that either \(x=3\) or/and \(y=0\). (2) says that \(x^3<0\), thus x is not 3, therefore y must be 0 > xy = 0.
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30 Dec 2016, 07:50
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, thank you so much for such an amazing post, so so so helpful. A quick query, regarding statement 2 here: (2) \(\frac{1}{n} > n\), Shouldn't it be true for all values of n such that n<1 (n#0) ? Eg: n =1/2: \(\frac{1}{(1/2)} > \frac{1}{2}\) : \(2 > \frac{1}{2}\) PS: It doesn't alter the final answer though.
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30 Dec 2016, 08:09
RR88 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, thank you so much for such an amazing post, so so so helpful. A quick query, regarding statement 2 here: (2) \(\frac{1}{n} > n\), Shouldn't it be true for all values of n such that n<1 (n#0) ? Eg: n =1/2: \(\frac{1}{(1/2)} > \frac{1}{2}\) : \(2 > \frac{1}{2}\) PS: It doesn't alter the final answer though. Yes, but the fact that it's true for all negative n's was enough to discard this statement. So, we did not need to find the actual range. Still if interested here it is: \(\frac{1}{ n }>n\) > multiply by \(n\) (we can safely do that since n>0): \(n*n < 1\). If \(n>0\), then we'll have \(n^2<1\) > \(1<n<1\). Since we consider the range when \(n>0\), then for this range we'll have \(0<n<1\). If \(n<0\), then we'll have \(n^2<1\) > \(n^2>1\). Which is true for any n from the range we consider. So, \(n*n < 1\) holds true for any negative value of n. Thus \(\frac{1}{ n }>n\) holds true if \(n<0\) and \(0<n<1\).
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Re: Inequality and absolute value questions from my collection [#permalink]
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25 Jul 2017, 13:48
Bunuel wrote: 8. a*b#0. Is a/b=a/b? (1) a*b=a*b (2) a/b=a/b a/b=a/b is true if and only a and b have the same sign, meaning a/b is positive.
(1) a*b=a*b, means a and b are both positive or both negative, as LHS is never negative (well in this case LHS is positive as neither a nor b equals to zero). Hence a/b is positive in any case. Hence a/b=a/b. Sufficient.
(2) a/b=a/b, from this we can not conclude whether they have the same sign or not. Not sufficient.
Answer: A. Hi Bunuel, From 2, can't we say, a/b>=0, in this case, both a & B will be either +ve or ve. Hence, sufficient? Pls pardon my ignorance.



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Re: Inequality and absolute value questions from my collection [#permalink]
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25 Jul 2017, 20:55
ManishKM1 wrote: Bunuel wrote: 8. a*b#0. Is a/b=a/b? (1) a*b=a*b (2) a/b=a/b a/b=a/b is true if and only a and b have the same sign, meaning a/b is positive.
(1) a*b=a*b, means a and b are both positive or both negative, as LHS is never negative (well in this case LHS is positive as neither a nor b equals to zero). Hence a/b is positive in any case. Hence a/b=a/b. Sufficient.
(2) a/b=a/b, from this we can not conclude whether they have the same sign or not. Not sufficient.
Answer: A. Hi Bunuel, From 2, can't we say, a/b>=0, in this case, both a & B will be either +ve or ve. Hence, sufficient? Pls pardon my ignorance. Absolute value of a number is nonnegative, so both a and b are >=0. So, naturally a/b>=0. But we cannot know whether a and b are positive or negative? Consider a = 1 and b = 2 OR a = 1 and b = 2.
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