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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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05 Sep 2012, 02:04
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. I do not wanna to be wrong but when we substitute x in xy/y>0 we should not have 1/2y>0 ??? and not 1/y>0.......then the meaning that y is positive eitherway not change. Let me know. Thanks
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Re: Inequality and absolute value questions from my collection [#permalink]
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05 Sep 2012, 03:53
carcass 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. I do not wanna to be wrong but when we substitute x in xy/y>0 we should not have 1/2y>0 ??? and not 1/y>0.......then the meaning that y is positive eitherway not change. Let me know. Thanks Check this post: inequalityandabsolutevaluequestionsfrommycollection8693960.html#p666191"I dropped 2, as (1/2y) > 0 and (1/y) >0 are absolutely the same (you can multiply both sides of inequality by 2 and you'll get 1/y>0). What is important that you can get that y>0 from either of them. Hope it's clear. "
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Re: Inequality and absolute value questions from my collection [#permalink]
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03 Oct 2012, 04:39
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!



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Re: Inequality and absolute value questions from my collection [#permalink]
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03 Oct 2012, 06:13
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 [#permalink]
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03 Oct 2012, 11:26
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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04 Oct 2012, 02:39
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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Last edited by carcass on 04 Oct 2012, 03:03, edited 2 times in total.



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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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04 Oct 2012, 03:20
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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Bunuel wrote: 11. Is x+y>xy? (1) x > y (2) xy < 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 xy? If and only when x and y have the same sign absolute value of x+y will always be more than absolute value of xy. As x+y when they have the same sign will contribute to each other and xy will not.
5+3=8 and 53=2 OR 53=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) xy < 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 viseversa is not right, meaning that x and y can have the same sign but x can be less than xy, but if x>xy the only possibility is x and y to have the same sign.)
Answer: B. Hi Bunuel  Can this solved in the below way? Is x+y>xy? Since both sides are +ve we can square both side of the inequality.... On squaring we get xy>0? statement 1 (1) x > y This is NS as xy can be opp sign as well as same sign (2) xy < x Squaring on both sides we get y^2 < 2xy Y cannot be zero otherwise the inequality cannot hold so Y^2 is +ve hence xy is +ve So we can answer the question xy>0 Is this fine Bunuel?



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11 Oct 2012, 02:22
Jp27 wrote: Bunuel wrote: 11. Is x+y>xy? (1) x > y (2) xy < 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 xy? If and only when x and y have the same sign absolute value of x+y will always be more than absolute value of xy. As x+y when they have the same sign will contribute to each other and xy will not.
5+3=8 and 53=2 OR 53=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) xy < 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 viseversa is not right, meaning that x and y can have the same sign but x can be less than xy, but if x>xy the only possibility is x and y to have the same sign.)
Answer: B. Hi Bunuel  Can this solved in the below way? Is x+y>xy? Since both sides are +ve we can square both side of the inequality.... On squaring we get xy>0? statement 1 (1) x > y This is NS as xy can be opp sign as well as same sign (2) xy < x Squaring on both sides we get y^2 < 2xy Y cannot be zero otherwise the inequality cannot hold so Y^2 is +ve hence xy is +ve So we can answer the question xy>0 Is this fine Bunuel? Yes, this approach is correct.
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23 Dec 2012, 03:40
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 [#permalink]
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23 Dec 2012, 04:29
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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23 Dec 2012, 04:37
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 [#permalink]
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23 Dec 2012, 04:40
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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25 Dec 2012, 04:19
Bunuel wrote: debayan222 wrote: Great collection Bunuel...Kudos.. Are these Qs. included in your signature or they exist as separate entity? merry Xmas...Happy Holidays. Yes, they are in Inequalities set. Thanks a lot Bunuel.. Well I guess, whatever Qs come from you directly to the forum, are included in you Sig. ? Hope I got you right..
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