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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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20 Sep 2013, 08:06



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20 Sep 2013, 20:46
Bunuel, I think I really need your help in resolving this for me. Please?? In the inequalities set that you posted, I've into this same situation with only problem 5 and 7. Actually my approach is not different. Looking at problem 7 Quote: 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.
In the above problem, when you are solving the option (2) you said as follows: x>2 and y<2 and hence as x is not equal to y, it has to be scenario B. My doubt is a similar one here: When we made Scenario B we know that Scenario B will hold only if: This will occur when either x or y is less then 2 and the other is more than 2.In such cases, shouldn't we compare [x>2 and y<2] from option (2) with the condition of Scenario B, before deciding that it belongs to Scenario B. Please please help me on this



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21 Sep 2013, 03:22
gmatter0913 wrote: Bunuel, I think I really need your help in resolving this for me. Please?? In the inequalities set that you posted, I've into this same situation with only problem 5 and 7. Actually my approach is not different. Looking at problem 7 Quote: 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.
In the above problem, when you are solving the option (2) you said as follows: x>2 and y<2 and hence as x is not equal to y, it has to be scenario B. My doubt is a similar one here: When we made Scenario B we know that Scenario B will hold only if: This will occur when either x or y is less then 2 and the other is more than 2.In such cases, shouldn't we compare [x>2 and y<2] from option (2) with the condition of Scenario B, before deciding that it belongs to Scenario B. Please please help me on this Please check easier solution here: inequalityandabsolutevaluequestionsfrommycollection86939160.html#p1111747
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Re: Inequality and absolute value questions from my collection [#permalink]
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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 1260 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 1239 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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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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Bunuel wrote: 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 Not sufficient. (2) x/y>1 > Draf line x/y=1. The solutions is the green region: Attachment: graph (1).png Not sufficient. (1)+(2) Intersection is the portion of the blue line which lies in the first quadrant. Sufficient. Answer: C. Hope it helps. Hey Bunuel, I am a bit confused. Shouldn't the green area in 3rd quadrant be above the line and below xaxis ? Lets take a point (0.5,1) in the green shaded region , then 0.5/1 = 1/2 <1..
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22 Sep 2013, 08:16
StormedBrain wrote: Bunuel wrote: StormedBrain wrote: 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 Not sufficient. (2) x/y>1 > Draf line x/y=1. The solutions is the green region: Attachment: graph (1).png Not sufficient. (1)+(2) Intersection is the portion of the blue line which lies in the first quadrant. Sufficient. Answer: C. Hope it helps. Hey Bunuel, I am a bit confused. Shouldn't the green area in 3rd quadrant be above the line and below xaxis ? Lets take a point (0.5,1) in the green shaded region , then 0.5/1 = 1/2 <1.. Yes, I marked the wrong region. Edited. Thank you. +1.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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22 Sep 2013, 08:18
Bunuel wrote: 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? Thanks Bunuel . I dedicated whole day today to inequality . Lots of learning.. thanks again.
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23 Sep 2013, 13:22
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.
Bunuel, why can't we open the first modulus as well? x+2 and x2? thanks



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24 Sep 2013, 00:10
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.
Hello Bunuel, what about (x1) ^ 2 < 1 means (x1) is for sure less than 1 , as square of less than 1 can only occur if the original number is less than 1?



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25 Sep 2013, 07:16
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. 1+2 also insufficient when a is ve .. consider a 1 ..
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25 Sep 2013, 08:21
Bunuel wrote: a cannot be a negative number because each statement says that a=(nonnegative value). yeah .. correct ..haha, how did I miss that ..
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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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14 Oct 2013, 00:02
madn800 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.
Also insufficient as x,y, and a could be 0 Why did you assume that ALL COULD be zero. Because they could be 0, why not? And IF x=y=a=0, then the answer would be NO but if they take some other values then the answer would be YES.
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Re: Inequality and absolute value questions from my collection [#permalink]
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14 Oct 2013, 20:10
12. Is r=s? (1) s<=r<=s (2) r>=s
From (1) take a number line
s0s
So, 5 cases are possible r0s (or) sr0s (or) s0rs (or) srs (r can be zero also) (or) s0r The last case is the correct answer, but we have other 4 cases.INSUFFICIENT.
From (2) r>=s >> r>=s and r>=s >> r>=s and r<=s
take number line (s)0+s Still we cannot say wether r=s. INSUFFICIENT.
NOW combine the two(1+2) we get (s)0+s i.e., r can lie anywhere on the line and still we CANNOT say wether r=s. Hence, E.




Re: Inequality and absolute value questions from my collection
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