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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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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.
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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. just to chime in here your thanks for all this..it's really useful



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6. If x and y are integer, is y > 0? (1) x +1 > 0 (2) xy > 0 (1) x+1>0 > x>1. As x is an integer x can take the following values 0,1,2,... But we know nothing about y. Not sufficient. (2) xy>0. x and y have the same sign (both positive OR both negative) and neither x nor y is zero. Not sufficient. (1)+(2) x is positive, as from (1) it's 0,1,2.. and from (2) x is not zero. Hence xy to be positive y also must be positive. Sufficient. Answer: C.
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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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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.
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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.
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12. Is r=s?(1) s<=r<=s (2) r>=s This one is tough. (1) s<=r<=s, we can conclude two things from this statement: A. s is either positive or zero, as s<=s; B. r is in the range (s,s) inclusive, meaning that r can be s as well as s. But we don't know whether r=s or not. Not sufficient. (2) r>=s, clearly insufficient. (1)+(2) s<=r<=s, s is not negative, r>=s > r>=s or r<=s. This doesn't imply that r=s, from this r can be s as well. Consider: s=5, r=5 > 5<=5<=5 5>=5 s=5, r=5 > 5<=5<=5 5>=5 Both statements are true with these values. Hence insufficient. Answer: E.
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Bunuel wrote: 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.
(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. would you think this is a 700+ question?



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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. Just curious if my thinking is correct. on the 2nd part I get y = 8 and y =14 Then I substituted the values into the first equation: 3x^24=10 the answer will never give 10/3 do the same for 14 3x^24=12 x = 0 using my methodology I also got C, but is my thinking correct?



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Awesome stuff Bunuel! Hats off to you dude. +5 from me.



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Awesome, not only have u put the question, but solution to all the problems. I am learning a lot. Thanks to Bunuel. Bunuel, more questions please.



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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 2<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 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 2<x<2 is true. Not sufficient.
Answer: E. Bunuel, two questions: shouldnt x1 < 1 be 0<x<2? and not 2<x<2? secondly, how does this happen: x(x2)<=0 > 0<=x<=2? does this not translate into x<=0 or x<=2? thank you very much for all the questions and solutions.



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tihor wrote: Bunuel, two questions: shouldnt x1 < 1 be 0<x<2? and not 2<x<2? secondly, how does this happen: x(x2)<=0 > 0<=x<=2? does this not translate into x<=0 or x<=2?
thank you very much for all the questions and solutions. Thank you very much for this catch. +1. There was a typo. So you are right with the first one: x1 < 1 means 0<x<2. Already edited the post. As for the second one: x(x2)<=0 means 0<=x<=2, if you plug the values from this range you'll get the values less than or equal to 0. If you plug the values less than 0 or more than 2 you'll get only positive values. x(x2) is "smiling" parabola, and the intersections with Xaxis are at x=0 and x=2, the range between will be below Xaxis. Hope it helps.
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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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