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Inequality and absolute value questions from my collection [#permalink]
16 Nov 2009, 11:33
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Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 03:15
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gmat620 wrote: Bunuel, thanks for the questions. Please provide the OA's too. It would be great if you can provide them soon. I am having my GMAT this week, so kinda tensed and impatient. Also, I am yet to give my MGMAT CAT's, so tell me whether should I solve the questions on the forum because if the questions are from the MGMAT CAT's or Gmat Prep then it may overestimate my result. I would appreciate your response. Thanks once again. These questions are from various sources. Couple of questions might be from MGMAT CAT or Gmat Prep, but not more than that. I'll provide OA in a day or two, after discussions. Tell me if you want the answers for the specific questions earlier than that and I'll mail you.
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Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 07:38
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3) I) (x+y)^2=9a x^2+y^2=9a-2xy NS II) (x-y)^2=a x^2+y^2=a+2xy NS Together 2(x^2+y^2)=10a x^2+y^2=5a If either x or y are larger than 0, the stem would be true, but if they’re both zero the stem is false, hence E
4) I don’t get the two clues; they seem to be mutually exclusive
5) I) 3|x^2-4|=y-2 either y=3x^2-10 or y=14-3x^2 NS II) |3-y|=11 either y=-8 or y=14 NS Together -8=3x^2-10 so 3x^2=2 ok 14=3x^2-10 so 3x^2=28 ok, hence E
6) I) x+1>0 so x={0, 1, 2, …} NS II) xy>0 so x and y have the same sign and none of them is zero NS Together, x={1, 2, 3, ..} and y has the same sign, hence C
7) |x+2|=|y+2| either x+2=y+2 or x+2=-y-2 (the other two combinations can be transformed into these by multiplying by -1) Reordering: x-y=0 or x+y=-4 I)xy<0, hence x and y have different signs and none of them is zero. The only possibility is x+y=-4 S II) x>2, y<2 hence x#y. The only possibility is x+y=-4 S, therefore D
8)a*b#0, hence a and b are both non-zero I) |a*b|=a*b a and b have the same sign and the stem is always true S II) |a|/|b|=|a/b| this is true regardless of the values of a and b, and nothing can be said about the stem NS, therefore A
9) I) –n=|-n| n<=0 NS II) n^2=16 n=+/-4 NS Together n=-4 therefore C
10)n#0 I) n^2>16, so |n|>4 S II) 1/|n|>n true for n<-1 NS, therefore A
11) Plugging in numbers I get B, but there’s no rime or reason to my solution
12) I) –s<=r<=s obviously NS. Since s>=-s, s is either positive or zero II)|r|>=s obviously NS Together: I) tells us that s>=0; II) tells us that r>=s or r<=-s. The only case in which I and II are simultaneously satisfied is r=s, therefore C
13) x=(0:2) with 0 and 2 excluded I) (x-1)^2<=1, hence x=[0:2] with 0 and 2 included, hence NS II) x^2-1>0 x<-1 or x>1. For x=1.5 the stem is true, for x=3 it is false, hence NS Together, for x=1.5 the stem is true, for x=2 it is false, hence E
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Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 10:07
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Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 10:34
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Bunuel wrote: 5. What is the value of y? (1) 3|x^2 -4| = y - 2 (2) |3 - y| = 11
Statement 1: Two equations, two unknowns... INSUFFICIENT Statement 2: |3 - y| = 11 (3-y)=11 or (3-y)=-11 y=-8, 14 INSUFFICIENT Statements 1 and 2: y must be 14 because 3|x^2 -4| can never be a negative value (no matter what you plug in for x, you will get a positve value because of the absolute value signs). SUFFICIENT ANSWER: C.
Last edited by h2polo on 17 Nov 2009, 10:54, edited 1 time in total.
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Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 13:47
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Quote: 2. If y is an integer and y = |x| + x, is y = 0? (1) x < 0 (2) y < 1 Question Stem gives us :
(a) If x > 0 ; y = 2x (b) If x < 0 ; y = 0 St. (1) : x < 0
Sufficient. St. (2) : y < 1
Since y is an integer and y cannot be less than 0 (question stem part b) therefore y must be 0. Sufficient. Answer : D
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Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 13:54
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Quote: 3. Is x^2 + y^2 > 4a? (1) (x + y)^2 = 9a (2) (x – y)^2 = a St. (1) : (x + y)^2 = 9a
x^2 + y^2 + 2xy = 9a Insufficient. St. (2) : (x - y)^2 = a
x^2 + y^2 - 2xy = a Insufficient. St. (1) and (2) together : x^2 + y^2 = 5a
When either x or y is not 0, question stem holds true. When x and y are both 0, question stem is false. Hence insufficient. Answer : E
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Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 14:43
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Quote: 7. |x+2|=|y+2| what is the value of x+y? (1) xy<0 (2) x>2 y<2 Question stem :
Note: Since the equations is symmetrical, there will only be two distinct cases. However, for the sake of explanation, I have illustrated all 4. (a) When both x and y are greater than - 2 ; x + 2 = y + 2 ; x = y (b) When both x and y are less than - 2 ; - x - 2 = - y - 2 ; x = y (c) When x is less than -2 and y is greater than -2 ; - x - 2 = y + 2 ; x + y = - 4 (d) When x is greater than -2 and y is less than -2 ; x + 2 = - y - 2 ; x + y = - 4 St. (1) : xy < 0
This implies that one is negative and the other is positive. Therefore, in order for xy to be less than 0, x cannot be equal to y. Thus in order to satisfy the question stem, it can only be cases (c) and (d). Thus Sufficient. St. (2) : x > 2 ; y < 2
Again, this implies that x and y cannot be equal. Thus, in order to satisfy the question stem it can only be cases (c) and (d). Thus Sufficient. Answer : D
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Re: Inequality and absolute value questions from my collection [#permalink]
20 Nov 2009, 14:45
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tihor wrote: Bunuel, two questions: shouldnt |x-1| < 1 be 0<x<2? and not -2<x<2? secondly, how does this happen: x(x-2)<=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: |x-1| < 1 means 0<x<2. Already edited the post. As for the second one: x(x-2)<=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(x-2) is "smiling" parabola, and the intersections with X-axis are at x=0 and x=2, the range between will be below X-axis. Hope it helps.
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Re: Inequality and absolute value questions from my collection [#permalink]
21 May 2010, 03:06
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ManishS wrote: I am a little confused on this one . Can the answer be E??
From A: 2x-2y=1 => x-y= 0.5 INSF
From B x/y > 1 => x > y INSF
From A & B x-y =0.5 and x > y
If x = -0.5 and y = -1 then x > y and x - y = (-0.5) - (-1) = -0.5 + 1 = 0.5 Hence both x and y can be negative
If x= 1 and y = 0.5 then x > y and x- y = 1 -0.5 = 0.5 Hence both x and y can be positive
Ans = E ?? Problem with your solution is that the red part is not correct. \frac{x}{y}>1 does not mean that x>y. If both x and y are positive, then x>y, BUT if both are negative, then x<y. From (2) \frac{x}{y}>1, we can only deduce that x and y have the same sigh (either both positive or both negative). When we consider two statement together:From (1): 2x-2y=1 --> x=y+\frac{1}{2}From (2): \frac{x}{y}>1 --> \frac{x}{y}-1>0 --> \frac{x-y}{y}>0 --> substitute x from (1) --> \frac{y+\frac{1}{2}-y}{y}>0--> \frac{1}{2y}>0 ( we can drop 2 as it won't affect anything here and write as I wrote \frac{1}{y}>0, but basically it's the same) --> \frac{1}{2y}>0 means y is positive, and from (2) we know that if y is positive x must also be positive. OR: as y is positive and as from (1) x=y+\frac{1}{2}, x=positive+\frac{1}{2}=positive, hence x is positive too. Answer: C. Hope it's clear.
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Re: Inequality and absolute value questions from my collection [#permalink]
20 Jan 2012, 14:30
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Re: Inequality and absolute value questions from my collection [#permalink]
23 Feb 2012, 12:00
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shankar245 wrote: Quote: 4. Are x and y both positive? (1) 2x-2y=1 (2) x/y>1
(1) 2x-2y=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. 2x-2y=1 --> y=x-1/2, 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.
In the first statement from 2x-2y=1 --> we can sat x-y=1/2 So it cud be 8.5-8 or 0.25 - (-0.25) HOw can we say both x and y are positive? similarily statement 2 x/y>1 =>x>y how can we be sure x and y have the same sign we can have 8>7 or 8>-8 Bunel can you pls xplain..or am i missing sumtin fundamental? For (1) we are not saying that x and y are positive, we saying that if they were both positive then it would mean that the line y=x-1/2 is only in I quadrant, which is not possible. So from (1) we cannot say that x and y are both positive. Or consider following numbers: x=2, y=1.5 and x=-2, y=-2.5. For (2). First of all you can not multiply x/y>1 by y and write x>y, since you don't know the sign of y: if it's positive then x>y but if it's negative then when multiplying by negative number you should flip the sing of the inequality and write x<y. Never multiply or divide inequality by a variable (or by an expression with variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.Next, since x/y>1 then x/y>0, which means that x and y are either both positive or both negative, hence they have the same sign. Hope it's clear.
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Re: Inequality and absolute value questions from my collection [#permalink]
25 Jun 2012, 02:03
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kuttingchai wrote: Bunuel wrote: 13. Is |x-1| < 1? (1) (x-1)^2 <= 1 (2) x^2 - 1 > 0
Last one.
Is |x-1| < 1? Basically the question asks is 0<x<2 true?
(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=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. Can you please explain : --> how did u derive --> "x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive" ??? Here is what i did Question: |x-1| < 1 critical point x>1 or x<1 when x>1 then (x-1)<1 x<2 when x<1 then -(x-1)<1 -x<0 therefore x>0 to prove 0<x<2 --? [understood this] (A)(x-1)^2 <= 1 x^2 - 2x + 1 <= 1 x (x-2) <= 0 [can you please explain after this who did you derive 0<=x<=2 ===?????] my thoughts [what m i doing wrong??] when (x = 0) then x-2<=0 therefoe x<=2 when (x-2 = 0) then x<=0 ????? i am confused here ????? (B) x^2-1>0 therefore we have x>1 or x<-1 - not sufficient [understood this] Thank you Check the following links: x2-4x-94661.html#p731476 (check this one first) inequalities-trick-91482.htmleverything-is-less-than-zero-108884.html?hilit=extreme#p868863xy-plane-71492.html?hilit=solving%20quadratic#p841486data-suff-inequalities-109078.htmlrange-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535Hope it helps.
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Re: Inequality and absolute value questions from my collection [#permalink]
25 Aug 2012, 07:26
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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^2-6x+9)=0 --> y*(x-3)^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) y-x=3. If y is not 0, x must be 3 and y-x 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:
y-x=3 --> x=y-3 --> y*(x-3)^2=y*(y-3-3)^2=y(y-6)^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.
The answer to this one is C right? B alone is not sufficient.
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Re: Inequality and absolute value questions from my collection [#permalink]
04 Oct 2012, 04:00
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Re: Inequality and absolute value questions from my collection [#permalink]
04 Oct 2012, 04:20
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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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Re: Inequality and absolute value questions from my collection [#permalink]
22 Feb 2013, 01:27
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JJ2014 wrote: Bunuel wrote: 5. What is the value of y? (1) 3|x^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 --> 3-y=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^2-4|=x^2-4 when x^2-4>0; |x^2-4|=-(x^2-4) when x^2-4<=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 3|x^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]
28 Feb 2013, 06:42
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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: inequality-and-absolute-value-questions-from-my-collection-86939-200.html#p652806OA for this question is D, not E. Discussed here: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653783 and here: inequality-and-absolute-value-questions-from-my-collection-86939-160.html#p1111747Hope it helps.
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Re: Inequality and absolute value questions from my collection [#permalink]
28 Feb 2013, 08:36
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Question 1:
6xy = x^2 y + 9y
y(x^2 -6x +9) = 0
y(x-3)^2 = 0
either y =0, or x=3
statement 1: y-x =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]
16 Nov 2009, 13:08
Bunuel wrote: 13. Is |x-1| < 1? (1) (x-1)^2 <= 1 (2) x^2 - 1 > 0
I'm getting B for this one 1. (x-1)^2 <= 1 x can be 0 which would make the question no or x can be 1/2 which would make the answer yes so 1 is insufficient 2. x^2 - 1 > 0 means x^2>1 so x<-1 or x>1 both of which make the question no so sufficient
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Re: Inequality and absolute value questions from my collection [#permalink]
16 Nov 2009, 13:19
Bunuel wrote:
12. Is r=s? (1) -s<=r<=s (2) |r|>=s
I'm getting c 1. s can be 3 and r can be 3 which makes question yes s can be 3 and r can be 2 which makes question no insufficient 2. r can be 3 and s can be 3 makes question yes r can be 3 s can be 2 makes question no insufficient combining: |r|>=s means r>=s or r<=-s and -s<=r<=s means -s<=r and r<=s now we have -s<=r and -s>=r so -s = r or s = r r>=s and r<=s so s = r
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Re: Inequality and absolute value questions from my collection
[#permalink]
16 Nov 2009, 13:19
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