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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) \(x^3< 0\)

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]

Re: Inequality and absolute value questions from my collection [#permalink]

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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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Bunuel I have very clear the 1 statement but not so much the secon one : 1/|n|< n this is true only for negative value. So we could have i. e. : -1 or -6 so insuff but how we you arrive to this conclusion:

Given: 1/|n| > n. Now, 1/|n| is always positive (because of the absolute value in the denominator), so if n is negative then this inequality will always hold true: 1/|n|=positive>n=negative.

Hope it's clear.

P.S. Complete solution for 1/|n| > n is n<0 or 0<n<1.
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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 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.

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.

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.

Re: Inequality and absolute value questions from my collection [#permalink]

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11 Jul 2013, 21:42

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Hi johncoffey , My two cents- for (1) - it is always useful to start out by factoring an expression if possible, especially when there is a variable in common ("y" in this example). Even though it does make sense to isolate the expression "xy" that we are being asked for- note that in this case that would give us more unknowns on the RHS. Hope tht helps.

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 n|n| < 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.

Re: Inequality and absolute value questions from my collection [#permalink]

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22 Sep 2013, 08:07

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Bunuel wrote:

StormedBrain wrote:

Bunuel wrote:

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.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> 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) 2x-2y=1 --> draw line y=x-1/2:

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 x-axis ?

Lets take a point (-0.5,-1) in the green shaded region , then -0.5/-1 = 1/2 <1..
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Re: Inequality and absolute value questions from my collection [#permalink]

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08 Sep 2015, 11:30

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reto wrote:

Bunuel wrote:

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.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.

\(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\)

If you substitute x, the new fraction should look as follows: \(\frac{0.5}{y}>0\) since \(x=y+\frac{1}{2}\) ?

Yes, you are correct.

After substituting for x, you get, \(\frac{0.5}{y} >0\) ---> \(\frac{1}{2y} >0\) ---> \(\frac{1}{y} >0\) ---> y >0

(2) (x – y)^2 = a x^2 + y^2 - 2xy = a x^2 + y^2 = a +2xy

So we can conclude that: 9a - 2xy = a + 2xy 8a = 4xy 4a = 2xy

Hence, in the first conclusion is: x^2 + y^2 = 9a - 4a x^2 + y^2 = 5a

Answer is C (Together, they can answer the initial question).

Am I wrong, or right?

You are correct till \(x^2+y^2=5a\) but what if x=y=0 giving you a=0. In this case, \(x^2+y^2\) will be = 4a and NOT > 4a. This is the reason why E is the correct answer.

If you were given "is \(x^2+y^2 \geq 4a\) instead of just >4a", then yes, you would have marked C as the correct answer but you are asked >4a which may or may not be true.

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.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> 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 \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) ? Any help is appreciated. Thank you.

This is explained couple of times on the previous pages: \(\frac{x}{y}>1\)

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: math-absolute-value-modulus-86462.html

correct 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|=x and if x<0 then |x|=-x

am i wrong? please reply!!!

Yes, you are wrong.

We can say that when x<=0, then |x| is also equal to -x:

(2) (x – y)^2 = a --> x^2-2xy+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.

Option E is correct only because of Zero? If we put any integer value then LHS > RHS every time. But if x,y and z are Zero then the Q doesn't hold true. So, Zero is the only option that warrants E?

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.

Re: Inequality and absolute value questions from my collection [#permalink]

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16 Nov 2009, 20:07

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.