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Help me to not be bad at abs value and inequalities... [#permalink]

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12 Mar 2012, 17:28

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I’m taking my test this Friday and am feeling OK. I’m shooting for a 47-49 quant to get 750 but I feel like quant beats me up a bit and depending on my luck with questions there could be variance.

Absolute value inequalities are my biggest weakness and I would like to beef up my proficiency. I seem to get half of the difficult questions right because I evaluate properly once I have the formulas, but it feels like whether I flip the sign correctly etc. is completely random. I’m not missing the tricks, I’m messing up the structure. I’m at a loss because I’ve researched all of the problems on the site (went through the modulus training etc) and the basic stuff is super easy. However, somewhere between basic and advanced I get lost. I’m hoping that if I post a few of my structural questions here, that you quant geniuses can tell me why I suck . I’m a fairly quick study and this perplexes me, so I’m sure it will help a ton of us less quanty types out there.

Most of these are from Bunnuel’s sets. Speaking of him, I’ve never been so appreciative towards somebody who destroys my confidence on a daily basis

Problem 1: |1/(x-2)| >= 4

Question 1: So the two cases to evaluate are whether the bracketed portion is positive or negative right? I get confused as to where 0 should go. Do I need to evaluate the critical point cases (x<2 and x>=2) instead of just distributing for the two answers, or do I evaluate and then work the critical point into the constraint? I guess I don’t understand the “positive” and “negative” case.

Question 2: Some of the quant masters change the equation to this |x-2| =< ¼. I don’t understand

Question 3: if you have a complex set of operations inside the bracket, where do you attribute the negative when testing that case? Do you assign the negative to the denominator? What if there are two variables? For example –(2x + 1/(x-2)) or –(yx + 1/(x-2))

Problem 2: |x+y|>|x-y|?

Question: The factoring here is easy because MGMAT told me to just leave the left side the same for the positive test and distribute a negative for the negative. However, conceptually, the positive test for x is really x>-2 right, not x>0? In this case it seems like testing. With double brackets like this, both neg and both positive yield the same equation, and mixed yields the other (in which y<2 x>2 and vise versa), correct? What would I do with the 2nd Y if it wasn’t an abs value… multiple variable confuse me…

Problem 3: 1/|n| > n

Question 1: This is one of my fundamental problems. If N is positive n^2>1, but if it is negative, do I do 1/-n>-n and multiply through, or is it 1/-n>n? I think it is the latter. If so, I get 1<–n^2 then I go n^2<-1 which isn’t possible and I realize I have no idea what I am doing.

Problem 4: |x-y| < |x|

Question 1: A fine example of my confusion. If I try to multiply out the scenarios I get: x-y<x -y<0y>0 and –x+y<x y<2x. I’m pretty sure I’m doing something wrong. Again if I test X as negative, do I need to make the x on the other side negative as well?

Problem 4: |x-1| < 1

Question: I can get to “is 0<x<2 true?” but then I get confused by the “<1” is that supposed to be a critical point? Lower bound for the positive case? Going crazy on these…

And finally, are there any shortcuts for this? For example, if I figure out the positive one case can I just swap the sign? I’m trying to simplify but having difficulty. Thanks a ton in advance for any help! Hopefully, I’m not the only one to run into these difficulties and others can benefit from the responses.
_________________

Question 3: if you have a complex set of operations inside the bracket, where do you attribute the negative when testing that case? Do you assign the negative to the denominator? What if there are two variables? For example –(2x + 1/(x-2)) or –(yx + 1/(x-2))

Missed this question before though I don't really understand your question very well. The negative sign comes right before the expression which has mod around it.

alphastrike wrote:

Problem 3: 1/|n| > n

Question 1: This is one of my fundamental problems. If N is positive n^2>1, but if it is negative, do I do 1/-n>-n and multiply through, or is it 1/-n>n? I think it is the latter. If so, I get 1<–n^2 then I go n^2<-1 which isn’t possible and I realize I have no idea what I am doing.

Single variable questions first. 1/|n| > n Multiply both sides by |n|. It must be positive (cannot be 0 since it is in the denominator) and the inequality will not change. n*|n| < 1

Case 1: If n > 0, substitute n in place of |n| n*n < 1 n^2 - 1 < 0 -1 < n < 1 But since n must be positive, 0 < n < 1 is the correct range.

Case 2: If n < 0, substitute -n in place of |n| n*(-n) < 1 -n^2 < 1 1 + n^2 > 0 No matter what the value of n (as long as it is real), this inequality will always hold. So all negative values of n will satisfy the inequality.

We get 0 < n < 1 or (- infinity) < n < 0

Let's see where you went wrong: "or is it 1/-n>n? I think it is the latter. If so, I get 1<–n^2 then I go n^2<-1 which isn’t possible "

Let's say we don't move |n|. That's ok. In case 2, if n < 0, you get 1/-n>n. Now how do you get 1<–n^2? Mind you, n is negative so -n is positive. When you multiply the inequality by -n (or take -n on the other side), the inequality sign does not change. you get 1>-n^2 or n^2 + 1 > 0 which is true for all values of n.
_________________

It will take a lot of time to answer all your queries and I am a little short so I will start right now and take them one by one through the day as and when I can steal some time...

alphastrike wrote:

Speaking of him, I’ve never been so appreciative towards somebody who destroys my confidence on a daily basis

alphastrike wrote:

Problem 1: |1/(x-2)| >= 4

Question 1: So the two cases to evaluate are whether the bracketed portion is positive or negative right? I get confused as to where 0 should go. Do I need to evaluate the critical point cases (x<2 and x>=2) instead of just distributing for the two answers, or do I evaluate and then work the critical point into the constraint? I guess I don’t understand the “positive” and “negative” case.

Question 2: Some of the quant masters change the equation to this |x-2| =< ¼. I don’t understand

I will answer Question 2 first:

\(\frac{1}{|x-2|} >= 4\)

[highlight]Inequalities are similar to equations - you can add/subtract something on both sides and the inequality doesn't change. You can multiply/divide both sides by a positive number and the inequality doesn't change. But if you multiply/divide both sides by a negative number, you have to flip the sign.[/highlight]

|x-2| is positive since mod is always non negative (it cannot be 0 i.e. x cannot be 2 because it is in the denominator and 0 cannot be in the denominator). You can multiply both sides of the inequality by |x-2| and the inequality will not change.

You get \(\frac{1}{|x-2|}* |x-2| >= 4|x-2|\)

Now you can divide both sides by 4 and the inequality will not change since 4 is positive

You get \(\frac{1}{4} >= |x-2|\) OR \(|x-2| <= \frac{1}{4}\)

Define Mod: |x| = x if x is positive or 0 |x| = -x if x is negative

You cannot solve an equation/inequality if you have something like |x| in it. You need to have a simple x or -x etc. So how do you get rid of the mod? You use the definition to replace |x| by x/-x. If x is positive, you simply replace |x| by x. If x is negative, you replace |x| by -x. That is the reason you need to take two cases. You can put the equal sign anywhere with '>' or '<'. It doesn't matter. We generally put it with the '>' sign. Here, you will be given that x cannot be 2 since x-2 is in the denominator so 'where to put the equal to sign' problem doesn't arise!

\(|x-2| <= \frac{1}{4}\) To solve this inequality, you can take two cases:

Case 1: x-2 is positive i.e. x > 2. You can replace |x-2| by (x-2)

x - 2 <= 1/4 x <= 9/4 Since x must be greater than 2, you get 2 < x < 9/4

Case 2: x-2 is negative i.e. x < 2. You can replace |x-2| by -(x-2)

-(x-2) <= 1/4 x-2 >= -1/4 (We multiplied the inequality by -1 so we flipped the sign) x >= 7/4 Since x is less than 2, you get 7/4 < x < 2

So you get 2 ranges for x: 2 < x < 9/4 or 7/4 < x < 2

Short Cut: There is a 10 sec solution where you don't have to make the two cases but for that you need to understand how to interpret mods. Check out this link for that: http://www.veritasprep.com/blog/2011/01 ... edore-did/ _________________

Question: I can get to “is 0<x<2 true?” but then I get confused by the “<1” is that supposed to be a critical point? Lower bound for the positive case? Going crazy on these…

Right. |x-1| < 1 gives us the following range for x: 0 < x < 2 which you can get in 10 secs using the diagrammatic approach I explained in my post. Now what is this about < 1? x should lie between 0 to 2 and it can take any value between these 2 values but it cannot take the values 0 and 2 since the sign of the inequality is not "<=". It is only "<". x can definitely take the value 1 too. If you get such doubts, just plug in the value and see if it works. e.g. x = 1, |1-1| < 1 which gives 0 < 1 True so it works x = 2, |2-1| < 1 which gives 1 < 1 Not true so it doesn't work x = 0, |0-1| < 1 which gives 1 < 1 Not true so it doesn't work

As for questions with 2 variables, I will try and get to them too today but it is not very likely for you to see them on actual GMAT so I would instead spend more time with questions using a single variable only.
_________________

To clarify my question on the negative case in complex abs value equations, if i have say, |1/(x-2)|, do I make everything negative? -1/(-x+2)?

No.

|1/(x-2)| = -1/(x-2) OR 1/(2-x)

If you make both the terms negative, -1/(2-x), isn't this equal to 1/(x-2)? In that case, you haven't considered the negative term at all. You put a single negative sign. You can stick it to the numerator or denominator depending on your convenience but definitely not both.
_________________

I have tried solving the problems posed by alphastrike. Among the 5 problems, Karishma already addressed three. I'd like to make sure that I solve the remaining two correctly.

Problem 2: |x+y|>|x-y| Case 1: (x+y)>(x-y) We get y>0. Case 2: (-x-y)>(x-y) We get x>0. So, y>0 or x>0 for the inequality to hold.

Problem 4: |x-y|<|x| -x<(x-y)<x So, y>0 or x>(y/2) for the inequality to hold.

Did I solve the two problems correctly? Thanks!

When there are two absolute value signs, you need to consider each individually.

Problem 2: |x+y|>|x-y|

Case 1: x+y is positive and x-y is positive If you replace |x+y| with x + y, you have to ensure that x+ y >= 0. So x >= -y If you replace |x-y| with x - y, you have to ensure that x - y >= 0. So x >= y x+y > x-y y > 0 This means -y < 0 < y < x

Case 2: x+y is positive and x-y is negative If you replace |x+y| with x + y, you have to ensure that x+ y >= 0. So x >= -y If you replace |x-y| with -(x - y), you have to ensure that x - y < 0. So x < y x+y > -(x-y) x > 0 -y < 0 < x < y

These two cases imply that if x and y both are positive, the inequality will hold.

Case 3: x+y is negative and x-y is positive If you replace |x+y| with -(x + y), you have to ensure that x+ y < 0. So x < -y If you replace |x-y| with x - y, you have to ensure that x - y >= 0. So x >= y -(x+y) > x-y x < 0 y < x < 0 < -y

Case 4: x+y is negative and x-y is negative If you replace |x+y| with -(x + y), you have to ensure that x+ y < 0. So x < -y If you replace |x-y| with -(x - y), you have to ensure that x - y < 0. So x < y -(x+y) > -(x-y) y < 0

x < y < 0 < -y

These two cases imply that if x and y both are negative, the inequality will hold.

So either x and y both are positive or both are negative.

I thought that the problem only has two cases (i.e. same signs and different signs). I now understand my mistake. So, here is the solution for problem 4:

|x-y|<|x| Case 1: (x-y)<x, where x>=y and x>=0 x>y>0

Case 2: (-x+y)<(-x), where x<y and x<0 x<y<0

Case 1 and 2 imply that x and y have the same signs (both negative or both positive).

Case 3: (x-y)<(-x), where x>=y and x<0 y<x<(y/2)<0

Case 4: (-x+y)<x, where x<y and x>=0 y>x>(y/2)>0

Case 3 and 4 imply that x and y still have the same signs, and that x lies between y and (y/2).

Is the above solution and intepretation correct? Thank you Karishma!

"Case 1 and 2 imply that x and y have the same signs (both negative or both positive)."

Yes. If they are both positive, x > y. If they are both negative, x < y. This is important else the inequality does not hold. e.g. x = 2, y = 10 : Doesn't hold.

"Case 3 and 4 imply that x and y still have the same signs." Yes. If they are both positive, and x < y then, x > y/2. If they are both negative and x > y when x < y/2.

All in all, we come to know that x and y should have the same signs. If they are both positive, x > y/2 If they are both negative, x < y/2 _________________

I have tried solving the problems posed by alphastrike. Among the 5 problems, Karishma already addressed three. I'd like to make sure that I solve the remaining two correctly.

Problem 2: |x+y|>|x-y| Case 1: (x+y)>(x-y) We get y>0. Case 2: (-x-y)>(x-y) We get x>0. So, y>0 or x>0 for the inequality to hold.

Problem 4: |x-y|<|x| -x<(x-y)<x So, y>0 or x>(y/2) for the inequality to hold.

Re: Help me to not be bad at abs value and inequalities... [#permalink]

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13 Mar 2012, 14:26

VeritasPrepKarishma wrote:

It will take a lot of time to answer all your queries and I am a little short so I will start right now and take them one by one through the day as and when I can steal some time...

alphastrike wrote:

Problem 1: |1/(x-2)| >= 4

Question 1: So the two cases to evaluate are whether the bracketed portion is positive or negative right? I get confused as to where 0 should go. Do I need to evaluate the critical point cases (x<2 and x>=2) instead of just distributing for the two answers, or do I evaluate and then work the critical point into the constraint? I guess I don’t understand the “positive” and “negative” case.

Question 2: Some of the quant masters change the equation to this |x-2| =< ¼. I don’t understand

I will answer Question 2 first:

\(\frac{1}{|x-2|} >= 4\)

[highlight]Inequalities are similar to equations - you can add/subtract something on both sides and the inequality doesn't change. You can multiply/divide both sides by a positive number and the inequality doesn't change. But if you multiply/divide both sides by a negative number, you have to flip the sign.[/highlight]

|x-2| is positive since mod is always non negative (it cannot be 0 i.e. x cannot be 2 because it is in the denominator and 0 cannot be in the denominator). You can multiply both sides of the inequality by |x-2| and the inequality will not change.

You get \(\frac{1}{|x-2|}* |x-2| >= 4|x-2|\)

Now you can divide both sides by 4 and the inequality will not change since 4 is positive

You get \(\frac{1}{4} >= |x-2|\) OR \(|x-2| <= \frac{1}{4}\)

Define Mod: |x| = x if x is positive or 0 |x| = -x if x is negative

You cannot solve an equation/inequality if you have something like |x| in it. You need to have a simple x or -x etc. So how do you get rid of the mod? You use the definition to replace |x| by x/-x. If x is positive, you simply replace |x| by x. If x is negative, you replace |x| by -x. That is the reason you need to take two cases. You can put the equal sign anywhere with '>' or '<'. It doesn't matter. We generally put it with the '>' sign. Here, you will be given that x cannot be 2 since x-2 is in the denominator so 'where to put the equal to sign' problem doesn't arise!

\(|x-2| <= \frac{1}{4}\) To solve this inequality, you can take two cases:

So you get 2 ranges for x: 2 < x < 9/4 or 7/4 < x < 2

Short Cut: There is a 10 sec solution where you don't have to make the two cases but for that you need to understand how to interpret mods. Check out this link for that: http://www.veritasprep.com/blog/2011/01 ... edore-did/

Thanks, this all makes perfect sense. I know the shortcut well and have been using it. The trick to solving the long way is to adjust the resulting range by adding the value that x needs to be to make the equation positive or negative.
_________________

Re: Help me to not be bad at abs value and inequalities... [#permalink]

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13 Mar 2012, 14:40

VeritasPrepKarishma wrote:

It will take a lot of time to answer all your queries and I am a little short so I will start right now and take them one by one through the day as and when I can steal some time...

alphastrike wrote:

Speaking of him, I’ve never been so appreciative towards somebody who destroys my confidence on a daily basis

alphastrike wrote:

Problem 1: |1/(x-2)| >= 4

Question 1: So the two cases to evaluate are whether the bracketed portion is positive or negative right? I get confused as to where 0 should go. Do I need to evaluate the critical point cases (x<2 and x>=2) instead of just distributing for the two answers, or do I evaluate and then work the critical point into the constraint? I guess I don’t understand the “positive” and “negative” case.

Question 2: Some of the quant masters change the equation to this |x-2| =< ¼. I don’t understand

I will answer Question 2 first:

\(\frac{1}{|x-2|} >= 4\)

[highlight]Inequalities are similar to equations - you can add/subtract something on both sides and the inequality doesn't change. You can multiply/divide both sides by a positive number and the inequality doesn't change. But if you multiply/divide both sides by a negative number, you have to flip the sign.[/highlight]

|x-2| is positive since mod is always non negative (it cannot be 0 i.e. x cannot be 2 because it is in the denominator and 0 cannot be in the denominator). You can multiply both sides of the inequality by |x-2| and the inequality will not change.

You get \(\frac{1}{|x-2|}* |x-2| >= 4|x-2|\)

Now you can divide both sides by 4 and the inequality will not change since 4 is positive

You get \(\frac{1}{4} >= |x-2|\) OR \(|x-2| <= \frac{1}{4}\)

Define Mod: |x| = x if x is positive or 0 |x| = -x if x is negative

You cannot solve an equation/inequality if you have something like |x| in it. You need to have a simple x or -x etc. So how do you get rid of the mod? You use the definition to replace |x| by x/-x. If x is positive, you simply replace |x| by x. If x is negative, you replace |x| by -x. That is the reason you need to take two cases. You can put the equal sign anywhere with '>' or '<'. It doesn't matter. We generally put it with the '>' sign. Here, you will be given that x cannot be 2 since x-2 is in the denominator so 'where to put the equal to sign' problem doesn't arise!

\(|x-2| <= \frac{1}{4}\) To solve this inequality, you can take two cases:

Case 1: x-2 is positive i.e. x > 2. You can replace |x-2| by (x-2)

x - 2 <= 1/4 x <= 9/4 Since x must be greater than 2, you get 2 < x < 9/4

Case 2: x-2 is negative i.e. x < 2. You can replace |x-2| by -(x-2)

-(x-2) <= 1/4 x-2 >= -1/4 (We multiplied the inequality by -1 so we flipped the sign) x >= 7/4 Since x is less than 2, you get 7/4 < x < 2

So you get 2 ranges for x: 2 < x < 9/4 or 7/4 < x < 2

Short Cut: There is a 10 sec solution where you don't have to make the two cases but for that you need to understand how to interpret mods. Check out this link for that: http://www.veritasprep.com/blog/2011/01 ... edore-did/

Thanks again. I understand all of this well now. For some reason I didnt get that concept from the books, but your explanation cleared things up nicely

To clarify my question on the negative case in complex abs value equations, if i have say, |1/(x-2)|, do I make everything negative? -1/(-x+2)?
_________________

Re: Help me to not be bad at abs value and inequalities... [#permalink]

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14 Mar 2012, 07:09

Thanks for your explanations! This was my first exposure to your Quarter Wit, Quarter Wisdom blog and I'm so impressed with the explanations that have already provided me with several revolations this morning.

Re: Help me to not be bad at abs value and inequalities... [#permalink]

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14 Mar 2012, 13:25

VeritasPrepKarishma wrote:

alphastrike wrote:

To clarify my question on the negative case in complex abs value equations, if i have say, |1/(x-2)|, do I make everything negative? -1/(-x+2)?

No.

|1/(x-2)| = -1/(x-2) OR 1/(2-x)

If you make both the terms negative, -1/(2-x), isn't this equal to 1/(x-2)? In that case, you haven't considered the negative term at all. You put a single negative sign. You can stick it to the numerator or denominator depending on your convenience but definitely not both.

Great explanation once again. Thanks. So lets say |1/(x-2)| = x. How would this work? I now know how to distribute the negative (chose either numerator or denominator) but how to i multiply it out? Am I assuming the "x" i negative, or that the whole expression in the bracket is negative? If i attribute the negative like this -1/(x-2) do i just multiply x by (x-2) without flipping the sign because i put the negative sign with the 1, whereas if I attribute the negative to the bottom i need to flip? Also, in the positive case i just multiply (x-2)x, right?

Thanks again Karishma. I'm using up my whole box of kudos on you
_________________

Great explanation once again. Thanks. So lets say |1/(x-2)| = x. How would this work? I now know how to distribute the negative (chose either numerator or denominator) but how to i multiply it out? Am I assuming the "x" i negative, or that the whole expression in the bracket is negative? If i attribute the negative like this -1/(x-2) do i just multiply x by (x-2) without flipping the sign because i put the negative sign with the 1, whereas if I attribute the negative to the bottom i need to flip? Also, in the positive case i just multiply (x-2)x, right?

Thanks again Karishma. I'm using up my whole box of kudos on you

|1/(x-2)| < x (Note: There should be an inequality for you to worry about how to multiply so I am putting '<' here instead of '=' that you used)

Focus on the mod. You need to remove the mod sign since you can't work with it in your equation. Notice that 1 is also in the mod but it can be taken out since mod 1 is 1.

1/|x-2| < x I would suggest you to multiply the inequality by |x-2| on both sides right away so that you don't have to worry about it later. Since |x-2| is positive, you can multiply it without worrying about changing signs. Still, if you do not do that, following is how you can do it later.

You have to consider the case where x-2 is positive/negative. That's all.

Case 1: x-2 is positive i.e. x > 2 1/(x-2) < x Since (x-2) is positive, you can multiply the inequality by x-2 without changing it. 1 < x(x-2) x^2 - 2x - 1 > 0 solve it in the usual way and accept values greater than 2. In GMAT, you will be able to make factors here.

Case 2: x-2 is negative i.e. x < 2 1/-(x-2) < x (you have to change the sign of only the mod term) Now you can do many things here: Method 1: -1/(x-2) < x -1 > x(x-2) Multiply both sides by (x-2). Since (x-2) is negative, you need to flip the sign. (x-1)^2 < 0 A square cannot be less than 0 so not true for any values of x.

Method 2: 1/-(x-2) < x 1/(x-2) > -x (Multiply both sides by -1 so flip the sign) 1 < -x(x-2) Multiply both sides by (x-2). Since (x-2) is negative, you need to flip the sign. x(x-2) + 1 < 0 (x-1)^2 < 0 A square cannot be less than 0 so not true for any values of x.

Method 3: 1/-(x-2) < x 1 < -(x-2)x (x-2) is negative so -(x-2) is positive so when you multiply both sides by it, the inequality doesn't change. (x-1)^2 < 0 A square cannot be less than 0 so not true for any values of x.
_________________

Re: Help me to not be bad at abs value and inequalities... [#permalink]

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03 Mar 2016, 02:45

Hi,

I have tried solving the problems posed by alphastrike. Among the 5 problems, Karishma already addressed three. I'd like to make sure that I solve the remaining two correctly.

Problem 2: |x+y|>|x-y| Case 1: (x+y)>(x-y) We get y>0. Case 2: (-x-y)>(x-y) We get x>0. So, y>0 or x>0 for the inequality to hold.

Problem 4: |x-y|<|x| -x<(x-y)<x So, y>0 or x>(y/2) for the inequality to hold.

I have tried solving the problems posed by alphastrike. Among the 5 problems, Karishma already addressed three. I'd like to make sure that I solve the remaining two correctly.

Problem 2: |x+y|>|x-y| Case 1: (x+y)>(x-y) We get y>0. Case 2: (-x-y)>(x-y) We get x>0. So, y>0 or x>0 for the inequality to hold.

Problem 4: |x-y|<|x| -x<(x-y)<x So, y>0 or x>(y/2) for the inequality to hold.

Help me to not be bad at abs value and inequalities... [#permalink]

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04 Mar 2016, 22:24

VeritasPrepKarishma wrote:

Now try problem 4 again.

Thank you for the above explanation!

I thought that the problem only has two cases (i.e. same signs and different signs). I now understand my mistake. So, here is the solution for problem 4:

|x-y|<|x| Case 1: (x-y)<x, where x>=y and x>=0 x>y>0

Case 2: (-x+y)<(-x), where x<y and x<0 x<y<0

Case 1 and 2 imply that x and y have the same signs (both negative or both positive).

Case 3: (x-y)<(-x), where x>=y and x<0 y<x<(y/2)<0

Case 4: (-x+y)<x, where x<y and x>=0 y>x>(y/2)>0

Case 3 and 4 imply that x and y still have the same signs, and that x lies between y and (y/2).

Is the above solution and intepretation correct? Thank you Karishma!

gmatclubot

Help me to not be bad at abs value and inequalities...
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04 Mar 2016, 22:24

Best Schools for Young MBA Applicants Deciding when to start applying to business school can be a challenge. Salary increases dramatically after an MBA, but schools tend to prefer...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...