Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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) \(x^3< 0\)

liarish wrote: Hi Bunuel, I have read all the responses to Q4. But I am still confused why C is the answer. Here is how I solved it.

4. Are x and y both positive? (1) 2x-2y=1 (2) x/y>1

1) Insufficient . First reduced equation to x-y=0.5 . Plugged in 2 positive and 2 negative values. I chose x=3, y=2.5 => 3-2.5=0.5 works.. then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5. works again. So x and y can be both +ve and -ve. So 1) is Insufficient. 2) x/y>1. Just tells us that both x and y have same sign. both are -ve or both are +ve. So Insufficient.

Now Combining, Picking the same values used in 1) x=3, y=2.5. both signs positive and x-y=0.5. works then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5 both signes negative. works as well. So we still don't know if both signs are +ve or -ve. So my answer is E.

Could you please take a look at my solution and tell me where I am going wrong? That would be a big help. Thanks a ton!

If x= -1 and y=-1.5, then x/y=2/3<1, so these values don't satisfy the second statement.

This question is also discussed here: are-x-and-y-both-positive-1-2x-2y-1-2-x-y-93964.html

Hope it helps.

Great.. I get it now.. Thanks Bunuel.

I am also stuck at Q10 :

10. If n is not equal to 0, is |n| < 4 ? (1) n^2 > 16 (2) 1/|n| > n

We need to see if |n|<4 (this means -4<n<4)

1) n^2>16 => n<-4 and n>4 So from n<-4, |n|<|-4| = |n|<4 (works) But n>4 does not work. Doesn't that make 1) Insufficient?

Could you please tell me what I am doing wrong here ??

If n<-4, then n, for example can be -4.5 --> |-4.5|=4.5>4, so |n|<4 doesn't hold true.

If n is not equal to 0, is |n| < 4 ?

Question basically asks whether \(-4<n<4\), so whether \(n\) is some number from this range.

(1) n^2>16. This implies that either \(n>4\) or \(n<-4\). No number from these ranges is between -4 and 4, thus the answer to the question whether \(-4<n<4\) is NO. Since we have a definite answer then this statement is sufficient.

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

Show Tags

04 Oct 2012, 03:53

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:
_________________

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

Show Tags

04 Oct 2012, 04:13

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 those information and we are not sure of course of - 4 < n < 4.

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

Show Tags

11 Oct 2012, 03:11

Bunuel wrote:

11. Is |x+y|>|x-y|? (1) |x| > |y| (2) |x-y| < |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 |x-y|? If and only when x and y have the same sign absolute value of x+y will always be more than absolute value of x-y. As x+y when they have the same sign will contribute to each other and x-y will not.

5+3=8 and 5-3=2 OR -5-3=-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) |x-y| < |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 vise-versa is not right, meaning that x and y can have the same sign but |x| can be less than |x-y|, but if |x|>|x-y| the only possibility is x and y to have the same sign.)

Answer: B.

Hi Bunuel - Can this solved in the below way?

Is |x+y|>|x-y|?

Since both sides are +ve we can square both side of the inequality.... On squaring we get xy>0?

statement 1

(1) |x| > |y|

This is NS as xy can be opp sign as well as same sign

(2) |x-y| < |x|

Squaring on both sides we get y^2 < 2xy Y cannot be zero otherwise the inequality cannot hold so Y^2 is +ve hence xy is +ve So we can answer the question xy>0

To answer this question you should visualize it. We have comparison of two absolute values. Ask yourself when |x+y| is more then than |x-y|? If and only when x and y have the same sign absolute value of x+y will always be more than absolute value of x-y. As x+y when they have the same sign will contribute to each other and x-y will not.

5+3=8 and 5-3=2 OR -5-3=-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) |x-y| < |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 vise-versa is not right, meaning that x and y can have the same sign but |x| can be less than |x-y|, but if |x|>|x-y| the only possibility is x and y to have the same sign.)

Answer: B.

Hi Bunuel - Can this solved in the below way?

Is |x+y|>|x-y|?

Since both sides are +ve we can square both side of the inequality.... On squaring we get xy>0?

statement 1

(1) |x| > |y|

This is NS as xy can be opp sign as well as same sign

(2) |x-y| < |x|

Squaring on both sides we get y^2 < 2xy Y cannot be zero otherwise the inequality cannot hold so Y^2 is +ve hence xy is +ve So we can answer the question xy>0

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

Show Tags

23 Dec 2012, 04:40

Bunuel wrote:

2. If y is an integer and y = |x| + x, is y = 0? (1) x < 0 (2) y < 1

Note: as \(y=|x|+x\) then \(y\) is never negative. For \(x>{0}\) then \(y=x+x=2x\) and for \(x\leq{0}\) then (when x is negative or zero) then \(y=-x+x=0\).

(1) \(x<0\) --> \(y=|x|+x=-x+x=0\). Sufficient.

(2) \(y<1\), as we concluded y is never negative, and we are given that \(y\) is an integer, hence \(y=0\). Sufficient.

Answer: D.

Hi Bunuel,

Thanks for the explanation to the above Q.

Regarding st 1 i.e X less than zero then [m]y=|x|+x = -x+x=0,

1. we know any value in modulus is positive then ideally the above should be interpreted as [m]y=|x|+x--> [m]y=x-x=0. 2.Also if from St 1 if we x<0 then [m]y=|x|+x= -x-x=-2x

3. Where as we also know that |x|= -x for X<0 and |x|= x for X>/ 0

So can you please tell me where am I going wrong with the concept.

Thanks Mridul
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

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

Show Tags

23 Dec 2012, 05:29

Bunuel wrote:

9. Is n<0? (1) -n=|-n| (2) n^2=16

(1) -n=|-n|, means that either n is negative OR n equals to zero. We are asked whether n is negative so we can not be sure. Not sufficient.

(2) n^2=16 --> n=4 or n=-4. Not sufficient.

(1)+(2) n is negative OR n equals to zero from (1), n is 4 or -4 from (2). --> n=-4, hence it's negative, sufficient.

Answer: C.

Hello Bunuel,

I got A as the answer to the Q.

From St1, we have -n=|-n|---> -n=n (As Mod value is +ve)---> we have 2n=0 or -2n=0. In both case we can say that n=0 and hence Ans should be A.

From your explanation, it is very clear that either n<0 or n=0. Could you tell me what was your approach to this Question. I mean did you assume values of 1. n as less than zero, 2. ngreater than zero and 3. n equal to zero

and check under which condition the St1 holds true.

If so, would this be a standard way of doing a modulus Question because clearly I just considered only 1 of the above conditions here.

Your inputs please

Thanks Mridul
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

2. If y is an integer and y = |x| + x, is y = 0? (1) x < 0 (2) y < 1

Note: as \(y=|x|+x\) then \(y\) is never negative. For \(x>{0}\) then \(y=x+x=2x\) and for \(x\leq{0}\) then (when x is negative or zero) then \(y=-x+x=0\).

(1) \(x<0\) --> \(y=|x|+x=-x+x=0\). Sufficient.

(2) \(y<1\), as we concluded y is never negative, and we are given that \(y\) is an integer, hence \(y=0\). Sufficient.

Answer: D.

Hi Bunuel,

Thanks for the explanation to the above Q.

Regarding st 1 i.e X less than zero then y=|x|+x = -x+x=0,

1. we know any value in modulus is positive then ideally the above should be interpreted as y=|x|+x--> y=x-x=0. 2.Also if from St 1 if we x<0 then y=|x|+x= -x-x=-2x

3. Where as we also know that |x|= -x for X<0 and |x|= x for X>/ 0

So can you please tell me where am I going wrong with the concept.

Thanks Mridul

Absolute value properties:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\);

So, if \(x<0\), then \(|x|=-x\) and \(y=|x|+x=-x+x=0\).

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

Show Tags

25 Dec 2012, 05:19

Bunuel wrote:

debayan222 wrote:

Great collection Bunuel...Kudos.. Are these Qs. included in your signature or they exist as separate entity?

merry Xmas...Happy Holidays.

Yes, they are in Inequalities set.

Thanks a lot Bunuel.. Well I guess, whatever Qs come from you directly to the forum, are included in you Sig. ? Hope I got you right..
_________________

(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.

I got C. Can you please explain why this is incorrect?

Statement 1 gives x^2+2xy+y^2 = 9a, and I rewrote it to 9a-2xy = x^2+y^2 Statement 2 gives x^2-2xy+y^2=a, and i rewrote this to x^2+y^2= a +2xy

Together, I have 9a-2xy = a+2xy, which leads to 8a = 4xy, but 8a is also equal to x^2+y^2? ... this last part must be wrong?
_________________

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

Show Tags

21 Feb 2013, 21:23

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.
_________________

(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.

I got C. Can you please explain why this is incorrect?

Statement 1 gives x^2+2xy+y^2 = 9a, and I rewrote it to 9a-2xy = x^2+y^2 Statement 2 gives x^2-2xy+y^2=a, and i rewrote this to x^2+y^2= a +2xy

Together, I have 9a-2xy = a+2xy, which leads to 8a = 4xy, but 8a is also equal to x^2+y^2? ... this last part must be wrong?

The red part is not right. If you sum the two equations you'll get 2(x^2+y^2)=10a --> x^2+y^2=5a.

(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.

I got C. Can you please explain why this is incorrect?

Statement 1 gives x^2+2xy+y^2 = 9a, and I rewrote it to 9a-2xy = x^2+y^2 Statement 2 gives x^2-2xy+y^2=a, and i rewrote this to x^2+y^2= a +2xy

Together, I have 9a-2xy = a+2xy, which leads to 8a = 4xy, but 8a is also equal to x^2+y^2? ... this last part must be wrong?

The red part is not right. If you sum the two equations you'll get 2(x^2+y^2)=10a --> x^2+y^2=5a.

So I understand that you summed the equations and got that answer. But why is setting them equal to each other wrong in this case? I'm trying to figure out what concept I'm missing so that I don't end up doing it again.
_________________

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

Show Tags

22 Feb 2013, 07:26

Bunuel wrote:

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.

So I understand that you summed the equations and got that answer. But why is setting them equal to each other wrong in this case? I'm trying to figure out what concept I'm missing so that I don't end up doing it again.

What are you trying to get when setting "them" equal? Anyway, you won't be able to solve two equations with three unknowns.
_________________