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) \(x^3< 0\)

What is the value of y? 1. y>=2 2. If y<3 then the value of y is -8, if y>3 then the value of y is 14 (think we have two groups here)

When using Option 1 in Option 2,

You are looking at comparing with the result of the groups (-8 or 14). As y>=2, it has to be 14.

I am looking at the conditions of the groups ( y< 3 or y>3). As y>=2 I cannot decide the group. Hence Option 1 and Option 2 together are not sufficient. What is wrong in this approach?

Can you please use the other approach? It's much easier.
_________________

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

Show Tags

20 Sep 2013, 19:46

Bunuel,

I think I really need your help in resolving this for me. Please??

In the inequalities set that you posted, I've into this same situation with only problem 5 and 7. Actually my approach is not different.

Looking at problem 7

Quote:

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=-y-2 --> 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 vise-versa 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.

In the above problem, when you are solving the option (2) you said as follows:

x>2 and y<2 and hence as x is not equal to y, it has to be scenario B.

My doubt is a similar one here: When we made Scenario B we know that Scenario B will hold only if: This will occur when either x or y is less then -2 and the other is more than -2.

In such cases, shouldn't we compare [x>2 and y<2] from option (2) with the condition of Scenario B, before deciding that it belongs to Scenario B.

I think I really need your help in resolving this for me. Please??

In the inequalities set that you posted, I've into this same situation with only problem 5 and 7. Actually my approach is not different.

Looking at problem 7

Quote:

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=-y-2 --> 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 vise-versa 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.

In the above problem, when you are solving the option (2) you said as follows:

x>2 and y<2 and hence as x is not equal to y, it has to be scenario B.

My doubt is a similar one here: When we made Scenario B we know that Scenario B will hold only if: This will occur when either x or y is less then -2 and the other is more than -2.

In such cases, shouldn't we compare [x>2 and y<2] from option (2) with the condition of Scenario B, before deciding that it belongs to Scenario B.

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

Show Tags

21 Sep 2013, 21:21

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 ?
_________________

The First and Last time !!!

BKPL - Below Kudos Poverty Line .....Need your help.

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 [ 6.13 KiB | Viewed 996 times ]

Not sufficient.

(2) x/y>1 --> Draf line x/y=1. The solutions is the green region:

Attachment:

graph (2).png [ 5.95 KiB | Viewed 975 times ]

Not sufficient.

(1)+(2) Intersection is the portion of the blue line which lies in the first quadrant. Sufficient.

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]

Show Tags

22 Sep 2013, 07:07

1

This post received KUDOS

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..
_________________

The First and Last time !!!

BKPL - Below Kudos Poverty Line .....Need your help.

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

Show Tags

22 Sep 2013, 07:18

Bunuel wrote:

StormedBrain wrote:

Bunuel wrote:

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.

Thus 1/|n| > n holds true if n<0 and 0<n<1.

Does this make sense?

Thanks Bunuel .

I dedicated whole day today to inequality . Lots of learning.. thanks again.
_________________

The First and Last time !!!

BKPL - Below Kudos Poverty Line .....Need your help.

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

Show Tags

23 Sep 2013, 12:22

Bunuel wrote:

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=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.

Bunuel, why can't we open the first modulus as well? x+2 and -x-2? thanks

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.

Hello Bunuel, what about (x-1) ^ 2 < 1 means (x-1) is for sure less than 1 , as square of less than 1 can only occur if the original number is less than 1?

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.

Hello Bunuel, what about (x-1) ^ 2 < 1 means (x-1) is for sure less than 1 , as square of less than 1 can only occur if the original number is less than 1?

Sorry, don't get what you mean. Can you please elaborate? Thank you.
_________________

(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.

1+2 also insufficient when a is -ve .. consider a -1 ..
_________________

(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.

1+2 also insufficient when a is -ve .. consider a -1 ..

a cannot be a negative number because each statement says that a=(non-negative value).
_________________

(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.

Also insufficient as x,y, and a could be 0 Why did you assume that ALL COULD be zero.

(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.

Also insufficient as x,y, and a could be 0 Why did you assume that ALL COULD be zero.

Because they could be 0, why not? And IF x=y=a=0, then the answer would be NO but if they take some other values then the answer would be YES.
_________________

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

Show Tags

14 Oct 2013, 19:10

12. Is r=s? (1) -s<=r<=s (2) |r|>=s

From (1) take a number line

-s---------------0-------------s

So, 5 cases are possible r----------------0---------------s (or) -s------r-------0---------------s (or) -s--------------0-------r-------s (or) -s--------------r----------------s (r can be zero also) (or) -s--------------0---------------r The last case is the correct answer, but we have other 4 cases.INSUFFICIENT.

From (2) |r|>=s ----->> r>=s and -r>=s ----->> r>=s and r<=-s

take number line ------(-s)-------------0--------------+s------ Still we cannot say wether r=s. INSUFFICIENT.

NOW combine the two(1+2) we get ------(-s)-------------0--------------+s------ i.e., r can lie anywhere on the line and still we CANNOT say wether r=s. Hence, E.

gmatclubot

Re: Inequality and absolute value questions from my collection
[#permalink]
14 Oct 2013, 19:10

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

[rss2posts title=The MBA Manual title_url=https://mbamanual.com/2016/11/22/mba-vs-mim-guest-post/ sub_title=MBA vs. MiM :3qa61fk6]Hey, guys! We have a great guest post by Abhyank Srinet of MiM-Essay . In a quick post and an...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...