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 350,000 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:

Inequality and absolute value questions from my collection [#permalink]
16 Nov 2009, 10:33

81

This post received KUDOS

Expert's post

212

This post was BOOKMARKED

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

Re: Inequality and absolute value questions from my collection [#permalink]
23 Aug 2013, 11:59

Hi Bunuel, sorry to disturb you.

In problem number 3:

3. Is x^2 + y^2 > 4a? (1) (x + y)^2 = 9a (2) (x – y)^2 = a

You've given answer E. But I remember, correct me if I am wrong, that x^2 + y^2 is always >= to 2xy. It this is true, the answer should be A, because the first statement would be sufficient. x^2 + y^2 would be >= 4,5a

Re: Inequality and absolute value questions from my collection [#permalink]
25 Aug 2013, 06:20

Expert's post

Recobita wrote:

Hi Bunuel, sorry to disturb you.

In problem number 3:

3. Is x^2 + y^2 > 4a? (1) (x + y)^2 = 9a (2) (x – y)^2 = a

You've given answer E. But I remember, correct me if I am wrong, that x^2 + y^2 is always >= to 2xy. It this is true, the answer should be A, because the first statement would be sufficient. x^2 + y^2 would be >= 4,5a

Re: Inequality and absolute value questions from my collection [#permalink]
08 Sep 2013, 03:33

I agree with what you said. But I dont with the interpretation.

In B, you showed how x and y should have same sign, for if not, inequality in B doesnt hold. Similarily we see for inequality in A to be true, x,y should be of same sign. I took (x,y) as ( 3,2) , (1,0), (-1,0) , (-2,-1) . In this set, the inequality in A holds in 1st and last case, clearly , x,y should be same sign. '0' is not a positive number so 1,0 is ruled out. What wrong I did?

Re: Inequality and absolute value questions from my collection [#permalink]
08 Sep 2013, 04:25

Expert's post

ygdrasil24 wrote:

I agree with what you said. But I dont with the interpretation.

In B, you showed how x and y should have same sign, for if not, inequality in B doesnt hold. Similarily we see for inequality in A to be true, x,y should be of same sign. I took (x,y) as ( 3,2) , (1,0), (-1,0) , (-2,-1) . In this set, the inequality in A holds in 1st and last case, clearly , x,y should be same sign. '0' is not a positive number so 1,0 is ruled out. What wrong I did?

Which question are you talking about? _________________

Re: Inequality and absolute value questions from my collection [#permalink]
08 Sep 2013, 04:35

Bunuel wrote:

ygdrasil24 wrote:

I agree with what you said. But I dont with the interpretation.

In B, you showed how x and y should have same sign, for if not, inequality in B doesnt hold. Similarily we see for inequality in A to be true, x,y should be of same sign. I took (x,y) as ( 3,2) , (1,0), (-1,0) , (-2,-1) . In this set, the inequality in A holds in 1st and last case, clearly , x,y should be same sign. '0' is not a positive number so 1,0 is ruled out. What wrong I did?

Re: Inequality and absolute value questions from my collection [#permalink]
08 Sep 2013, 04:38

Expert's post

ygdrasil24 wrote:

Bunuel wrote:

ygdrasil24 wrote:

I agree with what you said. But I dont with the interpretation.

In B, you showed how x and y should have same sign, for if not, inequality in B doesnt hold. Similarily we see for inequality in A to be true, x,y should be of same sign. I took (x,y) as ( 3,2) , (1,0), (-1,0) , (-2,-1) . In this set, the inequality in A holds in 1st and last case, clearly , x,y should be same sign. '0' is not a positive number so 1,0 is ruled out. What wrong I did?

Re: Inequality and absolute value questions from my collection [#permalink]
08 Sep 2013, 04:44

In B, you showed how x and y should have same sign, for if not, inequality in B doesnt hold. Similarily we see for inequality in A to be true, x,y should be of same sign. I took (x,y) as ( 3,2) , (1,0), (-1,0) , (-2,-1) . In this set, the inequality in A holds in 1st and last case, clearly , x,y should be same sign. '0' is not a positive number so 1,0 is ruled out. What wrong I did?[/quote]

In that case I don't understand what you mean. Please elaborate.[/quote]

Question asks Is |x+y|>|x-y|? From analysis we know this holds good when x,y have same sign ? from B we concluded that and found x,y to have same sign.right ? For me answer should be D, because from statement I, we are sure x and y should have same sign. (x,y) = (3,2) or (-2,-1) . So we can answer the question . I took (x,y) as ( 3,2) , (1,0), (-1,0) , (-2,-1) . In this set, the inequality in A holds in 1st and last case, clearly , x,y should be same sign. '0' is not a positive number so 1,0 is ruled out. So we are sure of the sign of x,y the same way we were in B. So this should be able to answer the main equality of the question. What wrong I did?[/quote]

Re: Inequality and absolute value questions from my collection [#permalink]
08 Sep 2013, 04:50

Expert's post

ygdrasil24 wrote:

Question asks Is |x+y|>|x-y|? From analysis we know this holds good when x,y have same sign ? from B we concluded that and found x,y to have same sign.right ? For me answer should be D, because from statement I, we are sure x and y should have same sign. (x,y) = (3,2) or (-2,-1) . So we can answer the question . I took (x,y) as ( 3,2) , (1,0), (-1,0) , (-2,-1) . In this set, the inequality in A holds in 1st and last case, clearly , x,y should be same sign. '0' is not a positive number so 1,0 is ruled out. So we are sure of the sign of x,y the same way we were in B. So this should be able to answer the main equality of the question. What wrong I did?

So, you question is why (1) is not sufficient???

(1) |x| > |y|. This does NOT mean that x and y have the same sign. Consider this: if x=2 and y=-1, then |x+y|=1<3=|x-y| and we have a NO answer. _________________

Re: Inequality and absolute value questions from my collection [#permalink]
20 Sep 2013, 05:23

Quote:

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.

I guess the answer is E here. Because what we get from 1 is that y>=2. However, when we try to use that in (2) we are still hanging between y<3 and y>=3. Hence we have y=-8 and y=14.

Re: Inequality and absolute value questions from my collection [#permalink]
20 Sep 2013, 05:27

Expert's post

gmatter0913 wrote:

Quote:

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.

I guess the answer is E here. Because what we get from 1 is that y>=2. However, when we try to use that in (2) we are still hanging between y<3 and y>=3. Hence we have y=-8 and y=14.

Kindly let me know if I am missing something.

y cannot be -8 (y cannot be lees than 2) because in this case the first statement would be violated: absolute value cannot equal to a negative number. If y=-8, then 3|x^2 -4| = y - 2 = -8 - 2 = -10, which cannot be true.

Re: Inequality and absolute value questions from my collection [#permalink]
20 Sep 2013, 05:59

Yes, it makes sense.

But, what is wrong with my approach?

My approach is as follows:

Looking at option 1

3 *|x^2 - 4| = y - 2

LHS is +ve or 0 Therefore, y>=2 Not Sufficient to answer the value of y.

Looking at option 2

|3-y|=11 is the given information

We have two cases here. Case 1: When 3-y<0 (y>3) => the given information |3-y|=11 becomes -(3-y)=11 => So, y=14 Case 2: When 3-y>0 (y<3) => the given information |3-y|=11 becomes 3-y=11 => So, y=-8

This option says that if you know that y < 3 then the value of y is -8 y>3 then the value of y is 14

This is still insufficient information because we still don't know whether y>3 or y<3

Combining option 1 and 2

From option we know that y>=2. So, let us see if this helps to boil down between y>3 or y<3.

We are still not sure as Option 1 says 'y' could be >3 or <3 (y>=2)

Please help me understand where I have gone wrong??

Re: Inequality and absolute value questions from my collection [#permalink]
20 Sep 2013, 06:12

Expert's post

gmatter0913 wrote:

Yes, it makes sense.

But, what is wrong with my approach?

My approach is as follows:

Looking at option 1

3 *|x^2 - 4| = y - 2

LHS is +ve or 0 Therefore, y>=2 Not Sufficient to answer the value of y.

Looking at option 2

|3-y|=11 is the given information

We have two cases here. Case 1: When 3-y<0 (y>3) => the given information |3-y|=11 becomes -(3-y)=11 => So, y=14 Case 2: When 3-y>0 (y<3) => the given information |3-y|=11 becomes 3-y=11 => So, y=-8

This option says that if you know that y < 3 then the value of y is -8 y>3 then the value of y is 14

This is still insufficient information because we still don't know whether y>3 or y<3

Combining option 1 and 2

From option we know that y>=2. So, let us see if this helps to boil down between y>3 or y<3.

We are still not sure as Option 1 says 'y' could be >3 or <3 (y>=2)

Please help me understand where I have gone wrong??

From (1) y>=2, From (2) y=-8 or y=14.

Question: what does y equals to??? _________________

Re: Inequality and absolute value questions from my collection [#permalink]
20 Sep 2013, 06:26

Sorry to bother you on this silly problem, but I am still not convinced.

I am rephrasing the problem removing all the Math, so that I can focus on the part I am getting confused at.

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

The problem is I am not sure whether I should use the information from Option 1 (y>=2) to find out which group (from option 2) it belongs to or should I say y>=2 hence it has to be 14.

What is wrong in using the information to solve which group?

Re: Inequality and absolute value questions from my collection [#permalink]
20 Sep 2013, 06:42

Expert's post

gmatter0913 wrote:

Sorry to bother you on this silly problem, but I am still not convinced.

I am rephrasing the problem removing all the Math, so that I can focus on the part I am getting confused at.

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

The problem is I am not sure whether I should use the information from Option 1 (y>=2) to find out which group (from option 2) it belongs to or should I say y>=2 hence it has to be 14.

What is wrong in using the information to solve which group?

Even if we use your approach: is y<3??? NO. So, we have the second group --> y=14. _________________

Re: Inequality and absolute value questions from my collection [#permalink]
20 Sep 2013, 06:47

That's not true, right?

Because Option 1 says y>=2 (y can be equal to 2 also)

If I say y=2 then it belongs to 1st group (If y<3 then the value of y is -8) If I say y>3 then it belongs to 2nd group (If y>3 then the value of y is 14)

Re: Inequality and absolute value questions from my collection [#permalink]
20 Sep 2013, 06:50

Expert's post

gmatter0913 wrote:

That's not true, right?

Because Option 1 says y>=2 (y can be equal to 2 also)

If I say y=2 then it belongs to 1st group (If y<3 then the value of y is -8) If I say y>3 then it belongs to 2nd group (If y>3 then the value of y is 14)

What I meant is:

If y is from the first group it's -8, if in the second it's 14. (2) says y>=2, so it's in the second group, so y=14. _________________

Re: Inequality and absolute value questions from my collection [#permalink]
20 Sep 2013, 07:04

That is exactly where my confusion is..

Quote:

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?

gmatclubot

Re: Inequality and absolute value questions from my collection
[#permalink]
20 Sep 2013, 07:04

Interested in applying for an MBA? In the fourth and final part of our live QA series with guest expert Chioma Isiadinso, co-founder of consultancy Expartus and former admissions...