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

78

This post received KUDOS

Expert's post

196

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]
24 Jun 2012, 12:45

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.

I solved it this way. I hope i am assuming it correct

is x=y= +ve = ?

(A) 2x-2y=1

y=x-1/2 equation of line,

So, y=0 then x=1/2 x=0 then y=-1/2

thus, the line passes through points (1/2,0) & (0,-1/2) i.e - I, III & IV quadrant

(+,+) or (-,-) or (+,-) values of x & y

Not sufficient infromation

(B)

x/y > 1 if y>0 x>y if y<0 x<y not sufficient infomration

(C)

combine (A) and (B)

if y>0 i.e line passes only through Ist quadrant & x>y because y= x - 1/2

x>y ==> therefore x=y=+ve satisfies only first quadrant

condition y<0 from (B) is not true for x<y because

when line pass through IV the quadrant x is +ve and y is -ve, so we can ignore this condition as it doesnot satisfy (A) and (B) combined

Therefore left with only y>0 and x>y where line passes thr' Ist quadrant: C

Re: Inequality and absolute value questions from my collection [#permalink]
24 Jun 2012, 17:41

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.

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.

Vote for D [Need help in understanding part (B) below]

I solved it this way. I really find it difficult to open mod sign, I hope i am doing it correctly

(A) xy < 0

So we have (x = +ve and y = -ve) or (x = -ve and y = +ve)

|x + 2| = |y + 2|

critical point (x > -2 & y < -2) or (x < -2 & y > -2)

if (x > -2 & y < -2) then

(x + 2) = -(y + 2) x + 2 = -y -2 x + y = -4

sufficient

(B) x > 2 and y < 2

when y < -2 (x + 2) = - (y + 2) x + y = -4

when 2 > y > -2 we get x - y = 0 - Cannot be true statement [Not sure if this is correct. Need help here]

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]

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]
09 Aug 2012, 16:06

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.

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.

Hi Bunuel,

I am getting E and just cannot understand D. Please see my solution below - I used number picking.

A. xy<0, x=+ and y=- For this condition choosing different values of x and y (x=2,y=-6: x=3, y=-7)satisfies the given condition in modulus. Hence x=y can be different value or x=- and y=+ - This condition doesn't satisfy the modulus condiotion

B- x>2 and y<2 - As per the above stmt 1 - condition 1, there can be various values for x and y, hence x+y is different.

Hence E. I know I am going wrong some where, please help.

Re: Inequality and absolute value questions from my collection [#permalink]
10 Aug 2012, 07:26

Bunuel wrote:

jayaddula wrote:

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.

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.

Hi Bunuel,

I am getting E and just cannot understand D. Please see my solution below - I used number picking.

A. xy<0, x=+ and y=- For this condition choosing different values of x and y (x=2,y=-6: x=3, y=-7)satisfies the given condition in modulus. Hence x=y can be different value or x=- and y=+ - This condition doesn't satisfy the modulus condiotion

B- x>2 and y<2 - As per the above stmt 1 - condition 1, there can be various values for x and y, hence x+y is different.

Hence E. I know I am going wrong some where, please help.

thanks jay

In your example, both pairs give the same value for x+y: 2-6=-4 and 3-7=-4.

We can solve this question in another way:

7. |x+2|=|y+2| what is the value of x+y?

Square both sides: \(x^2+4x+4=y^2+4y+4\) --> \(x^2-y^2+4x-4y=0\) --> \((x+y)(x-y)+4(x-y)=0\) --> \((x-y)(x+y+4)=0\) --> either \(x=y\) or \(x+y=-4\).

(1) xy<0 --> the first case is not possible, since if \(x=y\), then \(xy=x^2\geq{0}\), not \(<0\) as given in this statement, hence we have the second case: \(x+y=-4\). Sufficient.

(2) x>2 and y<2. This statement implies that \(x\neq{y}\), therefore \(x+y=-4\). Sufficient.

Answer: D.

Hope it's clear.

Thanks Bunuel. Its a shame that i did all the work and completely misread the question.

Re: Inequality and absolute value questions from my collection [#permalink]
26 Aug 2012, 00:23

dvinoth86 wrote:

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.

Don't forget that \(y*(x-3)^2=0.\) B alone is definitely sufficient. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Inequality and absolute value questions from my collection [#permalink]
05 Sep 2012, 02:04

Expert's post

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.

I do not wanna to be wrong but when we substitute x in x-y/y>0 we should not have 1/2y>0 ??? and not 1/y>0.......then the meaning that y is positive eitherway not change.

Re: Inequality and absolute value questions from my collection [#permalink]
05 Sep 2012, 03:53

Expert's post

carcass 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.

I do not wanna to be wrong but when we substitute x in x-y/y>0 we should not have 1/2y>0 ??? and not 1/y>0.......then the meaning that y is positive eitherway not change.

"I dropped 2, as (1/2y) > 0 and (1/y) >0 are absolutely the same (you can multiply both sides of inequality by 2 and you'll get 1/y>0). What is important that you can get that y>0 from either of them.

Re: Inequality and absolute value questions from my collection [#permalink]
03 Oct 2012, 04:39

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!

Re: Inequality and absolute value questions from my collection [#permalink]
03 Oct 2012, 06:13

Expert's post

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.

Re: Inequality and absolute value questions from my collection [#permalink]
03 Oct 2012, 11:26

Quote:

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 ??

Re: Inequality and absolute value questions from my collection [#permalink]
04 Oct 2012, 02:39

Expert's post

liarish wrote:

Quote:

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]
04 Oct 2012, 02:53

Expert's post

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]
04 Oct 2012, 03:13

Expert's post

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]
11 Oct 2012, 02:11

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

Re: Inequality and absolute value questions from my collection [#permalink]
11 Oct 2012, 02:22

Expert's post

Jp27 wrote:

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

Re: Inequality and absolute value questions from my collection [#permalink]
23 Dec 2012, 03:40

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.”

gmatclubot

Re: Inequality and absolute value questions from my collection
[#permalink]
23 Dec 2012, 03:40

How the growth of emerging markets will strain global finance : Emerging economies need access to capital (i.e., finance) in order to fund the projects necessary for...

One question I get a lot from prospective students is what to do in the summer before the MBA program. Like a lot of folks from non traditional backgrounds...