Inequality and absolute value questions from my collection : GMAT Data Sufficiency (DS) - Page 3
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 18 Jan 2017, 18:56

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Inequality and absolute value questions from my collection

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7077

Kudos [?]: 93118 [96] , given: 10552

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

### Show Tags

16 Nov 2009, 10:33
96
KUDOS
Expert's post
438
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) $$x^3< 0$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-20.html#p653690

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-20.html#p653695

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653697

4. Are x and y both positive?
(1) $$2x-2y=1$$
(2) $$\frac{x}{y}>1$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653709

5. What is the value of y?
(1) $$3|x^2 -4| = y - 2$$
(2) $$|3 - y| = 11$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653731

6. If x and y are integer, is y > 0?
(1) $$x +1 > 0$$
(2) $$xy > 0$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653740

7. $$|x+2|=|y+2|$$ what is the value of x+y?
(1) $$xy<0$$
(2) $$x>2$$, $$y<2$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653783 AND inequality-and-absolute-value-questions-from-my-collection-86939-160.html#p1111747

8. $$a*b \neq 0$$. Is $$\frac{|a|}{|b|}=\frac{a}{b}$$?
(1) $$|a*b|=a*b$$
(2) $$\frac{|a|}{|b|}=|\frac{a}{b}|$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653789

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653792

10. If n is not equal to 0, is |n| < 4 ?
(1) $$n^2 > 16$$
(2) $$\frac{1}{|n|} > n$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653796

11. Is $$|x+y|>|x-y|$$?
(1) $$|x| > |y|$$
(2) $$|x-y| < |x|$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653853

12. Is r=s?
(1) $$-s \leq r \leq s$$
(2) $$|r| \geq s$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653870

13. Is $$|x-1| < 1$$?
(1) $$(x-1)^2 \leq 1$$
(2) $$x^2 - 1 > 0$$

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653886

Official answers (OA's) and detailed solutions are in my posts on pages 2 and 3.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7077

Kudos [?]: 93118 [1] , given: 10552

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

### Show Tags

04 Oct 2012, 03:00
1
KUDOS
Expert's post
carcass wrote:
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:

Given: 1/|n| > n. Now, 1/|n| is always positive (because of the absolute value in the denominator), so if n is negative then this inequality will always hold true: 1/|n|=positive>n=negative.

Hope it's clear.

P.S. Complete solution for 1/|n| > n is n<0 or 0<n<1.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7077

Kudos [?]: 93118 [1] , given: 10552

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

### Show Tags

04 Oct 2012, 03:20
1
KUDOS
Expert's post
carcass wrote:
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 this information and we are not sure of course of - 4 < n < 4.

Correct ???

Thanks

1/|n| > n --> 2 cases:

If n<0, then |n|=-n, so we'll have that 1/-n>n --> multiply by n and flip the sign (since we consider negative n): -1<n^2 --> which holds true for any n from this range, so for any negative n.

If n>0, then |n|=n, so we'll have that 1/n>n --> multiply by positive n, this time: 1>n^2 --> -1<n<1, since we consider n>0, then finally we'll get 0<n<1.

So, 1/|n| > n holds true for n<0 and 0<n<1.

Hope it's clear.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7077

Kudos [?]: 93118 [1] , given: 10552

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

### Show Tags

22 Feb 2013, 00:27
1
KUDOS
Expert's post
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.

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.

Hope it's clear.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7077

Kudos [?]: 93118 [1] , given: 10552

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

### Show Tags

28 Feb 2013, 05:42
1
KUDOS
Expert's post
piealpha wrote:
7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2

The solution seem confusing to me as I see four cases:
a] x<-2, y<-2
b]x>-2, y>-2
c] x<-2, y>-2
d]x>-2, y<-2

case [a] and [b] support x=y while case [c] and [d] support x+y=-4

when xy<0, the case [c]or[d] always do not apply, for example: x=-3 and y=3 would come under case[c] and x=-1 and y=3 would come under case [b] , so it is insufficient.

when x>2 , y<2, we have a case [b] with x=3, y=-1 and a case [d] with x=3,y=-3. So insufficient

when we combine(1)+(2) , we have a case as shown above , it is also insufficient.

So my answer choice would be E.

Can somebody help if I am wrong.

Links to OA's and solutions are given in the original post: inequality-and-absolute-value-questions-from-my-collection-86939-200.html#p652806

OA for this question is D, not E. Discussed here: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653783 and here: inequality-and-absolute-value-questions-from-my-collection-86939-160.html#p1111747

Hope it helps.
_________________
Intern
Joined: 10 Aug 2012
Posts: 19
Location: India
Concentration: General Management, Technology
GPA: 3.96
Followers: 0

Kudos [?]: 6 [1] , given: 15

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

### Show Tags

28 Feb 2013, 07:36
1
KUDOS
Question 1:

$$6xy = x^2 y + 9y$$

$$y(x^2 -6x +9) = 0$$

$$y(x-3)^2 = 0$$

either y =0, or x=3

statement 1: y-x =3
If y= 0, xy =0, irrespective of x
If x=3, y =6, xy= 18

So, A & D are not correct

statement 2:

$$x^3 < 0 => x <0$$

=> x is not equal to 3 so y=0, and xy = 0

Intern
Joined: 05 May 2013
Posts: 27
GMAT 1: 730 Q50 V39
GRE 1: 1480 Q800 V680
Followers: 0

Kudos [?]: 22 [1] , given: 5

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

### Show Tags

11 Jul 2013, 20:42
1
KUDOS
Hi johncoffey ,
My two cents- for (1) - it is always useful to start out by factoring an expression if possible, especially when there is a variable in common ("y" in this example). Even though it does make sense to isolate the expression "xy" that we are being asked for- note that in this case that would give us more unknowns on the RHS.
Hope tht helps.
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7077

Kudos [?]: 93118 [1] , given: 10552

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

### Show Tags

22 Sep 2013, 05:00
1
KUDOS
Expert's post
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.

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?
_________________
Intern
Joined: 10 Aug 2013
Posts: 20
Followers: 0

Kudos [?]: 9 [1] , given: 17

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

### Show Tags

22 Sep 2013, 07:07
1
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.

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.

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.

Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2654
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Followers: 116

Kudos [?]: 1338 [1] , given: 789

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

### Show Tags

08 Sep 2015, 10:30
1
KUDOS
Expert's post
reto 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.

$$\frac{x}{y}>1$$ --> $$\frac{x-y}{y}>0$$ --> substitute x --> $$\frac{1}{y}>0$$ --> $$y$$ is positive, and as $$x=y+\frac{1}{2}$$, $$x$$

If you substitute x, the new fraction should look as follows: $$\frac{0.5}{y}>0$$ since $$x=y+\frac{1}{2}$$ ?

Yes, you are correct.

After substituting for x, you get, $$\frac{0.5}{y} >0$$ ---> $$\frac{1}{2y} >0$$ ---> $$\frac{1}{y} >0$$ ---> y >0
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2654
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Followers: 116

Kudos [?]: 1338 [1] , given: 789

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

### Show Tags

04 Oct 2015, 12:34
1
KUDOS
Expert's post
RafaelPina wrote:
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

(1) (x + y)^2 = 9a
x^2 + y^2 + 2xy = 9a
x^2 + y^2 = 9a - 2xy

(2) (x – y)^2 = a
x^2 + y^2 - 2xy = a
x^2 + y^2 = a +2xy

So we can conclude that:
9a - 2xy = a + 2xy
8a = 4xy
4a = 2xy

Hence, in the first conclusion is:
x^2 + y^2 = 9a - 4a
x^2 + y^2 = 5a

Am I wrong, or right?

You are correct till $$x^2+y^2=5a$$ but what if x=y=0 giving you a=0. In this case, $$x^2+y^2$$ will be = 4a and NOT > 4a. This is the reason why E is the correct answer.

If you were given "is $$x^2+y^2 \geq 4a$$ instead of just >4a", then yes, you would have marked C as the correct answer but you are asked >4a which may or may not be true.

Hope this helps.
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7077

Kudos [?]: 93118 [1] , given: 10552

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

### Show Tags

05 Nov 2015, 00:28
1
KUDOS
Expert's post
Johnbreeden85 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.

How did you figure out that $$\frac{x}{y}>1$$ --> $$\frac{x-y}{y}>0$$ ? Any help is appreciated. Thank you.

This is explained couple of times on the previous pages:
$$\frac{x}{y}>1$$

$$\frac{x}{y}-1>0$$

$$\frac{x}{y}-\frac{y}{y}>0$$

$$\frac{x-y}{y}>0$$.

Hope it's clear.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7077

Kudos [?]: 93118 [1] , given: 10552

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

### Show Tags

22 Feb 2016, 23:55
1
KUDOS
Expert's post
nishantdoshi wrote:
Bunuel wrote:
nishantdoshi wrote:

hey bunuel
can you please clear my doubt?
in statement 1 you've written either n is negative OR n equals to zero but as per my knowledge shouldn't n be negative only because I've read it in many post that are on absolute value, here's a link: math-absolute-value-modulus-86462.html

correct me if i'm wrong!

An absolute value cannot be negative but it CAN be 0. For this particular case 0 fits:
-n=|-n| --> -0 = |-0| --> 0 = 0.

if x>=0 then |x|=x
and if x<0 then |x|=-x

am i wrong?

Yes, you are wrong.

We can say that when x<=0, then |x| is also equal to -x:

|0| = -0.
_________________
VP
Joined: 05 Mar 2008
Posts: 1473
Followers: 11

Kudos [?]: 261 [0], given: 31

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

### Show Tags

16 Nov 2009, 12:08
Bunuel wrote:

13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

I'm getting B for this one

1. (x-1)^2 <= 1
x can be 0 which would make the question no
or x can be 1/2 which would make the answer yes
so 1 is insufficient

2. x^2 - 1 > 0
means x^2>1
so x<-1 or x>1
both of which make the question no
so sufficient
VP
Joined: 05 Mar 2008
Posts: 1473
Followers: 11

Kudos [?]: 261 [0], given: 31

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

### Show Tags

16 Nov 2009, 12:19
Bunuel wrote:

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

I'm getting c

1. s can be 3 and r can be 3 which makes question yes
s can be 3 and r can be 2 which makes question no
insufficient

2. r can be 3 and s can be 3 makes question yes
r can be 3 s can be 2 makes question no
insufficient

combining:
|r|>=s means
r>=s or r<=-s

and -s<=r<=s means
-s<=r and r<=s

now we have -s<=r and -s>=r so -s = r or s = r
r>=s and r<=s so s = r
VP
Joined: 05 Mar 2008
Posts: 1473
Followers: 11

Kudos [?]: 261 [0], given: 31

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

### Show Tags

16 Nov 2009, 14:33
10. If n is not equal to 0, is |n| < 4 ?
(1) n^2 > 16
(2) 1/|n| > n

because in number 2 n can be negative or a fraction
Intern
Joined: 19 Oct 2009
Posts: 44
Followers: 0

Kudos [?]: 60 [0], given: 4

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

### Show Tags

16 Nov 2009, 19:07
Bunuel, thanks for the questions. Please provide the OA's too. It would be great if you can provide them soon. I am having my GMAT this week, so kinda tensed and impatient. Also, I am yet to give my MGMAT CAT's, so tell me whether should I solve the questions on the forum because if the questions are from the MGMAT CAT's or Gmat Prep then it may overestimate my result. I would appreciate your response. Thanks once again.
Manager
Joined: 13 Aug 2009
Posts: 203
Schools: Sloan '14 (S)
Followers: 3

Kudos [?]: 103 [0], given: 16

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

### Show Tags

16 Nov 2009, 20:39
Quality questions as always... Thanks Bunuel! +1
SVP
Joined: 29 Aug 2007
Posts: 2492
Followers: 67

Kudos [?]: 734 [0], given: 19

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

### Show Tags

16 Nov 2009, 21:46
lagomez wrote:
Bunuel wrote:

13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

I'm getting B for this one

1. (x-1)^2 <= 1
x can be 0 which would make the question no
or x can be 1/2 which would make the answer yes
so 1 is insufficient

2. x^2 - 1 > 0
means x^2>1
so x<-1 or x>1
both of which make the question no
so sufficient

(1) (x-1)^2 <= 1
x is 0 to 2.
If x = 2, yes.
If x < 2, No.

(2) x^2 - 1 > 0
x cannot be -1 to 1 i.e. x<-1 or x>1. NSF.

From 1 and 2: x is >1 but <=2. NSF..

E.
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Manager
Joined: 13 Aug 2009
Posts: 203
Schools: Sloan '14 (S)
Followers: 3

Kudos [?]: 103 [0], given: 16

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

### Show Tags

17 Nov 2009, 04:18
Bunuel wrote:
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 I reduced the given equation (divided out the y) and solved for x:
6*x*y = x^2*y + 9*y
6*x = x^2 + 9
0 = x^2 - 6*x + 9
0 = (x-3)^2
x = 3

Statement 1:

y-x=3
y-3=3
y=6
xy=3*6=18

SUFFICIENT

Statement 2:

x^3<0

We have no idea what the value of y is from this statement. The only thing that made me look twice was the face that if x^3 is true, then x should be a negative value... did I calculate the value of x incorrectly above?

INSUFFICIENT

Manager
Joined: 13 Aug 2009
Posts: 203
Schools: Sloan '14 (S)
Followers: 3

Kudos [?]: 103 [0], given: 16

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

### Show Tags

17 Nov 2009, 04:34
Bunuel wrote:
2. If y is an integer and y = |x| + x, is y = 0?
(1) x < 0
(2) y < 1

Another way of looking at the problem is to ask, is x<0? Because if it is, then we know that y is zero. The only case in which y will not be zero is if x is positive.

Statement 1:

SUFFICIENT

Statement 2:

y<1

Because y is an integer, it must be one of the following values: 0, -1, -2, -3...

BUT |x| + x can never be a negative value. The lowest value that it can be is 0.

Hence, y can never be negative and the only possible value it can be then is 0.

SUFFICIENT

Re: Inequality and absolute value questions from my collection   [#permalink] 17 Nov 2009, 04:34

Go to page   Previous    1   2   3   4   5   6   7   8   9   10   11  ...  22    Next  [ 427 posts ]

Similar topics Replies Last post
Similar
Topics:
2 If z and x are integers with absolute values greater than 1, is z^x 3 09 Feb 2016, 08:54
237 The annual rent collected by a corporation from a certain 55 02 Jan 2010, 14:10
19 The annual rent collected by a corporation from a certain 14 29 Dec 2009, 04:55
66 Collection of 12 DS questions 78 17 Oct 2009, 17:45
87 Collection of 8 DS questions 50 13 Oct 2009, 19:16
Display posts from previous: Sort by