GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Aug 2018, 09:35

### 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

# Which of the following expressions is defined for all

Author Message
Intern
Joined: 05 Apr 2008
Posts: 23
Which of the following expressions is defined for all  [#permalink]

### Show Tags

06 Sep 2008, 10:31
Which of the following expressions is defined for all integer values of z, such that (z^2) <9?
I. (z+1)/z
II. (z-4)/(z^2-4z+4)
III. 18/(z^2-4z-5)

a) None
b) I only
c) II only
d) III only
e) II and III

*** What confuses me is z<3 and z<-3. Am I taking the right approach?

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Director
Joined: 12 Jul 2008
Posts: 511
Schools: Wharton

### Show Tags

06 Sep 2008, 10:40
HVD1975 wrote:
Which of the following expressions is defined for all integer values of z, such that (z^2) <9?
I. (z+1)/z
II. (z-4)/(z^2-4z+4)
III. 18/(z^2-4z-5)

a) None
b) I only
c) II only
d) III only
e) II and III

*** What confuses me is z<3 and z<-3. Am I taking the right approach?

I get A

The question is asking which expressions, I, II, and/or III, are defined. Each of these expressions are undefined when their denominators are 0.

The prompt gives the restriction that z^2 < 9, which means -3 < z < 3

I. The expression is undefined when z = 0.

II. The expression is undefined when z^2-4z+4 = 0
(z-2)^2 = 0
z = 2

III. The expression is undefined when z^2-4z-5 = 0
(z-5)(z+1) = 0
z = 5 or -1
Director
Joined: 23 Sep 2007
Posts: 760

### Show Tags

06 Sep 2008, 10:41
HVD1975 wrote:
Which of the following expressions is defined for all integer values of z, such that (z^2) <9?
I. (z+1)/z
II. (z-4)/(z^2-4z+4)
III. 18/(z^2-4z-5)

a) None
b) I only
c) II only
d) III only
e) II and III

*** What confuses me is z<3 and z<-3. Am I taking the right approach?

E

if z = 0, equation 1 is undefined
VP
Joined: 05 Jul 2008
Posts: 1328

### Show Tags

06 Sep 2008, 10:59
HVD1975 wrote:
Which of the following expressions is defined for all integer values of z, such that (z^2) <9?
I. (z+1)/z
II. (z-4)/(z^2-4z+4)
III. 18/(z^2-4z-5)

a) None
b) I only
c) II only
d) III only
e) II and III

*** What confuses me is z<3 and z<-3. Am I taking the right approach?

z ^ 2 < 9 is a quadratic inequality

z ^2 -9 < 0

(z-3) < 0 & (z+3 ) > 0 means z<3 and z > -3 --> -3 < z < 3

or

(z-3) > 0 & (z+3) <0 z> 3 or z < -3 means z is between [- infinity, 3] and [3, Infinity]

I am guessing that we ignore the second one because this graph is a parabola and we are only concerned with the points at which it cuts the x axis. Hence we ignore the second set.

The rest of the solution is already explained.
Intern
Joined: 05 Apr 2008
Posts: 23

### Show Tags

06 Sep 2008, 11:16
But why -3<z<3 and not z<-3 and z<3

1. If z<3 then z could be 2,1,0,-1,-2,-3,-4,-5, etc
2. If z<-3 then z could be -4,-5,-6, etc

Shouldn't z<-3 cancels the other integers less than 3 (2,1,0,etc)?
Director
Joined: 12 Jul 2008
Posts: 511
Schools: Wharton

### Show Tags

06 Sep 2008, 12:08
HVD1975 wrote:
But why -3<z<3 and not z<-3 and z<3

1. If z<3 then z could be 2,1,0,-1,-2,-3,-4,-5, etc
2. If z<-3 then z could be -4,-5,-6, etc

Shouldn't z<-3 cancels the other integers less than 3 (2,1,0,etc)?

The prompt says z^2 < 9

If z < -3, then z^2 > 9.
VP
Joined: 17 Jun 2008
Posts: 1474

### Show Tags

07 Sep 2008, 23:52
Very good question. Initially, I thought something is missing in the question..
Senior Manager
Joined: 31 Jul 2008
Posts: 269

### Show Tags

08 Sep 2008, 14:46
A

divide by 0 is never defined !
Manager
Joined: 15 Jul 2008
Posts: 205

### Show Tags

08 Sep 2008, 15:39

x^2 < 9 ... only -2,-1,0,1,2 statisfy this condition.

and 0 does not satisfy the first option. so that is not defined for all x^2<9

2 and 3 are good. e.

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: Math Question &nbs [#permalink] 08 Sep 2008, 15:39
Display posts from previous: Sort by

# Which of the following expressions is defined for all

Moderator: chetan2u

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.