It is currently 23 Oct 2017, 14:12

Close

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
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.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Algebra - JSQ91

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Senior Manager
Senior Manager
User avatar
Joined: 25 Nov 2006
Posts: 333

Kudos [?]: 70 [0], given: 0

Schools: St Gallen, Cambridge, HEC Montreal
Algebra - JSQ91 [#permalink]

Show Tags

New post 07 Feb 2009, 16:59
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Image

Kudos [?]: 70 [0], given: 0

Senior Manager
Senior Manager
avatar
Joined: 30 Nov 2008
Posts: 484

Kudos [?]: 362 [0], given: 15

Schools: Fuqua
Re: Algebra - JSQ91 [#permalink]

Show Tags

New post 07 Feb 2009, 17:36
IMO E.

from stmt 1,

\(sqrt(xz)\) is an integer ==> product of x and z is a square number like 4,9,16 etc

let \(x = 2, z = 8 ==> sqrt(xz) = sqrt(16)\)= which is an integer. But \(sqrt(z)\) which is \(sqrt(8)\) is not a integer.
let \(x= 4, z = 9 ==> sqrt(xz) = sqrt(36) =\)which is an integer. And \(sqrt(z)\) which is \(sqrt(9)\) is an integer.

Hence stmt 1 alone is not sufficient to say if z is an integer or not.

From stmt 2,

\(x = z^2. ==> sqrt(xz) = sqrt(z^3)\)

If z is anything other than 1 then sqrt(z) is not an integer
But if z = 1 then \(sqrt(z) = 1\)which is an integer.

Hence stmt 2 alone is not sufficient to say if z is an integer or not.

Togerther also, does not provide enough information.

Kudos [?]: 362 [0], given: 15

SVP
SVP
User avatar
Joined: 07 Nov 2007
Posts: 1792

Kudos [?]: 1065 [0], given: 5

Location: New York
Re: Algebra - JSQ91 [#permalink]

Show Tags

New post 07 Feb 2009, 18:12
agree with E.

same explanation.
_________________

Your attitude determines your altitude
Smiling wins more friends than frowning

Kudos [?]: 1065 [0], given: 5

Manager
Manager
avatar
Joined: 04 Jan 2009
Posts: 237

Kudos [?]: 13 [0], given: 0

Re: Algebra - JSQ91 [#permalink]

Show Tags

New post 07 Feb 2009, 19:09
mrsmarthi wrote:
IMO E.

from stmt 1,

\(sqrt(xz)\) is an integer ==> product of x and z is a square number like 4,9,16 etc

let \(x = 2, z = 8 ==> sqrt(xz) = sqrt(16)\)= which is an integer. But \(sqrt(z)\) which is \(sqrt(8)\) is not a integer.
let \(x= 4, z = 9 ==> sqrt(xz) = sqrt(36) =\)which is an integer. And \(sqrt(z)\) which is \(sqrt(9)\) is an integer.

Hence stmt 1 alone is not sufficient to say if z is an integer or not.

From stmt 2,

\(x = z^2. ==> sqrt(xz) = sqrt(z^3)\)

If z is anything other than 1 then sqrt(z) is not an integer
But if z = 1 then \(sqrt(z) = 1\)which is an integer.

Hence stmt 2 alone is not sufficient to say if z is an integer or not.

Togerther also, does not provide enough information.

I agree with you that both by themselves not sufficient.
Now, if taken together we have the following:
z*sqrt(z) is integer. Z is integer known already.
if z=1, sqrt z is integer.
if z=2, sqrtz is noninteger. But B is not satisfied.
If z=4, sqrtz is integer. and this satisfies both a and B.
Both the statements are satisfied for a and b to obe true.

Hence, C.
_________________

-----------------------
tusharvk

Kudos [?]: 13 [0], given: 0

SVP
SVP
User avatar
Joined: 07 Nov 2007
Posts: 1792

Kudos [?]: 1065 [0], given: 5

Location: New York
Re: Algebra - JSQ91 [#permalink]

Show Tags

New post 07 Feb 2009, 19:32
tusharvk wrote:
mrsmarthi wrote:
IMO E.

from stmt 1,

\(sqrt(xz)\) is an integer ==> product of x and z is a square number like 4,9,16 etc

let \(x = 2, z = 8 ==> sqrt(xz) = sqrt(16)\)= which is an integer. But \(sqrt(z)\) which is \(sqrt(8)\) is not a integer.
let \(x= 4, z = 9 ==> sqrt(xz) = sqrt(36) =\)which is an integer. And \(sqrt(z)\) which is \(sqrt(9)\) is an integer.

Hence stmt 1 alone is not sufficient to say if z is an integer or not.

From stmt 2,

\(x = z^2. ==> sqrt(xz) = sqrt(z^3)\)

If z is anything other than 1 then sqrt(z) is not an integer
But if z = 1 then \(sqrt(z) = 1\)which is an integer.

Hence stmt 2 alone is not sufficient to say if z is an integer or not.

Togerther also, does not provide enough information.

I agree with you that both by themselves not sufficient.
Now, if taken together we have the following:
z*sqrt(z) is integer. Z is integer known already.
if z=1, sqrt z is integer.
if z=2, sqrtz is noninteger. But B is not satisfied.
If z=4, sqrtz is integer. and this satisfies both a and B.
Both the statements are satisfied for a and b to obe true.

Hence, C.


good catch..

sqrt(z^3) = z*sqrt(z) = integer.. --> sqrt(z) must be integer

how did I miss that?
_________________

Your attitude determines your altitude
Smiling wins more friends than frowning

Kudos [?]: 1065 [0], given: 5

Senior Manager
Senior Manager
avatar
Joined: 30 Nov 2008
Posts: 484

Kudos [?]: 362 [0], given: 15

Schools: Fuqua
Re: Algebra - JSQ91 [#permalink]

Show Tags

New post 07 Feb 2009, 19:35
tusharvk wrote:
mrsmarthi wrote:
IMO E.

from stmt 1,

\(sqrt(xz)\) is an integer ==> product of x and z is a square number like 4,9,16 etc

let \(x = 2, z = 8 ==> sqrt(xz) = sqrt(16)\)= which is an integer. But \(sqrt(z)\) which is \(sqrt(8)\) is not a integer.
let \(x= 4, z = 9 ==> sqrt(xz) = sqrt(36) =\)which is an integer. And \(sqrt(z)\) which is \(sqrt(9)\) is an integer.

Hence stmt 1 alone is not sufficient to say if z is an integer or not.

From stmt 2,

\(x = z^2. ==> sqrt(xz) = sqrt(z^3)\)

If z is anything other than 1 then sqrt(z) is not an integer
But if z = 1 then \(sqrt(z) = 1\)which is an integer.

Hence stmt 2 alone is not sufficient to say if z is an integer or not.

Togerther also, does not provide enough information.

I agree with you that both by themselves not sufficient.
Now, if taken together we have the following:
z*sqrt(z) is integer. Z is integer known already.
if z=1, sqrt z is integer.
if z=2, sqrtz is noninteger. But B is not satisfied.
If z=4, sqrtz is integer. and this satisfies both a and B.
Both the statements are satisfied for a and b to obe true.

Hence, C.


Tusharvk, I will have to agree with you. By using both the eq, we definitely know that \(sqrt(z)\)is an integer. Hence together is sufficient. here is my approach.

From stmt 2, we know that \(x = z^2 ==> sqrt(xz) = sqrt(z^3) ==> z sqrt(z).\)

from stmt 1, we know that \(sqrt(xz)\) is an integer ==> \(z sqrt(z)\) is an integer. And this possible only when \(sqrt(z)\) is an integer.

I strongly believe that skill of reasoning is what is needed to master the DS questions in addition to the basic maths. But is there any how can help me know, whether we can cross verify the approach to confirm the correctness? :roll:

Kudos [?]: 362 [0], given: 15

SVP
SVP
User avatar
Joined: 29 Aug 2007
Posts: 2472

Kudos [?]: 845 [0], given: 19

Re: Algebra - JSQ91 [#permalink]

Show Tags

New post 07 Feb 2009, 20:41
mrsmarthi wrote:
IMO E.

from stmt 1,

\(sqrt(xz)\) is an integer ==> product of x and z is a square number like 4,9,16 etc

let \(x = 2, z = 8 ==> sqrt(xz) = sqrt(16)\)= which is an integer. But \(sqrt(z)\) which is \(sqrt(8)\) is not a integer.
let \(x= 4, z = 9 ==> sqrt(xz) = sqrt(36) =\)which is an integer. And \(sqrt(z)\) which is \(sqrt(9)\) is an integer.

Hence stmt 1 alone is not sufficient to say if z is an integer or not.

From stmt 2,

\(x = z^2. ==> sqrt(xz) = sqrt(z^3)\)

If z is anything other than 1 then sqrt(z) is not an integer
But if z = 1 then \(sqrt(z) = 1\)which is an integer.

Hence stmt 2 alone is not sufficient to say if z is an integer or not.

Togerther also, does not provide enough information.


You mixed st 1 in st 2: \(x = z^2. ==> sqrt(xz) = sqrt(z^3)\)
_________________

Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html
Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html
Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html


GT

Kudos [?]: 845 [0], given: 19

Senior Manager
Senior Manager
avatar
Joined: 30 Nov 2008
Posts: 484

Kudos [?]: 362 [0], given: 15

Schools: Fuqua
Re: Algebra - JSQ91 [#permalink]

Show Tags

New post 07 Feb 2009, 20:50
GMAT TIGER wrote:
You mixed st 1 in st 2:


Hum....ok now I got the mistake. well Let me see if I can do it correctly in similar example. Thank you. :)

Kudos [?]: 362 [0], given: 15

Re: Algebra - JSQ91   [#permalink] 07 Feb 2009, 20:50
Display posts from previous: Sort by

Algebra - JSQ91

  post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

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®.