Find all School-related info fast with the new School-Specific MBA Forum

It is currently 22 Oct 2014, 22:54

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Is x/m(m^2 + n^2 + k^2) = xm + yn + zk? (1) z/k = x/m (2)

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Senior Manager
Senior Manager
avatar
Joined: 26 Jul 2005
Posts: 315
Location: Los Angeles
Followers: 1

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

Is x/m(m^2 + n^2 + k^2) = xm + yn + zk? (1) z/k = x/m (2) [#permalink] New post 27 Oct 2005, 15:35
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions
Is x/m(m^2 + n^2 + k^2) = xm + yn + zk?

(1) z/k = x/m
(2) x/m = y/n
Manager
Manager
avatar
Joined: 20 Mar 2005
Posts: 201
Location: Colombia, South America
Followers: 1

Kudos [?]: 6 [0], given: 0

 [#permalink] New post 27 Oct 2005, 15:49
I think is C

x/m(m^2 + n^2 + z^2) = xm + yn + zk

that is equal to

xm + xn^2/m + xz^2/n = xm + yn + zk

from I: x/m = y/n then xn = ym and from there (xn*n)/m is equal to
ym*n/m equal to yn

so insuff

from ii, do the same stuff as i and you get zk

so insuff

taking both it works so C
Senior Manager
Senior Manager
avatar
Joined: 26 Jul 2005
Posts: 315
Location: Los Angeles
Followers: 1

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

 [#permalink] New post 28 Oct 2005, 14:54
Right again - OA is C.
Intern
Intern
avatar
Joined: 14 Sep 2010
Posts: 24
Followers: 0

Kudos [?]: 6 [0], given: 4

Re: Is x/m(m^2 + n^2 + k^2) = xm + yn + zk? (1) z/k = x/m (2) [#permalink] New post 18 Nov 2011, 02:39
Is x/m(m^2 + n^2+ k^2) = xm + yn + zk? 

xm^2 + xn^2 + xk^2 = xmm +ynm +zkm?

Is xn^2 + xk^2 = ynm + zkm?

(1) z/k = x/m (xk = mz)

xk^2 = (xk)k = (mzk)

xn^2 + xk^2 = ynm + zkm =

xn^2 = ynm?
xn = my?

(2) x/m = y/n (xn = my)

xn^2 = (xn)n = (my)n

xn^2 + xk^2 = ynm + zkm =

xk^2 = zkm?
xk = mz?

Put (1) and (2) together 

xn^2 + xk^2 = ynm + zkm =

(my)n + (mz)k = ynm + zkm

Both statements used = Answer C

Posted from my mobile device Image
Senior Manager
Senior Manager
avatar
Joined: 13 May 2011
Posts: 324
WE 1: IT 1 Yr
WE 2: Supply Chain 5 Yrs
Followers: 19

Kudos [?]: 136 [0], given: 11

Re: Is x/m(m^2 + n^2 + k^2) = xm + yn + zk? (1) z/k = x/m (2) [#permalink] New post 18 Nov 2011, 04:19
is xm^2+xn^2+xk^2=xm^2+ynm+zkm
i. e. xn^2+xk^2=ynm+zkm
statement 1: xk=zm
xn^2=ynm Not Sufficient
Statement 2: ym=xn
xk^2=zkm Not Sufficient
Together: xn^2+xk^2=xn^2+xk^2
Sufficient
Re: Is x/m(m^2 + n^2 + k^2) = xm + yn + zk? (1) z/k = x/m (2)   [#permalink] 18 Nov 2011, 04:19
    Similar topics Author Replies Last post
Similar
Topics:
4 Experts publish their posts in the topic If kmn#0, is x/m*(m^2+n^2+k^2)=xm+yn+zk? Bunuel 5 23 Feb 2014, 06:41
21 Experts publish their posts in the topic Is x/m*(m^2+n^2+k^2)=xm+yn+zk? heyholetsgo 12 19 Aug 2010, 02:29
x/m) (m^2 + n^2 + k^2) = xm+yn+zk? 1. z/k = x/m 2. x/m = y/n vksunder 2 11 Sep 2008, 11:55
What is the product of A and B? 1) A= 2/xm 2) B=xm OA=C but boubi 4 06 Jul 2007, 05:26
What is the product of A & B? A = 2/xm B =24xm praveen_rao7 13 17 Mar 2005, 18:16
Display posts from previous: Sort by

Is x/m(m^2 + n^2 + k^2) = xm + yn + zk? (1) z/k = x/m (2)

  Question banks Downloads My Bookmarks Reviews Important topics  


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

Powered by phpBB © phpBB Group and phpBB SEO

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®.