m06 #06 : Retired Discussions [Locked]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 16 Jan 2017, 14:46

### 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

# m06 #06

Author Message
Intern
Joined: 13 Oct 2008
Posts: 16
Followers: 0

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

### Show Tags

12 Nov 2008, 13:39
2
This post was
BOOKMARKED
If y(u - c) = 0 and j(u - k) = 0, which of the following must be true, assuming c < k?

(A) yj < 0
(B) yj > 0
(C) yj = 0
(D) j = 0
(E) y = 0

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Expl given
---------

From the first equation of the setup we learn that either y = 0 or u = c. If y = 0, then yj = 0. If u = c then from the second equation j = 0 as c < k. Taking two equations combined, we will arrive to the conclusion that yj = 0.

How is colored portion true ?? pls explain.
Director
Joined: 10 Sep 2007
Posts: 947
Followers: 8

Kudos [?]: 286 [2] , given: 0

### Show Tags

12 Nov 2008, 14:28
2
KUDOS
Given facts
1) y(u - c) = 0
2) j(u - k) = 0,
3) c < k

Now your question why this statement is true u = c then from the second equation j = 0 as c < k.

Given in 3 that c<k.
If u = c,
Then u - k cannot be equal to 0 because that also means c - k and as c < k so c - k is not zero.

Now to make statement 2 true. j has to be zero.

Which again means yj = 0
CIO
Joined: 02 Oct 2007
Posts: 1218
Followers: 94

Kudos [?]: 911 [0], given: 334

### Show Tags

26 Nov 2008, 08:17
Thanks, guys. We'll have to edit the original explanation to be more precise. +1, abhijit_sen.
_________________

Welcome to GMAT Club!

Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?

Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.

GMAT Club Premium Membership - big benefits and savings

Senior Manager
Joined: 01 Feb 2010
Posts: 267
Followers: 1

Kudos [?]: 54 [0], given: 2

### Show Tags

17 Feb 2010, 07:12
mbaobsessed wrote:
If y(u - c) = 0 and j(u - k) = 0, which of the following must be true, assuming c < k?

(A) yj < 0
(B) yj > 0
(C) yj = 0
(D) j = 0
(E) y = 0

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Expl given
---------

From the first equation of the setup we learn that either y = 0 or u = c. If y = 0, then yj = 0. If u = c then from the second equation j = 0 as c < k. Taking two equations combined, we will arrive to the conclusion that yj = 0.

How is colored portion true ?? pls explain.

In this from first equation either j=0 or u=k
and from second equation either y=0 or u=c
now it is given c < k so u=k & u=c cannot be true at same time.
Thts why if j=0 then u=c.
else if y=0 then u=k.
SVP
Joined: 16 Nov 2010
Posts: 1672
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 33

Kudos [?]: 514 [0], given: 36

### Show Tags

23 Feb 2011, 03:49
Two cases from 1 - y = 0 or c = u

Two cases from 2 - j = 0 or u = k

=> c != k so either c = u or k = u but not both

if y = 0, then c != u

so u = k, but yj = 0

if c = u, then j = 0, and k != u, so j = 0, => yj = 0

so either y = 0 or j = 0 always

_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Director
Joined: 21 Dec 2009
Posts: 591
Concentration: Entrepreneurship, Finance
Followers: 18

Kudos [?]: 661 [0], given: 20

### Show Tags

24 Feb 2011, 13:37
How do we arrive at saying: if y(u-c)=0, then u = c?
it's like saying if y(x) = 0, then x = 0; I don't think so.
x (expression inside the braces) can be any value, but the
equation (independent variables) reduce to zero.

Am i missing something?
_________________

KUDOS me if you feel my contribution has helped you.

Manager
Joined: 02 Feb 2012
Posts: 204
Location: United States
GMAT 1: 770 Q49 V47
GPA: 3.08
Followers: 1

Kudos [?]: 6 [0], given: 104

### Show Tags

25 Feb 2012, 12:33
gmatbull, the way I understood/processed the question was as follows:

y(u-c) = 0 so lets assume u-c = 0 or u = c

c < k

therefore u-k < 0

for j(u-k) = 0 to be true j = 0

therefore jy=0

You can also switch the assumption around and you will find that y = 0 and the answer still holds.

Hope it helped a bit.
Intern
Joined: 12 Dec 2010
Posts: 5
Schools: Pittsburg State University
Followers: 0

Kudos [?]: 0 [0], given: 1

### Show Tags

28 Feb 2012, 14:15
Yekrut wrote:
gmatbull, the way I understood/processed the question was as follows:

y(u-c) = 0 so lets assume u-c = 0 or u = c

c < k

therefore u-k < 0

for j(u-k) = 0 to be true j = 0

therefore jy=0

You can also switch the assumption around and you will find that y = 0 and the answer still holds.

Hope it helped a bit.

I got the right answer "C" but reading your thought process certainly helped me frame what I was thinking lol
Math Expert
Joined: 02 Sep 2009
Posts: 36520
Followers: 7066

Kudos [?]: 92911 [5] , given: 10528

### Show Tags

26 Feb 2013, 06:24
5
KUDOS
Expert's post
If y(u - c) = 0 and j(u - k) = 0, which of the following must be true, assuming c < k?

A. yj < 0
B. yj > 0
C. yj = 0
D. j = 0
E. y = 0

$$y(u - c) = 0$$ --> $$u=c$$ or $$y=0$$;
$$j(u - k) = 0$$ --> $$u=k$$ or $$j=0$$;

Now, the first option ($$u=c$$ and $$u=k$$) cannot be simultaneously correct for both equations because if it is, then it would mean that $$u=c=k$$, but we are given that $$c<k$$. So, only one can be correct so either $$y=0$$ or $$j=0$$, which makes $$yj = 0$$ is always true.

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

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

### Show Tags

27 Feb 2013, 07:31
1
KUDOS
Given y(u-c) = 0 and j(u-k) =0 and c <k.

y(u-c)=0 => either y=0 or u-c=0 => u=c
j(u-k)=0 => either j=0 or u-k=0 => u=k

from above two eqn c=k but c <k
so y =0 and j=0

yj=0

Manager
Joined: 20 Oct 2013
Posts: 78
Location: United States
Concentration: General Management, Real Estate
Followers: 0

Kudos [?]: 7 [0], given: 15

### Show Tags

21 Apr 2014, 07:52
Since c<k, u-c and u-k cannot be both 0 -> At least one of y and j is 0 -> yj=0
Re: m06 #06   [#permalink] 21 Apr 2014, 07:52
Similar topics Replies Last post
Similar
Topics:
4 m06 q30 12 18 Nov 2008, 18:44
46 M06 Q9 20 12 Nov 2008, 13:52
11 m06 Q5 19 12 Nov 2008, 13:36
34 M06 #08 26 04 Nov 2008, 00:59
1 M06 #19 6 24 Sep 2008, 20:16
Display posts from previous: Sort by

# m06 #06

Moderator: Bunuel

 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®.