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

It is currently 20 Aug 2018, 09:37

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

If xyz 0, is x (y + z) = 0? (1) |y + z| = |y| + |z| (2) |x +

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
VP
VP
User avatar
Joined: 30 Jun 2008
Posts: 1004
If xyz 0, is x (y + z) = 0? (1) |y + z| = |y| + |z| (2) |x +  [#permalink]

Show Tags

New post 24 Oct 2008, 03:33
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

HideShow timer Statistics

If xyz ≠ 0, is x (y + z) = 0?
(1) |y + z| = |y| + |z|
(2) |x + y| = |x| + |y|

I do not have an OA :(

--== 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.

_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

VP
VP
avatar
Joined: 17 Jun 2008
Posts: 1474
Re: DS : MODULUS  [#permalink]

Show Tags

New post 24 Oct 2008, 05:26
I would say A.

From the question x is not zero. Hence, x(y+z) will be zero only if y and z are of different sign.

From stmt1: y and z are of the same sign. Else, |y+z| will not equal |y| + |z|. Hence, sufficient.

From stmt2: x and y are of the same sign. But, nothing is known of z. Hence, insufficient.
Manager
Manager
avatar
Joined: 12 Oct 2008
Posts: 98
Re: DS : MODULUS  [#permalink]

Show Tags

New post 24 Oct 2008, 06:34
well, it is not A... I picked A too.
VP
VP
User avatar
Joined: 30 Jun 2008
Posts: 1004
Re: DS : MODULUS  [#permalink]

Show Tags

New post 24 Oct 2008, 07:38
linau1982 wrote:
well, it is not A... I picked A too.


it is not A ?? I had got A too ....

can you explain your approach ?
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Manager
Manager
avatar
Joined: 12 Oct 2008
Posts: 98
Re: DS : MODULUS  [#permalink]

Show Tags

New post 24 Oct 2008, 07:44
I have no idea :-D I had this one in one of my practice test, picked A, wrong... I'll hold on to OA for a while.
Retired Moderator
User avatar
Joined: 18 Jul 2008
Posts: 893
Re: DS : MODULUS  [#permalink]

Show Tags

New post 24 Oct 2008, 09:17
I'll join in with the A, even though it's wrong.
Senior Manager
Senior Manager
avatar
Joined: 18 Jun 2007
Posts: 272
Re: DS : MODULUS  [#permalink]

Show Tags

New post 24 Oct 2008, 20:03
amitdgr wrote:
If xyz ≠ 0, is x (y + z) = 0?
(1) |y + z| = |y| + |z|
(2) |x + y| = |x| + |y|

I do not have an OA :(


Why A?
from question we know that y and z has to be of different signs.
from st.1, how can we assume that y and z are of different signs?
take y = 2 and z=3.
=> |2+3|= |2| + |3| = 5

And if y = -2 and z = -2
=> |-2 -2| = |-2| + |-2|
=> 4 = 4 .....No

and with different signs, st1 will result in Yes.
statement 1 insufficient

IMO anwer is E
VP
VP
User avatar
Joined: 30 Jun 2008
Posts: 1004
Re: DS : MODULUS  [#permalink]

Show Tags

New post 24 Oct 2008, 20:42
rishi2377 wrote:
amitdgr wrote:
If xyz ≠ 0, is x (y + z) = 0?
(1) |y + z| = |y| + |z|
(2) |x + y| = |x| + |y|

I do not have an OA :(


Why A?
from question we know that y and z has to be of different signs.
from st.1, how can we assume that y and z are of different signs?
take y = 2 and z=3.
=> |2+3|= |2| + |3| = 5

And if y = -2 and z = -2
=> |-2 -2| = |-2| + |-2|
=> 4 = 4 .....No

and with different signs, st1 will result in Yes.
statement 1 insufficient

IMO anwer is E


rishi :) haven't you proved that y and z have to be of the same sign in your solution using the statement 1 ?

that means A is just sufficient to prove that y and z are not of different signs and thus x (y + z) is NOT zero
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Senior Manager
Senior Manager
avatar
Joined: 18 Jun 2007
Posts: 272
Re: DS : MODULUS  [#permalink]

Show Tags

New post 24 Oct 2008, 21:29
amitdgr wrote:
rishi2377 wrote:
amitdgr wrote:
If xyz ≠ 0, is x (y + z) = 0?
(1) |y + z| = |y| + |z|
(2) |x + y| = |x| + |y|

I do not have an OA :(


Why A?
from question we know that y and z has to be of different signs.
from st.1, how can we assume that y and z are of different signs?
take y = 2 and z=3.
=> |2+3|= |2| + |3| = 5

And if y = -2 and z = -2
=> |-2 -2| = |-2| + |-2|
=> 4 = 4 .....No

and with different signs, st1 will result in Yes.
statement 1 insufficient

IMO anwer is E


rishi :) haven't you proved that y and z have to be of the same sign in your solution using the statement 1 ?

that means A is just sufficient to prove that y and z are not of different signs and thus x (y + z) is NOT zero



lol yes..silly mistake. perhaps I am thinking too much or too little.


takes a similar question.

If /x/ + /y/ = /x + y/ then which of the following must be true?
(A) x + y > 0
(B) x + y < 0
(C) x - y > 0
(D) xy > 0
(E) xy < 0

ans is D... cause only thing above question suggests is that both x and y are of same digit....I got this question right and made a silly mistake in the topic question.. Gee! :oops:
Director
Director
avatar
Joined: 14 Aug 2007
Posts: 692
Re: DS : MODULUS  [#permalink]

Show Tags

New post 24 Oct 2008, 23:27
A is sufficient to prove that y and z have same sign.

A for me.
VP
VP
avatar
Joined: 17 Jun 2008
Posts: 1474
Re: DS : MODULUS  [#permalink]

Show Tags

New post 25 Oct 2008, 04:09
rishi2377 wrote:
takes a similar question.

If /x/ + /y/ = /x + y/ then which of the following must be true?
(A) x + y > 0
(B) x + y < 0
(C) x - y > 0
(D) xy > 0
(E) xy < 0

ans is D... cause only thing above question suggests is that both x and y are of same digit....I got this question right and made a silly mistake in the topic question.. Gee! :oops:


How can we prove that for x = y = 0, D should be the answer? Somehow, I feel the question is not complete.
Senior Manager
Senior Manager
avatar
Joined: 18 Jun 2007
Posts: 272
Re: DS : MODULUS  [#permalink]

Show Tags

New post 25 Oct 2008, 19:46
scthakur wrote:
rishi2377 wrote:
takes a similar question.

If /x/ + /y/ = /x + y/ then which of the following must be true?
(A) x + y > 0
(B) x + y < 0
(C) x - y > 0
(D) xy > 0
(E) xy < 0

ans is D... cause only thing above question suggests is that both x and y are of same digit....I got this question right and made a silly mistake in the topic question.. Gee! :oops:


How can we prove that for x = y = 0, D should be the answer? Somehow, I feel the question is not complete.

scthakur, to me question is complete. as we discussed earlier, given equation can be right only if x and y are of same sign whether positive or negative and in both the casesxy would be > 0. here it is not that x = y = 0 but xy>0
VP
VP
avatar
Joined: 17 Jun 2008
Posts: 1474
Re: DS : MODULUS  [#permalink]

Show Tags

New post 26 Oct 2008, 13:47
rishi2377 wrote:
scthakur wrote:
rishi2377 wrote:
takes a similar question.

If /x/ + /y/ = /x + y/ then which of the following must be true?
(A) x + y > 0
(B) x + y < 0
(C) x - y > 0
(D) xy > 0
(E) xy < 0

ans is D... cause only thing above question suggests is that both x and y are of same digit....I got this question right and made a silly mistake in the topic question.. Gee! :oops:


How can we prove that for x = y = 0, D should be the answer? Somehow, I feel the question is not complete.

scthakur, to me question is complete. as we discussed earlier, given equation can be right only if x and y are of same sign whether positive or negative and in both the casesxy would be > 0. here it is not that x = y = 0 but xy>0



Rishi, I meant, the question shoulld also have a mention of xy <> 0. In the absence of this, D cannot be the right answer.
VP
VP
User avatar
Joined: 30 Jun 2008
Posts: 1004
Re: DS : MODULUS  [#permalink]

Show Tags

New post 26 Oct 2008, 20:59
scthakur wrote:
Rishi, I meant, the question shoulld also have a mention of xy <> 0. In the absence of this, D cannot be the right answer.


exactly!! GMAC does not (generally) leave questions ambiguous ....
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Intern
Intern
avatar
Joined: 14 Jul 2008
Posts: 12
Re: DS : MODULUS  [#permalink]

Show Tags

New post 27 Oct 2008, 04:52
I think it should be "D"
Combine two statements, X Y and Z have to be all negative or all position.
Therefore, X(Y+Z) can't equal to 0

--== 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: DS : MODULUS &nbs [#permalink] 27 Oct 2008, 04:52
Display posts from previous: Sort by

If xyz 0, is x (y + z) = 0? (1) |y + z| = |y| + |z| (2) |x +

  post reply Question banks Downloads My Bookmarks Reviews Important topics  

Events & Promotions

PREV
NEXT


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| 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®.