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

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Manager
Joined: 14 Jun 2008
Posts: 162
If xyz 0, is x (y + z) = 0? (1) |y + z| = |y| + |z| (2) |x [#permalink]

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24 Jul 2008, 11:15
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If xyz ≠ 0, is x (y + z) = 0?

(1) |y + z| = |y| + |z|
(2) |x + y| = |x| + |y|

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Current Student
Joined: 31 Aug 2007
Posts: 365
Re: PS : absolute numbers [#permalink]

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24 Jul 2008, 11:22
I'd say, A--no.
I'd like to see the OE though.
Current Student
Joined: 28 Dec 2004
Posts: 3309
Location: New York City
Schools: Wharton'11 HBS'12
Re: PS : absolute numbers [#permalink]

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24 Jul 2008, 11:38
i get A too..but what should i know..i am just a lousy 42er on Q..

rephrase the stem its asking is y=-z?

we know that xyz are not ZERO

1) if y and z are not zero then

|y+z|=|y|+|z| only if y=z..

therefore sufficient x(y-z) is not equal to zero..
Manager
Joined: 14 Jun 2008
Posts: 162
Re: PS : absolute numbers [#permalink]

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24 Jul 2008, 11:56
OA sayz C, and I disagree with it.

i got A as well.

same logic

for 1) to be true, either both y and z negative, or both +ive

thanx
SVP
Joined: 30 Apr 2008
Posts: 1850
Location: Oklahoma City
Schools: Hard Knocks
Re: PS : absolute numbers [#permalink]

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24 Jul 2008, 12:24
I'm not sure about C (haven't gotten there in my analysis yet, but I know it's not A, here is what i think:

Without the absolute value, y + z = 0 would have to mean that |y| = |z|. Say y = 3 and z = -3. Then |3-3|= |0|, but |3| + |-3| = 6. One way #1 can be true is if y or z = 0. The only way x(y+z) = 0 is if x or (y+z) = 0. We know x ≠ 0 because xyz =0 would be true but the stems says otherwise. The only way #1 can be true and we can answer that x(y+z) = 0 [implied is "all the time"], both y and z must = 0

Because we can find multiple situations that allow |y + z| = |y| + |z| and some of those situations make (y+z) = 0 and others do not, A is insufficient.

sset009 wrote:
If xyz ≠ 0, is x (y + z) = 0?

(1) |y + z| = |y| + |z|
(2) |x + y| = |x| + |y|

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a\$\$.

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Manager
Joined: 10 Apr 2008
Posts: 53
Re: PS : absolute numbers [#permalink]

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24 Jul 2008, 18:17
jallenmorris wrote:
I'm not sure about C (haven't gotten there in my analysis yet, but I know it's not A, here is what i think:

Without the absolute value, y + z = 0 would have to mean that |y| = |z|. Say y = 3 and z = -3. Then |3-3|= |0|, but |3| + |-3| = 6. One way #1 can be true is if y or z = 0. The only way x(y+z) = 0 is if x or (y+z) = 0. We know x ≠ 0 because xyz =0 would be true but the stems says otherwise. The only way #1 can be true and we can answer that x(y+z) = 0 [implied is "all the time"], both y and z must = 0

Because we can find multiple situations that allow |y + z| = |y| + |z| and some of those situations make (y+z) = 0 and others do not, A is insufficient.

sset009 wrote:
If xyz ≠ 0, is x (y + z) = 0?

(1) |y + z| = |y| + |z|
(2) |x + y| = |x| + |y|

Actually that is incorrect. The answer should be A.

xyz != 0. This means x!=0, y!=0 AND z!=0 (!= ---> not equal)

#1 |y+z| = |y| + |z|
This is true only if y = z OR y = 0 OR z = 0. We know the latter two are not true.
If y = -z, then there is an inequality. Try it with numbers.

So we know from #1 that y!=-z, therefore y+z !=0. So A says no.
So answer has to be A or D

#2 |x+y| = |x| + |y|
use the same logic above to find that x = y. This doesn't tell us anything about the original question and is irrelevant.

A.
Senior Manager
Joined: 19 Mar 2008
Posts: 350
Re: PS : absolute numbers [#permalink]

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24 Jul 2008, 21:26
Given xyz ≠ 0, so, x ≠ 0, y ≠ 0, z ≠ 0

The question asks if x(y+z) = 0?
As x cannot be 0
We have to see whether y+z = or ≠ 0

(1)
if y and z are of different signs (+ and -)

I y + z I ≠ |y| + |z|

So (1) only tells us that y and z are of same sign

Nothing can proof y + z = 0

So (1) alone is Not sufficient

(2) , similar to the above

it tells x and y of same sign.

So (2) alone is Not sufficient

(1) + (2) also not sufficient
Senior Manager
Joined: 06 Apr 2008
Posts: 401
Re: PS : absolute numbers [#permalink]

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24 Jul 2008, 21:38
thinkblue wrote:
jallenmorris wrote:
I'm not sure about C (haven't gotten there in my analysis yet, but I know it's not A, here is what i think:

Without the absolute value, y + z = 0 would have to mean that |y| = |z|. Say y = 3 and z = -3. Then |3-3|= |0|, but |3| + |-3| = 6. One way #1 can be true is if y or z = 0. The only way x(y+z) = 0 is if x or (y+z) = 0. We know x ≠ 0 because xyz =0 would be true but the stems says otherwise. The only way #1 can be true and we can answer that x(y+z) = 0 [implied is "all the time"], both y and z must = 0

Because we can find multiple situations that allow |y + z| = |y| + |z| and some of those situations make (y+z) = 0 and others do not, A is insufficient.

sset009 wrote:
If xyz ≠ 0, is x (y + z) = 0?

(1) |y + z| = |y| + |z|
(2) |x + y| = |x| + |y|

Actually that is incorrect. The answer should be A.

xyz != 0. This means x!=0, y!=0 AND z!=0 (!= ---> not equal)

#1 |y+z| = |y| + |z|
This is true only if y = z OR y = 0 OR z = 0. We know the latter two are not true.
If y = -z, then there is an inequality. Try it with numbers.

So we know from #1 that y!=-z, therefore y+z !=0. So A says no.
So answer has to be A or D

#2 |x+y| = |x| + |y|
use the same logic above to find that x = y. This doesn't tell us anything about the original question and is irrelevant.

A.

Hold on

|y + z| = |y| + |z|

If I chose y=7 and z=8 then above is true so statement "This is true only if y = z OR y = 0 OR z = 0" is wrong

Coming to Q answer should be E)

If I chose x=6, y=7 and z=8 then none of conditions satisfy even if you join both the statements
Intern
Joined: 15 Apr 2008
Posts: 38
Re: PS : absolute numbers [#permalink]

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24 Jul 2008, 23:13
I got E as well. I will post my analysis later after i get home from work. Excited to see the OA.. pls, sset009...
Manager
Joined: 10 Apr 2008
Posts: 53
Re: PS : absolute numbers [#permalink]

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25 Jul 2008, 01:02
@judokan and mmohindru

Yes |y+z| = |y| + |z| is true for all positive numbers and 0.
HOWEVER, the question isn't asking if y,z>=0
It is asking if y = -z.

This CANNOT true be with #1.
Basically A is saying (conclusively) that y != -z AND y,z !=0

If y != -z then Y+z !=0

Therefore #1 gives us sufficient information to answer the original question.
The answer does not have to be yes for the condition to be sufficient.
Senior Manager
Joined: 06 Apr 2008
Posts: 401
Re: PS : absolute numbers [#permalink]

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25 Jul 2008, 02:00
thinkblue wrote:
@judokan and mmohindru

Yes |y+z| = |y| + |z| is true for all positive numbers and 0.
HOWEVER, the question isn't asking if y,z>=0
It is asking if y = -z.

This CANNOT true be with #1.
Basically A is saying (conclusively) that y != -z AND y,z !=0

If y != -z then Y+z !=0

Therefore #1 gives us sufficient information to answer the original question.
The answer does not have to be yes for the condition to be sufficient.

Got it thinkblue thanks for the catch!
Current Student
Joined: 12 Jun 2008
Posts: 287
Schools: INSEAD Class of July '10
Re: PS : absolute numbers [#permalink]

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25 Jul 2008, 02:32
judokan wrote:
Given xyz ≠ 0, so, x ≠ 0, y ≠ 0, z ≠ 0

The question asks if x(y+z) = 0?
As x cannot be 0
We have to see whether y+z = or ≠ 0

(1)
if y and z are of different signs (+ and -)

I y + z I ≠ |y| + |z|

So (1) only tells us that y and z are of same sign

Nothing can proof y + z = 0

So (1) alone is Not sufficient

Look at the two blue bold parts in your message above.

y and z are of same sign, y≠0, z≠0

It is enough to prove that y+z≠0

If y+z=0, then y=-z : either y=z=0 (this is not the case here) or y and z are of different sign (that is not the case here) ==> y+z cannot be equal to 0

And since x cannot be equal to zero as well:

==> Answer is (A)
Manager
Joined: 14 Jun 2008
Posts: 162
Re: PS : absolute numbers [#permalink]

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25 Jul 2008, 07:35
OA says C, no explanation provided
Director
Joined: 01 Jan 2008
Posts: 604
Re: PS : absolute numbers [#permalink]

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25 Jul 2008, 07:39
sset009 wrote:
OA says C, no explanation provided

OA is obviously incorrect.
Manager
Joined: 14 Jun 2008
Posts: 162
Re: PS : absolute numbers [#permalink]

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25 Jul 2008, 07:43
jallenmorris wrote:

Because we can find multiple situations that allow |y + z| = |y| + |z| and some of those situations make (y+z) = 0 and others do not, A is insufficient.

the only way that the above statement is true if both are negative, or both are positive.
e.g y = -3, z = -2
|-5| = |-3| + |-2|

or if y = 3, z = 2
|5| = |3| + |2|

addition of two negative or positive numbers will take it furhter away from zero.
if the are different signs, the above equality will not hold true.
thereefore y+z 0

and we already know that

--== 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: PS : absolute numbers   [#permalink] 25 Jul 2008, 07:43
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