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

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

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New post 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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Re: PS : absolute numbers [#permalink]

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New post 24 Jul 2008, 11:22
I'd say, A--no.
I'd like to see the OE though.

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Re: PS : absolute numbers [#permalink]

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New post 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..

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Re: PS : absolute numbers [#permalink]

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New post 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

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Re: PS : absolute numbers [#permalink]

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New post 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|

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Re: PS : absolute numbers [#permalink]

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New post 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.

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Re: PS : absolute numbers [#permalink]

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New post 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

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Re: PS : absolute numbers [#permalink]

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New post 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

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Re: PS : absolute numbers [#permalink]

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New post 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...

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Re: PS : absolute numbers [#permalink]

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New post 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.

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Re: PS : absolute numbers [#permalink]

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New post 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!

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Re: PS : absolute numbers [#permalink]

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New post 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)

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Re: PS : absolute numbers [#permalink]

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New post 25 Jul 2008, 07:35
    OA says C, no explanation provided :(

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    Re: PS : absolute numbers [#permalink]

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


      OA is obviously incorrect.

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      Re: PS : absolute numbers [#permalink]

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      New post 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

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      Re: PS : absolute numbers   [#permalink] 25 Jul 2008, 07:43
      Display posts from previous: Sort by

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

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