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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 == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost 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.



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Re: PS : absolute numbers [#permalink]
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24 Jul 2008, 11:22
I'd say, Ano. I'd like to see the OE though.



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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(yz) is not equal to zero..



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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



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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 33= 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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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 33= 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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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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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 33= 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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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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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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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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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 0And 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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25 Jul 2008, 07:35
OA says C, no explanation provided



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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.



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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 repost 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
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