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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 Oct 2008, 02:33
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D
E

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

I do not have an OA :(
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Re: DS : MODULUS [#permalink]

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New post 24 Oct 2008, 04:26
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.

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Re: DS : MODULUS [#permalink]

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New post 24 Oct 2008, 05:34
well, it is not A... I picked A too.

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Re: DS : MODULUS [#permalink]

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New post 24 Oct 2008, 06:38
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 ?
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Re: DS : MODULUS [#permalink]

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New post 24 Oct 2008, 06:44
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.

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Re: DS : MODULUS [#permalink]

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New post 24 Oct 2008, 08:17
I'll join in with the A, even though it's wrong.

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Re: DS : MODULUS [#permalink]

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New post 24 Oct 2008, 19:03
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

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Re: DS : MODULUS [#permalink]

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New post 24 Oct 2008, 19:42
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
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Re: DS : MODULUS [#permalink]

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New post 24 Oct 2008, 20:29
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:

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Re: DS : MODULUS [#permalink]

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New post 24 Oct 2008, 22:27
A is sufficient to prove that y and z have same sign.

A for me.

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Re: DS : MODULUS [#permalink]

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New post 25 Oct 2008, 03:09
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.

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Re: DS : MODULUS [#permalink]

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New post 25 Oct 2008, 18:46
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

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Re: DS : MODULUS [#permalink]

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New post 26 Oct 2008, 12:47
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.

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Re: DS : MODULUS [#permalink]

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New post 26 Oct 2008, 19:59
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 ....
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Re: DS : MODULUS [#permalink]

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New post 27 Oct 2008, 03:52
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

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Re: DS : MODULUS   [#permalink] 27 Oct 2008, 03:52
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