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# If xyz ≠ 0, is x(y + z) >= 0?

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If xyz ≠ 0, is x(y + z) >= 0?  [#permalink]

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28 Oct 2010, 22:37
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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: If xyz ≠ 0, is x(y + z) >= 0?  [#permalink]

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08 Jan 2011, 15:39
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ajit257 wrote:
If xyz ≠ 0, is x (y + z) = 0?

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

Can some explain the concept of absolute values

I think the question should be:

If xyz ≠ 0, is x (y + z) >= 0?

xyz ≠ 0 means that neither of unknowns is 0.

(1) |y + z| = |y| + |z| --> either both $$y$$ and $$z$$ are positive or both are negative, because if they have opposite signs then $$|y+z|$$ will be less than $$|y|+|z|$$ (|-3+1|<|-3|+1). Not sufficient, as no info about $$x$$.

(2) |x + y| = |x| + |y| --> the same here: either both $$x$$ and $$y$$ are positive or both are negative. Not sufficient, as no info about $$z$$.

(1)+(2) Either all three are positive or all three are negative --> but in both cases the product will be positive: $$x(y+z)=positive*(positive+positive)=positive>0$$ and $$x(y+z)=negative*(negative+negative)=negative*negative=positive>0$$. Sufficient.

For theory on absolute values check: math-absolute-value-modulus-86462.html
For practice check: search.php?search_id=tag&tag_id=37 (DS), search.php?search_id=tag&tag_id=58 (PS), inequality-and-absolute-value-questions-from-my-collection-86939.html (700+ DS).

Hope it helps.
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Re: If xyz ≠ 0, is x(y + z) >= 0?  [#permalink]

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29 Oct 2010, 00:15
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Note that |a+b|=|a|+|b| means that the sign of a & b is the same ... either they are both positive or both negative.

In this question, we are given xyz is not 0, hence none of those numbers can be 0.

(1) sign(y)=sign(z) ... Insufficient ... the given product can still be positive of negative (take y=z=4, x=1 and x=-1)
(2) sign(x)=sign(y) ... Insufficient ... again the product may have eithe sign (take x=1, y=1 ... then take z=1 or z=-2)

(1+2) sign(x)=sign(y)=sign(z) ... whether they are all positive or all negative, the given product will always be positive ... Sufficient

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Re: If xyz ≠ 0, is x(y + z) >= 0?  [#permalink]

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Updated on: 29 Apr 2013, 05:03
1
If xyz ≠ 0, is x(y + z) >= 0?

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

Originally posted by ajit257 on 08 Jan 2011, 14:42.
Last edited by Bunuel on 29 Apr 2013, 05:03, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If xyz ≠ 0, is x(y + z) >= 0?  [#permalink]

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09 Jan 2011, 10:59
2
Thanks Brunuel but as written, wouldnt the answer be A?

If X(Y+Z)=0? and X does not equal 0, this is true only if Y= -Z and this is true only if either:

Y is negative and Z is possitive or Z is positive and Y is negative

If |Y+Z| = |y| + |Z|, if either of the two cases is true, |Y+Z| < |Y| + |Z| so B would be sufficient.

If |X+Y| = |X| + |Y|, this is true only if X and Y are both positive. This does not supply information about Z so insufficient. For instance, X and Y and Z can all be possiive which makes the question true or X and Y can be possitive and Z negative which makes it false.
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Re: If xyz ≠ 0, is x(y + z) >= 0?  [#permalink]

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09 Jan 2011, 11:09
2
abmyers wrote:
Thanks Brunuel but as written, wouldnt the answer be A?

If X(Y+Z)=0? and X does not equal 0, this is true only if Y= -Z and this is true only if either:

Y is negative and Z is possitive or Z is positive and Y is negative

If |Y+Z| = |y| + |Z|, if either of the two cases is true, |Y+Z| < |Y| + |Z| so B would be sufficient.

If |X+Y| = |X| + |Y|, this is true only if X and Y are both positive. This does not supply information about Z so insufficient. For instance, X and Y and Z can all be possiive which makes the question true or X and Y can be possitive and Z negative which makes it false.

Yes, in it's current form the answer is A:

If xyz ≠ 0, is x (y + z)=0?

xyz ≠ 0 means that neither of unknowns is 0, so as $$x\neq{0}$$ then $$x(y + z)=0$$ is true only if $$y+z=0$$. So the question is whether $$y+z=0$$ is true.

(1) |y + z| = |y| + |z| --> if $$y+z=0$$ is true then $$LHS=|y+z|=0$$ and in order RHS to equal to zero both $$y$$ and $$z$$ must be zero, but we are given that neither of unknowns is 0, so $$y+z\neq{0}$$. Sufficient.

(2) |x + y| = |x| + |y| --> insufficient, as no info about $$z$$.

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Re: If xyz ≠ 0, is x(y + z) >= 0?  [#permalink]

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11 Dec 2013, 07:32
I am bit confused here with x(y + z) >= 0? Greater equal to 0.

I know x(y + z) will always be greater than Zero, combining both options, but mind is saying as x(y + z) can not be equal to zero, in that case should I say x(y + z) is not >= 0 and answer is C as we concluded bcz it will always be greater than zero, but not >= to 0.

Thanks
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Re: If xyz ≠ 0, is x(y + z) >= 0?  [#permalink]

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11 Dec 2013, 07:56
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PiyushK wrote:
I am bit confused here with x(y + z) >= 0? Greater equal to 0.

I know x(y + z) will always be greater than Zero, combining both options, but mind is saying as x(y + z) can not be equal to zero, in that case should I say x(y + z) is not >= 0 and answer is C as we concluded bcz it will always be greater than zero, but not >= to 0.

Thanks

$$\geq{0}$$ translates to: more than or equal to 0. How can a number simultaneously be more than 0 AND equal to it?

The question asks is x(y + z) more than or equal to 0. We get that it's more than 0, thus we have an answer to our question.

Hope it's clear.
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Re: If xyz ≠ 0, is x(y + z) >= 0?  [#permalink]

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11 Dec 2013, 08:21
Got it Bunuel, actually I was thinking too much about the range of solution.

(super set)
{0,1,2,3,10000 infinity } solutions hold true for >=0
(subset)
{1,2,3,1000 infinity } each element also hold true for >=0, even if we don't have 0 in solution set, e.g 1>=0 (yes)
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Re: If xyz ≠ 0, is x(y + z) >= 0?  [#permalink]

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11 Feb 2014, 06:18
Bunuel wrote:
abmyers wrote:
Thanks Brunuel but as written, wouldnt the answer be A?

If X(Y+Z)=0? and X does not equal 0, this is true only if Y= -Z and this is true only if either:

Y is negative and Z is possitive or Z is positive and Y is negative

If |Y+Z| = |y| + |Z|, if either of the two cases is true, |Y+Z| < |Y| + |Z| so B would be sufficient.

If |X+Y| = |X| + |Y|, this is true only if X and Y are both positive. This does not supply information about Z so insufficient. For instance, X and Y and Z can all be possiive which makes the question true or X and Y can be possitive and Z negative which makes it false.

Yes, in it's current form the answer is A:

If xyz ≠ 0, is x (y + z)=0?

xyz ≠ 0 means that neither of unknowns is 0, so as $$x\neq{0}$$ then $$x(y + z)=0$$ is true only if $$y+z=0$$. So the question is whether $$y+z=0$$ is true.

(1) |y + z| = |y| + |z| --> if $$y+z=0$$ is true then $$LHS=|y+z|=0$$ and in order RHS to equal to zero both $$y$$ and $$z$$ must be zero, but we are given that neither of unknowns is 0, so $$y+z\neq{0}$$. Sufficient.

(2) |x + y| = |x| + |y| --> insufficient, as no info about $$z$$.

Hi Bunuel is this approach valid?

By property Is |x - y| >= |x| - |y|, and they are only equal when both x and y have the same signs.
In statement 1, we only know that z and y have the same signs but no info on x. Insuff. In Statement 2, we have that 'x' and 'y' have the same sign but no info on 'z'. Both together we know that all of them have the same sign. Thus the answer will always be >0. Sufficient. C
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Re: If xyz ≠ 0, is x(y + z) >= 0?  [#permalink]

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26 Feb 2014, 20:03
hi, is it possible to get to final answer using the following approach?

xy + xz >= 0

if x>0, then y >= -z
if x<0, then y <= -z

is this hypothesis of any use or is it too detailed for this problem? just wondering b/c I usually solve these type of problems using this approach but I was stuck on this one even after 4 mins.
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Re: If xyz ≠ 0, is x(y + z) >= 0?  [#permalink]

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26 Jun 2014, 04:50
What did I miss here? This was my first question I attempted to answer on this forum - looks like a great practice forum - but where / how do you see what the ABCDE multiple choice options are? I didn't see a drop down anywhere.... I think I missed something.....I tried to click on a letter to see the offered answer, but it selected it for me! Where do I view the possible answers assigned to the letters? Thanks in advance!
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Re: If xyz ≠ 0, is x(y + z) >= 0?  [#permalink]

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26 Jun 2014, 07:21
jamielinderart wrote:
What did I miss here? This was my first question I attempted to answer on this forum - looks like a great practice forum - but where / how do you see what the ABCDE multiple choice options are? I didn't see a drop down anywhere.... I think I missed something.....I tried to click on a letter to see the offered answer, but it selected it for me! Where do I view the possible answers assigned to the letters? Thanks in advance!

This is a data sufficiency question. Options for DS questions are always the same.

The data sufficiency problem consists of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the statements, plus your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise), you must indicate whether—

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

I suggest you to go through the following post ALL YOU NEED FOR QUANT.

Hope this helps.
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Re: If xyz ≠ 0, is x(y + z) >= 0?  [#permalink]

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27 Apr 2015, 06:20
Easiest to attack this question without writing anything down.

Stem: if xyz does not equal 0 then you know neither of the variables can equal zero.

Solve: is x(y+z) >= 0?

St (1) tells us y and z have the same signs. Whether they are both negative or both positive. NS.

St(2) tells up x and y have the same signs. Whether they are both negative or both positive. NS.

St(1) & St(2) provide sufficient information to answer the question. pos *pos*pos >= 0. neg*neg*neg>=0. Suff
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Re: If xyz ≠ 0, x (y + z) ≥ 0?  [#permalink]

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20 May 2015, 18:26
If xyz ≠ 0, x (y + z) ≥ 0?

(1) | y + z | = |y| + |z| insufficient as we do not know about x.

(2) | x + y | = |x| + |y|
insufficient as we do not know z.

Together We know x,y,z all will have same sign (- or +) .

So
x (y + z) ≥ 0

anyone knows why we need xyz ≠ 0?
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Re: If xyz ≠ 0, is x(y + z) >= 0?  [#permalink]

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24 Jun 2019, 17:07
monirjewel wrote:
If xyz ≠ 0, is x(y + z) >= 0?

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

Are x & (y+z) same sign?

1) y & z are same sign (Using Property 1). No info about x. Not sufficient.
2) x & y are same sign (Using Property 1). No info about y. Not sufficient.

1+2) x & y & z are all same sign.
+ (+ + +) => +
- (- + -) => +
Sufficient
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Re: If xyz ≠ 0, is x(y + z) >= 0?   [#permalink] 24 Jun 2019, 17:07
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