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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]  08 Jan 2011, 13:42
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If xyz ≠ 0, is x(y + z) >= 0?

(1) |y + z| = |y| + |z|
(2) |x + y| = |x| + |y|
[Reveal] Spoiler: OA

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Ajit

Last edited by Bunuel on 29 Apr 2013, 04:03, edited 1 time in total.
Renamed the topic and edited the question.
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Re: absolute values x (y + z) [#permalink]  08 Jan 2011, 14:39
Expert's post
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: absolute values x (y + z) [#permalink]  09 Jan 2011, 09:59
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: absolute values x (y + z) [#permalink]  09 Jan 2011, 10:09
Expert's post
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: absolute values x (y + z) [#permalink]  12 Jan 2011, 16:49
yes the answer is A as
from question stem we see none of the x,y,z are 0
for option A to be true y and z both have to be either positive or either negative,
[with x <> 0 information given in question] we always get x(y+z) as a non zero number.

Option A
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Re: If x, y, and z are nonzero numbers, is (x)(y + z) > 0? [#permalink]  27 Feb 2011, 17:37
Expert's post

dc123 wrote:
If x, y, and z are nonzero numbers, is (x)(y + z) > 0?
(1) |x + y| = |x| + |y|

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

(1) INSUFFICIENT: We learn from this statement that x and y have the same sign. They are either both positive or both negative. (To prove this try choosing values with opposite signs for x and y, for example 3 and -4.) This assures us that xy > 0, however since we know nothing about the sign of z, we can't answer the question.

(2) INSUFFICIENT: We learn from this statement that y and z have the same sign. They are either both positive or both negative. This assures us that yz > 0. However, we know nothing about the sign of x, so we can't answer the question.

so can someone prove this for me, not sure xy>1 in the in the end.

Thanks

Also:
 ! Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/No posting of PS/DS questions is allowed in the main Math forum.

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Re: If xyz ≠ 0, is x(y + z) >= 0? [#permalink]  11 Dec 2013, 06:32
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]  11 Dec 2013, 06:56
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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]  11 Dec 2013, 07:21
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: absolute values x (y + z) [#permalink]  11 Feb 2014, 05:18
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]  26 Feb 2014, 19:03
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.
Re: If xyz ≠ 0, is x(y + z) >= 0?   [#permalink] 26 Feb 2014, 19:03
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