If xyz ≠ 0, is x(y + z) = 0?

Question

If xyz ≠ 0, is x(y + z) >= 0?

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

shrouded1 · Answer

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

Answer is (c)

Bunuel · Answer

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. Answer: C. 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.

abmyers · Answer

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.

Bunuel · Answer

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
eq{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
eq{0} . Sufficient.

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

Answer: A.

PiyushK · Answer

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

Bunuel · Answer

\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.

saurabhbajpai · Answer

Can't we change
xy+xz>=0 to xy>=-xz to y>=-z. In that case statement one should be sufficient.

Posted from my mobile device

Bunuel · Answer

No, we cannot simply divide by x because we don't know its sign. If x is positive, then yes, we would get y ≥ -z. However, if x is negative, dividing by a negative value would flip the inequality sign, resulting in y ≤ -z. Remember, never reduce an inequality by a variable if its sign is unknown.

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If xyz ≠ 0, is x(y + z) >= 0?

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