## x, y and z nonzero numbers - DS (>700)

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GMAT_ClubMBA
Bot GMAT
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### x, y and z nonzero numbers - DS (>700)

If x, y and z are nonzero numbers, is $x(y+z) > 0$ ?

(1) $|x+y| = |x|+|y|$

(1) $|z+y| = |y|+|z|$
respuesta: show

Luke
Mensajes: 258
Registrado: 26 Sep 2011, 18:49

### Re: x, y and z nonzero numbers - DS (>700)

(1) This expression means that both x and y are >0. In that case, the value of the product depends on the value of z.
If z>0, (y+z) can be >0;<0;=0, depending on the value of z vs value of y. As we don't know anything abut z (only z<>0), this is not sufficient.

(2) This expression means that both y and z are >0. In that case, the product will be positive if X>0 and will be negative if x<0. As we only know x<>0, this is not sufficient.

Together, we know x,y and z are >0, so the product is always positive.

C is the answer

lox
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Registrado: 13 Feb 2011, 15:57
Alma mater: Matemáticas, Universidad Complutense

### Re: x, y and z nonzero numbers - DS (>700)

|-3 + (- 5)| = |-3| + |-5| :P

fredurst
Becario
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Registrado: 05 Sep 2012, 14:12
Alma mater: Universidad de Sevilla

### Re: x, y and z nonzero numbers - DS (>700)

abs(x+y)=abs(x)+abs(y) if x,y are both neg o positive. But the answer to this questions stays the same. C