Absolution!

[bhumika.k.shah]

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

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

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

Source : MGMAT CAT

[thephoenix]

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

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

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

Source : MGMAT CAT

IMO C

we need to prove x(y+z)>0----->x(y+z)=+ve number
it can happen if bth x and (y+z) are +ve or >0 OR bth x and (y+z) are -ve or <0 at same point of time
s1)----->either x and y are +ve or bth -ve-----> if x is >0--->y>0 or else x<0---->y<0
insuff
s2) same if z>0,y>0 or if z<0,y<0
insuff
combining
if x>0--->y>0---->z>0---->prodct>0
if x<0--->y<0---->z<0----->prod>0
hence suff

[bhumika.k.shah]

YEa but how ?

Could you pick #s and explain ?

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

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

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

Source : MGMAT CAT

IMO C

[thephoenix]

YEa but how ?

Could you pick #s and explain ?

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

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

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

Source : MGMAT CAT

IMO C

just take simple no's x=1 y=2 and z=3
s1)1+2=3 for bth side
but if u take one -ve and one positive value at same point of time then s1 does not holds
say x=-1 and y=2
then LHS=1 and RHS =3
in order to satisfy this statement bth x and y has to be of same sign
same is with s2)

[suryapal]

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

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

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

Source : MGMAT CAT

statement 1 : insufficient
according to statement we can conclude that both x and y are of the same sign .. either -ve or +ve i.e if the x and y are of different signs then scondtion given in statement 1 can not be satisfies ( pick some number and try )
but it don't say anything about z , so by taking different values of z ... (x) (y+z) could be greater than or less than zero ( pick some no. and try )
eg.
(2) (3-4) < 0
and
(2) (3+4) >0

similarly statement 2 is insufficient

but both statement states that x,y and z are of same sign .. either -ve or +ve

so if all positive then
(2) (3 +4) > 0

and if all negative
(-2) (-3-4) >0

so taking both statement together answer is always yes...

so the answer is C