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
(1) |x + y| = |x| + |y|
(2) |z + y| = |y| + |z|
Source : MGMAT CAT
IMO Cbhumika.k.shah wrote: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
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
