Baten80 wrote:

(x+y)/Z>0 is x<0?

(1) x<y

(2) z<0

Sol:

\(\frac{x+y}{z} > 0\)

Means,

If \(z>0\), \(x+y>0\)

If \(z<0\), \(x+y<0\)

1. x<y

x=1, y=2, z=3;

\(\frac{x+y}{z}=\frac{1+2}{3}=1\) AND x>0

x=-2, y=-1, z=-3;

\(\frac{x+y}{z}=\frac{-1-2}{-3}=1\) BUT x<0

Not Sufficient.

2. z<0

Means;

\(x+y<0\)

x=-10; y=1; x+y= -9; x<0

x=1; y=-10; x+y=-9; x>0

Not Sufficient.

Combining both;

From stem we know,

If \(z<0\), \(x+y<0\)-----1

From St1, \(x<y\) OR \(x-y<0\)-------2

Adding 1 and 2;

\(2x<0\)

OR

\(x<0\)

Sufficient.

Ans: "C"

_________________

~fluke

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