If x > y and z < 0, where $x, y, z \in Z$ and $x, y > 0,$ then which of the following are true ?

(A) $x.z> y.z$

(B) $\frac{x-y}{z}>\frac{y-x}{z}$

(C)$\frac{x+y}{z}>\frac{x-y}{z}$

(D) $x \div z < y \div z$

(E) $\frac{y-x}{z}< 0$

Choose the correct answer from the options given below :

(D) Only

The correct answer is Option 4: (D) Only

(A) $x.z> y.z$ :  Since z is negative, multiplying both sides of x > y by z reverses the inequality:  x·z < y·z . So, (A) is false.

(B) $\frac{x-y}{z}>\frac{y-x}{z}$ : 

 x − y > 0 , y − x < 0 and  Since z < 0

(x−y)/z = positive ÷ negative = negative
(y−x)/z = negative ÷ negative = positive

negative > positive → false.  Hence, (B) is false.

(C)$\frac{x+y}{z}>\frac{x-y}{z}$: 

Since x + y > x − y and z is negative, dividing by z reverses the inequality: (x+y)/z < (x−y)/z

So, (C) is false.

(D) $x \div z < y \div z$ :  Since z < 0, dividing x > y by z reverses the inequality: x/z < y/z

So, (D) is true.

(E) $\frac{y-x}{z}< 0$ :  Since y − x < 0 and z < 0: (y−x)/z > 0

So, (E) is false.