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 :
Answer & explanation
Correct answer: option 4
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.