If $x > y$ and $z yz$ (B) $xz \geq yz$ (C) $\dfrac{x}{z} > \dfrac{y}{z}$ (D) $\dfrac{x}{z} < \dfrac{y}{z}$

> [!TLDR] > Dividing both sides of $x>y$ by the negative number $z$ reverses the inequality, so $\dfrac{x}{z} [!FORMULA] > If $a>b$ and $c y$ and $z [!ANSWER] > (D) $\dfrac{x}{z}<\dfrac{y}{z}$

If x > y and z < 0, then : (A) xz > yz (B) xz ≥ yz (C) — CBSE Class 12 Applied Mathematics 2024 PYQ

Question

x > y

CBSECBSE Class XII Board 2024MCQ· 1mImportance★★★★★est

✓ Free question

Dividing both sides of $x>y$ by the negative number $z$ reverses the inequality, so $\dfrac{x}{z}<\dfrac{y}{z}$ .

a>b

Given $x>y$ and $z<0$ .
Multiplying by $z<0$ reverses the sign: $xz<yz$ — so options (A) and (B) are wrong.
Dividing by $z<0$ also reverses the sign: $\dfrac{x}{z}<\dfrac{y}{z}$ — so (C) is wrong and (D) is correct.

✓Final answer

\dfrac{x}{z}<\dfrac{y}{z}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set 465/S/WXYZ/41 markMCQ
$\dfrac{|x+1|}{x+1} > 0, x \in R$
›Reveal solutionSolution
$\dfrac{|x+1|}{x+1}>0$ holds precisely where $x+1>0$ , i.e. $x>-1$ , excluding $x=-1$ where it is undefined.

$t \ne 0$

Let $t = x+1$ ; the expression is $\dfrac{|t|}{t}$ .

If $t>0$ , $|t| = t$ , so $\dfrac{|t|}{t} = 1 > 0$ . ✓

If $t<0$ , $|t| = -t$ , so $\dfrac{|t|}{t} = -1 < 0$ . ✗

At $t=0$ the fraction is undefined, so $x=-1$ is excluded.

Thus the inequality needs $t>0 \Rightarrow x+1>0 \Rightarrow x>-1$ , giving $x \in (-1,\infty)$ .

✓Final answer
$x \in (-1, \infty)$
CBSE 2025Set 465/S/WXYZ/41 markMCQ
$|3x - 2| \leq \dfrac{1}{2}$
›Reveal solutionSolution
The Assertion's solution set $[\tfrac12,\tfrac56]$ is correct, but the Reason misstates the modulus rule (it describes $\ge$ , not $\le$ ), so (C).

$|x-a|\le r \iff a-r \le x \le a+r$

Solve the Assertion: $|3x-2|\le \tfrac12 \iff -\tfrac12 \le 3x-2 \le \tfrac12$ .

Add 2 throughout: $\tfrac32 \le 3x \le \tfrac52$ .

Divide by 3: $\tfrac12 \le x \le \tfrac56$ , i.e. solution set $\left[\tfrac12,\tfrac56\right]$ — Assertion is true.

Examine the Reason: it claims $|x-a|\le r \Leftrightarrow x\le a-r \text{ or } x\ge a+r$ . That "or" form is the rule for $|x-a|\ge r$ , not $\le r$ . So the Reason is false.

Assertion true, Reason false $\Rightarrow$ option (C).

✓Final answer
$(C) Assertion (A) is true, but Reason (R) is false.$
CBSE 2024Set 465/RQPS/41 markMCQ
$x > y$
›Reveal solutionSolution
Dividing both sides of $x>y$ by the negative number $z$ reverses the inequality, so $\dfrac{x}{z}<\dfrac{y}{z}$ .

$a>b$

Given $x>y$ and $z<0$ .

Multiplying by $z<0$ reverses the sign: $xz<yz$ — so options (A) and (B) are wrong.

Dividing by $z<0$ also reverses the sign: $\dfrac{x}{z}<\dfrac{y}{z}$ — so (C) is wrong and (D) is correct.

✓Final answer
$\dfrac{x}{z}<\dfrac{y}{z}$
CBSE 2023Set 465/EF1GH/41 markMCQ
$\dfrac{|x+1|}{x+1} > 0$
(a) $x \in [-1, \infty)$
(b) $x \in (-1, \infty)$
(c) $x \in (-\infty, -1)$
(d) $x \in (-\infty, -1]$
›Reveal solutionSolution
$\dfrac{|x+1|}{x+1}>0$ forces $x+1>0$ , i.e. $x\in(-1,\infty)$ .

$t\neq 0$

Let $t=x+1$ . The expression $\dfrac{|t|}{t}>0$ holds only when $t>0$ .

So $x+1>0 \Rightarrow x>-1$ .

At $x=-1$ the expression is $\tfrac{0}{0}$ , undefined, so $-1$ is excluded — the endpoint is open.

Hence the solution set is the open interval $(-1,\infty)$ ; option (a) wrongly includes $-1$ .

✓Final answer
$x\in(-1,\infty)$

Q.If $x > y$ and $z < 0$ , then : (A) $xz > yz$ (B) $xz \geq yz$ (C) $\dfrac{x}{z} > \dfrac{y}{z}$ (D) $\dfrac{x}{z} < \dfrac{y}{z}$

Concept understanding — Linear and Modulus Inequalities

Q.If x>yx > yx>y and z<0z < 0z<0, then : (A) xz>yzxz > yzxz>yz (B) xz≥yzxz \geq yzxz≥yz (C) xz>yz\dfrac{x}{z} > \dfrac{y}{z}zx​>zy​ (D) xz<yz\dfrac{x}{z} < \dfrac{y}{z}zx​<zy​

🎓Unlock everything free for 14 days

Q.If $x > y$ and $z < 0$ , then : (A) $xz > yz$ (B) $xz \geq yz$ (C) $\dfrac{x}{z} > \dfrac{y}{z}$ (D) $\dfrac{x}{z} < \dfrac{y}{z}$

Unlock everything free for 14 days