Question

Consider the following statements
Statement I The solution set of the inequality $\frac{3(x-2)}{5}\le \frac{5(2-x)}{3}$ is $(-\infty ,2)$.
Statement II The solution set of the inequality $\frac{1}{2}\left(\frac{3x}{5}+4\right)\ge \frac{1}{3}(x-6)$ is $(-\infty ,120]$.
Choose the correct option.

• A. Statement I is true
• B. Statement II is true
• C. Both are true
• D. Both are false

Answer 1

Answer: (B)

I. We have, $\frac{3(x-2)}{5}\le \frac{5(2-x)}{3}$
$\Rightarrow \frac{3x-6}{5}\le \frac{10-5x}{3}$
$\Rightarrow 9x-18\le 50-25x$
Transferring the terms $(-25x)$ to LHS and the term ( $-18)$ to $RHS$,
$9x+25x\le 50+18$
$\Rightarrow 34x\le 68$
$\Rightarrow x\le \frac{68}{34}\Rightarrow x\le 2$

$\therefore$ Solution set is $(-\infty ,2]$.
II. We have, $\frac{1}{2}\left(\frac{3x}{5}+4\right)\ge \frac{1}{3}(x-6)$
$\Rightarrow \frac{1}{2}\left(\frac{3x}{5}+\frac{4}{1}\right)\ge \frac{1}{3}(x-6)$
Taking LCM in LHS,
$\frac{1}{2}\left(\frac{3x+20}{5}\right)\ge \frac{1}{3}(x-6)$
$\Rightarrow \frac{3x+20}{10}\ge \frac{x-6}{3}$
$\Rightarrow 3(3x+20)\ge 10(x-6)$
$\Rightarrow 9x+60\ge 10x-60$
Transferring the term $10x$ to LHS and the term 60 to RHS,
$9x-10x\ge -60-60\Rightarrow -x\ge -120$
Multiplying both sides by $-1$,
$x\le 120$