Question

Consider the following statements
Statement I The solution set of the inequality $\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}$ is $(-\infty, 2)$.
Statement II The solution set of the inequality $\frac{1}{2}\left(\frac{3 x}{5}+4\right) \geq \frac{1}{3}(x-6)$ 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} \leq \frac{5(2-x)}{3} $
$ \Rightarrow \frac{3 x-6}{5} \leq \frac{10-5 x}{3} $
$ \Rightarrow 9 x-18 \leq 50-25 x$
Transferring the terms $(-25 x)$ to LHS and the term ( $-18)$ to $RHS$,
$9 x+25 x \leq 50+18$
$\Rightarrow 34 x \leq 68 $
$\Rightarrow x \leq \frac{68}{34} \Rightarrow x \leq 2$

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