MCQ
Write the solution of inequality $\frac{1}{5}\bigg(\frac{3\text{x}}{5}+4\bigg)\geq\frac{1}{3}(\text{x}-6).$
- A$\text{x}\leq\frac{105}{8}$
- B$\text{x}\geq\frac{105}{8}$
- C$\text{x}\geq120$
- D$\text{x}\leq120$
Solution:
$\frac{1}{5}\big(\frac{3\text{x}}{5}+4\big)\geq\frac{1}{3}(\text{x}-6).$
$\Rightarrow3\big(\frac{3\text{x}}{5}+4\big)\geq5\big(\text{x}-6\big)$
$\Rightarrow\big(\frac{9\text{x}}{5}+12\big)\geq5\text{x}-6$
$\Rightarrow(30+12)\geq-\frac{9\text{x}}{5}+5\text{x}$
$\Rightarrow42\geq\frac{-9\text{x}+25\text{x}}{5}$
$\Rightarrow42\geq\frac{16\text{x}}{5}$
$\Rightarrow\frac{42\times5}{16}\geq\text{x}$
$\text{x}\leq\frac{105}{8}$
Therefore option (1) is the correct answere.
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