Question

If $\frac{2x+3}{5} < \frac{4x-1}{2}$, then $x$ lies in the interval

• A. $\left[0, \frac{11}{16}\right)$
• B. $\left[\frac{11}{16}, \infty\right)$
• C. $\left(0, \frac{11}{16}\right)$
• D. $\left(-\infty, \frac{11}{16}\right)$
• E. $\left(\frac{11}{16}, \infty\right)$

Answer 1

Answer: (E)

Given, $\frac{2 x+3}{5}<\frac{4 x-1}{2}$
$\Rightarrow \frac{4 x-1}{2}-\frac{2 x+3}{5}>0$
$\Rightarrow \frac{20 x-5-(4 x+6)}{10}>0$
$\Rightarrow \frac{16 x-11}{10}>0$
$\Rightarrow 16 x-11>0$
$\Rightarrow x>\frac{11}{16}$