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