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If \(\frac{2 x+3}{5} < \frac{4 x-1}{2}\), then \(x\) lies in the interval
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The correct answer is:
\(\left(\frac{11}{16}, \infty\right)\)
\(\begin{aligned}
& \frac{2 x+3}{5} < \frac{4 x-1}{2} \Rightarrow-16 x < -11 \\
& \Rightarrow 16 x > 11 \Rightarrow x > \frac{11}{16} \\
& \text { Hence, } x \in\left(\frac{11}{16}, \infty\right)
\end{aligned}\)
& \frac{2 x+3}{5} < \frac{4 x-1}{2} \Rightarrow-16 x < -11 \\
& \Rightarrow 16 x > 11 \Rightarrow x > \frac{11}{16} \\
& \text { Hence, } x \in\left(\frac{11}{16}, \infty\right)
\end{aligned}\)
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