CAT 2022 Slot 3 Paper

Let $\frac{\mathit{x}}{\mathit{y}}$ be $\mathit{t}$
Therefore, $\mathit{c}=16\mathit{t}+\frac{49}{\mathit{t}}$
Applying $\mathrm{AM}>=\mathrm{GM}$
$\frac{\left(16\mathit{t}+\frac{49}{\mathit{t}}\right)}{2}\ge {\left(16\mathit{t}\times \frac{49}{\mathit{t}}\right)}^{\frac{1}{2}}$
$16\mathit{t}+\frac{49}{\mathit{t}}\ge 56$
When $\mathit{t}$ is positive then $\mathit{c}$ is greater than equal to $56$ .
When $\mathit{t}$ is negative then $\mathit{c}$ is less than equal to $-56$ .
Therefore $\mathit{c}\in \left(-\mathit{\infty },-56\right]\cup \left[56,\mathit{\infty }\right]$
As $-50$ is not in the range of $\mathit{c}$ so it is the answer