If x varies inversely as (y² - 1) and x is equal to 24 when y = 10, then the value of x when y = 5 is?

Solution (By Examveda Team)

$$\eqalign{ & x \propto \frac{1}{{{y^2} - 1}}{\text{ }}\left( {{\text{Given}}} \right) \cr & x = k \times \frac{1}{{{y^2} - 1}}\left( {k{\text{ is constant}}} \right) \cr & {\text{Now }}x = 24{\text{ when }}y = 10{\text{ given}} \cr & \Rightarrow 24 = k \times \frac{1}{{{{\left( {10} \right)}^2} - 1}} \cr & \Rightarrow 24 = \frac{k}{{99}} \cr & \Rightarrow k = 24 \times 99 \cr & x = ? \cr & y = 5 \cr & \Rightarrow x = 24 \times 99 \times \frac{1}{{25 - 1}} \cr & \Rightarrow x = 24 \times 99 \times \frac{1}{{24}} \cr & \Rightarrow x = 99 \cr} $$