Xavier Aptitude Test 2011 Paper

(e) Let the ladder be placed in the position $\mathit{A}\mathit{C}$ first.
After drawing out the base of the ladder the new position of the ladder becomes $\mathit{E}\mathit{D}.$
According to the question $\mathrm{}\mathit{A}\mathit{D}=\mathit{x},\mathrm{}\mathit{E}\mathit{C}=2\mathit{x}$
$\mathrm{}\mathit{B}\mathit{C}=7,\mathit{A}\mathit{C}=25$
$\therefore \mathit{A}\mathit{B}=24$
In $\mathit{\Delta }\mathit{D}\mathit{B}\mathit{E},\mathrm{}{\mathit{D}\mathit{B}}^2+\mathrm{}{\mathit{B}\mathit{E}}^2=\mathrm{}{\mathit{D}\mathit{E}}^2$
$\Rightarrow {\left(24-\mathit{x}\right)}^2+{\left(7+2\mathit{x}\right)}^2={25}^2$
$\Rightarrow \mathrm{}{24}^2-48\mathit{x}+{\mathit{x}}^2+7^2+28\mathit{x}+4{\mathit{x}}^2={25}^2$
$\Rightarrow \mathrm{}5{\mathit{x}}^2=20\mathit{x}\Rightarrow \mathit{x}=4\left(\because \mathit{x}\ne 0\right)$
$\therefore \mathrm{}\mathrm{}\mathit{E}\mathit{C}=8$None of the option includes 8 in the interval.