Management Aptitude Test 2015 Paper

(d) $\frac{\mathit{x}}{4}=\frac{3\mathit{y}}{5}=\frac{-\mathit{z}}{3}$ = $\frac{3\mathit{x}-6\mathit{y}-4\mathit{z}}{\mathit{k}}$
Taking first and second terms,
$\frac{\mathit{x}}{4}=\frac{3\mathit{y}}{5}$ ⇒ 5x = 12y ........(i)
Taking second and third terms,
$\frac{3\mathit{y}}{5}=\frac{-\mathit{z}}{3}$ ⇒ 9y = - 5z ..........(ii)
Taking second and last terms,
$\frac{3\mathit{y}}{5}$ = $\frac{3\mathit{x}-6\mathit{y}-4\mathit{z}}{\mathit{k}}$
On putting the values from Eqs. (i) and (ii), we get
$\frac{3\mathit{y}}{5}$ = $\frac{3\times \frac{12}{5}\mathit{y}-6\mathit{y}-4\times \frac{9}{5}\mathit{y}}{\mathit{k}}$
$\mathit{k}\times \frac{3\mathit{y}}{5}$ = $\frac{36}{5}\mathit{y}-6\mathit{y}-\frac{36}{5}\mathit{y}$
⇒ $\mathit{k}\times \frac{3\mathit{y}}{5}$ = $\frac{72}{5}\mathit{y}-6\mathit{y}$
⇒ $\frac{\mathit{k}\times 3\mathit{y}}{5}$ = $\frac{72\mathit{y}-30\mathit{y}}{5}$
$\mathit{k}\times \frac{3\mathit{y}}{5}=\frac{42}{5}\mathit{y}$
⇒ 3 x k = 42
∴ k = $\frac{42}{3}$ = 14