Management Aptitude Test 2014 Paper

(a ) Let the filling capacity be x ${\mathit{m}}^3$/min.
So, the discharging capacity = (x + 10) ${\mathit{m}}^3$/min
Time taken to the tank = $\frac{4800}{\mathit{x}}$ mjn
and time taken to discharge the fuel from tank
= $\frac{4800}{\mathit{x}+10}$ min
Now, according to the given conditions
⇒ $\frac{4800}{\mathit{x}}-\frac{4800}{\mathit{x}+10}$ = 16
⇒ $\frac{4800\mathit{x}+48000-4800\mathit{x}}{\mathit{x}\left(\mathit{x}+10\right)}$ = 16
⇒ 48000 = 16 ( ${\mathit{x}}^2$ + 10x)
⇒ 3000 = ${\mathit{x}}^2$ + 10x
⇒ ${\mathit{x}}^2$ + 10x - 3000 = 0
⇒ ${\mathit{x}}^2$ + 60x -50x - 3000 = 0
⇒ x(x + 60) - 50(x+ 60) = 0
⇒ (x - 50) (x + 60) = 0
⇒ x = 50, -60
∴ x = 50, because x cannot b e negative.
∴ Filling capacity of pump = 50 m3/min