Management Aptitude Test Feb 2012 Paper

(3) Here, BD = 6 cm

DC = 8 cm
∴ BC = $\sqrt{\left(6\right)2+{\left(8\right)}^2}$ = $\sqrt{36+64}$ = 10 cm
In ΔACD and ΔEOC
∠ADC =∠OEC = 90°
∠ACD = ∠OCE (common angle)
∠CAD = ∠EOC (remaining angle)
∴ ΔACD ~ ΔEOC
Also, AD = AE = 6 cm
(∵ the length of two tangents drawn from an external point to circle are equal)
∴ EC = AC - AE = 10 - 6 = 4 cm
In similar ΔACD and ΔEOC
$\frac{\mathit{D}\mathit{C}}{\mathit{A}\mathit{D}}=\frac{\mathit{E}\mathit{C}}{\mathit{O}\mathit{E}}$
⇒ $\mathit{O}\mathit{E}=\frac{\mathit{A}\mathit{D}\times \mathit{E}\mathit{C}}{\mathit{D}\mathit{C}}$ = $\frac{6\mathit{x}4}{8}$ = 3 cm
∴ Required of sphere = 3 cm
Now, volume of cone = $\frac{1}{3}\mathit{\pi }{\mathit{r}}^2\mathit{h}$ = $\frac{1}{3}$ x π x 36 x 8
And volume of sphere = $\frac{4}{3}\mathit{\pi }{\mathit{r}}^3$ = $\frac{4}{3}$ x π x 27
∴ Required fraction of water
= $\frac{\frac{4}{3}\times \mathit{\pi }\times 27}{\frac{1}{3}\times \mathit{\pi }\times 36\times 8}$ = $\frac{4\times 27}{36\times 8}=\frac{3}{8}$ = 3: 8