Management Aptitude Test Dec 2013 Paper

(d ) Here, radius of cylindrical tub = 12 cm
Depth of cylindrical tub = 20 cm
Let the radius of spherical iron ball be R cm.
According to the question,
π x ${\left(12\right)}^2$ x 20 + $\frac{4}{3}\mathit{\pi }{\left(\mathit{R}\right)}^3$ = π ${\left(12\right)}^2$ x (26.75)
⇒ $\frac{4}{3}{\left(\mathit{R}\right)}^3$ = ${\left(12\right)}^2$ x (26.75) - ${\left(12\right)}^2$ x 20
⇒ ${\mathit{R}}^3=\frac{144\times 6.75\times 3}{4}$
⇒ ${\mathit{R}}^3$ = 729
∴ R = 9 cm