If the radius of a sphere is increased by 16\(\frac{2}{3}\)%, its volume is increased by how much percent?

58.7%

16\(\frac{2}{3}\)% = \(\frac{1}{6}\)

New radius(R) = \(\frac{7}{6}\)  x old radius (r)

Volume  = \(\frac{4}{3}\) \(\pi \) R³

             = \(\frac{4}{3}\) \(\pi \) (\(\frac{7}{6}\) r)³

             = \(\frac{343}{216}\) × \(\frac{4}{3}\) \(\pi \) r³

             = 1.587 × original volume

⇒ Increase percent = 58.7%