If the radius of a sphere is increased by 16\(\frac{2}{3}\)%, its volume is increased by how much percent?
Answer & explanation
Correct answer: option 2
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%