WAEC Further Mathematics 2011 Paper

Concept:The volume of a sphere is proportional to the cube of its radius, so a small change in radius leads to a larger percentage change in volume.Explanation:Let the original radius be $r$ and the original volume be $V$.$V = \frac{4}{3}\pi r^{3}$.A $45\%$ increase in volume gives the new volume as $145\%$ of $V$, which is $1.45V$.Let the new radius be $R$.Then the new volume is also given by $\frac{4}{3}\pi R^{3}$.So, $\frac{4}{3}\pi R^{3} = 1.45V$.Substitute $V = \frac{4}{3}\pi r^{3}$:$\frac{4}{3}\pi R^{3} = 1.45 \times \frac{4}{3}\pi r^{3}$.Cancel $\frac{4}{3}\pi$ from both sides:$R^{3} = 1.45r^{3}$.Taking cube roots:$R = \sqrt[3]{1.45}\,r$.$\sqrt[3]{1.45} \approx 1.132$.Hence the new radius is about $113.2\%$ of the original radius.Therefore, the percentage increase in radius is $113.2\% - 100\% = 13.2\%$.Rounded to the nearest whole percent, this is approximately $13\%$.Answer:A. $13\%$