The slant height of a cone is increased by 10% and its radius is doubled. What is the percentage increase in the surface area of the cone?

Let the original radius of the cone be r and its slant height be l.

∴ The original surface area of the cone = πrl

New radius of the cone = 2r

New slant height of the cone = $\mathrm{l}+\frac{10}{100}\mathrm{l}=\frac{11}{10}\mathrm{l}$ 

∴ New surface area of the cone = $\mathrm{\pi }\left(2\mathrm{r}\right)\left(\frac{11}{10}\mathrm{l}\right)=\frac{11}{5}\mathrm{\pi rl}$ 

Percentage increase in the surface area of the cone 

$=\frac{{\displaystyle \frac{11}{5}\mathrm{\pi rl}-\mathrm{\pi rl}}}{\mathrm{\pi rl}}\times 100$

$=(\frac{11}{5}-1)\times 100$

$=120\%$

Thus, the percentage increase in the surface area of the cone is 120%.