Discussion
If the radius of sphere is increased by 9.09 %, then surface area of sphere is increased by how much percentage.
Answer: A
An increase of \( 9.09\% \) is a factor of \( \tfrac{1}{11} \), since \( \tfrac{1}{11} = 0.0909 \). The radius therefore becomes \( \tfrac{12}{11}r \).
Surface area varies as the square of the radius:
\[ \frac{S'}{S} = \left(\frac{12}{11}\right)^{2} = \frac{144}{121} = 1.1901 \]
\[ \text{increase} = 19.01\% \approx 19\% \]
Recognising 9.09% as \( \tfrac{1}{11} \) keeps the arithmetic exact; the decimal route gives 18.998% and invites a rounding error.
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