The ball

The ball has a radius of 2m. What percentage of the surface and volume is another sphere whose radius is 20% larger?

Correct answer:

p₁ =  44 %
p₂ =  72.8 %

Step-by-step explanation:

$$\begin{gathered} r_1=2\text{ m } \\ {r}_2=(1+20\mathrm{/}100)\cdot r_1=(1+20\mathrm{/}100)\cdot 2={\displaystyle \frac{12}{5}}=2.4 \\ {S}_1=4\pi \cdot r_1^2=4\cdot 3.1416\cdot 2^2\doteq 50.2655{\text{m}}^2 \\ {S}_2=4\pi \cdot r_2^2=4\cdot 3.1416\cdot 2.4^2\doteq 72.3823{\text{m}}^2 \\ {p}_1=100\cdot {\displaystyle \frac{S_2-S_1}{S_1}}=100\cdot {\displaystyle \frac{72.3823-50.2655}{50.2655}}=44 \\ \text{ Verifying Solution: } \\ q=1.2 \\ {p}_{11}=100\cdot (q^2-1)=100\cdot (1.2^2-1)=44\mathrm{\%} \\ {p}_1=p_{11}=44\mathrm{\%} \end{gathered}$$

$$\begin{gathered} V_1={\displaystyle \frac{4}{3}}\cdot \pi \cdot r_1^3={\displaystyle \frac{4}{3}}\cdot 3.1416\cdot 2^3\doteq 33.5103{\text{m}}^3 \\ {V}_2={\displaystyle \frac{4}{3}}\cdot \pi \cdot r_2^3={\displaystyle \frac{4}{3}}\cdot 3.1416\cdot 2.4^3\doteq 57.9058{\text{m}}^3 \\ {p}_2=100\cdot {\displaystyle \frac{V_2-V_1}{V_1}}=100\cdot {\displaystyle \frac{57.9058-33.5103}{33.5103}}={\displaystyle \frac{364}{5}}=72.8 \\ \text{ Verifying Solution: } \\ {p}_{21}=100\cdot (q^3-1)=100\cdot (1.2^3-1)={\displaystyle \frac{364}{5}}=72.8\mathrm{\%} \\ {p}_2=p_{21}={\displaystyle \frac{364}{5}}\mathrm{\%}=72.8\mathrm{\%} \end{gathered}$$

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