The cube

The cube has an edge of 12 dm. The second cube has an edge exactly 20% longer. How many % is more water in the second cube than in the first cube, if the first cube is full to 3/4 and the second to 3/8?

Correct result:

p =  -13.6 %

Solution:

$$\begin{gathered} a_1=12\text{ dm } \\ {a}_2=(1+{\displaystyle \frac{20}{100}})\cdot a_1=(1+{\displaystyle \frac{20}{100}})\cdot 12={\displaystyle \frac{72}{5}}=14.4\text{ dm } \\ {V}_1={\displaystyle \frac{3}{4}}\cdot a_1^3={\displaystyle \frac{3}{4}}\cdot 12^3=1296{\text{dm}}^3 \\ {V}_2={\displaystyle \frac{3}{8}}\cdot a_2^3={\displaystyle \frac{3}{8}}\cdot 14.4^3=1119.744{\text{dm}}^3 \\ p=100\cdot {\displaystyle \frac{V_2-V_1}{V_1}}=100\cdot {\displaystyle \frac{1119.744-1296}{1296}}=-{\displaystyle \frac{68}{5}}=-{\displaystyle \frac{68}{5}}\mathrm{\%}=-13.6\mathrm{\%} \end{gathered}$$

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