The cube

The first cube has an edge of 12 dm. The second cube has an edge exactly 20% longer. By what percentage does the second cube contain more water than the first, if the first cube is filled to 3/4 of its volume and the second to 3/8?

Final Answer:

p =  -13.6 %

Step-by-step explanation:

$$\begin{gathered} a_1=12\text{dm} \\ {a}_2=(1+\frac{20}{100})\cdot a_1=(1+\frac{20}{100})\cdot 12=\frac{72}{5}=14.4\text{dm} \\ {V}_1=\frac{3}{4}\cdot a_1^3=\frac{3}{4}\cdot 12^3=1296{\text{dm}}^3 \\ {V}_2=\frac{3}{8}\cdot a_2^3=\frac{3}{8}\cdot {\frac{72}{5}}^3=1119.744{\text{dm}}^3 \\ p=100\cdot \frac{V_2-V_1}{V_1}=100\cdot \frac{\frac{139968}{125}-1296}{1296}=-\frac{68}{5}\%=-13.6\% \end{gathered}$$

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