Original model

When drying, the modeling mass loses 36% of its original volume. If the cube's volume is 5.12 dm³ after drying, what was the original length of the edge?

Correct answer:

a =  2 dm³

Step-by-step explanation:

$$\begin{gathered} V_2=5.12{\text{dm}}^3 \\ r=100\%-36\%=1-\frac{36}{100}=0.64 \\ {V}_2=V_1\cdot r \\ {V}_1=V_2/r=5.12/0.64=8{\text{dm}}^3 \\ {V}_1=a^3 \\ a=\sqrt[3]{V_1}=\sqrt[3]{8}=2{\text{dm}}^3 \end{gathered}$$

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