Question

Three cubes of a metal whose edges are in the ratio 3 : 4 : 5 are melted and converted into a single cube whose diagonal is $12\sqrt{3}$ cm. Find the
edges of the three cubes. [CBSE 2013C]

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{edge}\mathrm{of}\mathrm{the}\mathrm{metal}\mathrm{cubes}\mathrm{be}3x,4x\mathrm{and}5x. \\ \mathrm{Let}\mathrm{the}\mathrm{edge}\mathrm{of}\mathrm{the}\mathrm{single}\mathrm{cube}\mathrm{be}a. \\ \mathrm{As}, \\ \mathrm{Diagonal}\mathrm{of}\mathrm{the}\mathrm{single}\mathrm{cube}=12\sqrt{3}\mathrm{cm} \\ \Rightarrow a\sqrt{3}=12\sqrt{3} \\ \Rightarrow a=12\mathrm{cm} \\ \mathrm{Now}, \\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{single}\mathrm{cube}=\mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{volumes}\mathrm{of}\mathrm{the}\mathrm{metallic}\mathrm{cubes} \\ \Rightarrow a^3={\left(3x\right)}^3+{\left(4x\right)}^3+{\left(5x\right)}^3 \\ \Rightarrow {12}^3=27x^3+64x^3+125x^3 \\ \Rightarrow 1728=216x^3 \\ \Rightarrow x^3=\frac{1728}{216} \\ \Rightarrow x^3=8 \\ \Rightarrow x=\sqrt[3]{8} \\ \Rightarrow x=2 \\ \mathrm{So},\mathrm{the}\mathrm{egdes}\mathrm{of}\mathrm{the}\mathrm{cubes}\mathrm{are}3\times 2=6\mathrm{cm},4\times 2=8\mathrm{cm}\mathrm{and}5\times 2=10\mathrm{cm}. \end{gathered}$$

Hence, the edges of the given three metallic cubes are 6 cm, 8 cm and 10 cm.