Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
Suppose that the side of the cube $=\mathrm{x} \mathrm{cm}$
Surface area of the cube $=6 \times(\text { side })^{2}=6 \times \mathrm{x}^{2}=6 \mathrm{x}^{2} \mathrm{~cm}^{2}$
i.e., the sum of the surface areas of three such cubes $=6 \mathrm{x}^{2}+6 \mathrm{x}^{2}+6 \mathrm{x}^{2}=18 \mathrm{x}^{2} \mathrm{~cm}^{2}$
Now, these three cubes area placed together to form a cuboid.
Then the length of the new cuboid will be 3 times the edge of the cube $=3 \times \mathrm{x}=3 \mathrm{x} \mathrm{cm}$
Breadth of the cuboid $=\mathrm{x} \mathrm{cm}$
Height of the cuboid $=\mathrm{x} \mathrm{cm}$
$\therefore$ Total surface area of the cuboid $=2 \times($ length $\times$ breadth $+$ breadth $\times$ height $+$ length $\times$ height $)$
$=2 \times(3 \mathrm{x} \times \mathrm{x}+\mathrm{x} \times \mathrm{x}+3 \mathrm{x} \times \mathrm{x})$
$=2 \times\left(3 \mathrm{x}^{2}+\mathrm{x}^{2}+3 \mathrm{x}^{2}\right)$
$=2 \times\left(7 \mathrm{x}^{2}\right)$
$=14 \mathrm{x}^{2} \mathrm{~cm}$
i.e., the ratio of the total surface area cuboid to the sum of the surface areas of the three cubes $=14 x^{2} \mathrm{~cm}^{2}: 18 x^{2} \mathrm{~cm}^{2}$
$=7: 9$
Hence, the ratio is $7: 9$.