If three equal cubes are placed adjacently in a row, then the ratio of the total surface area of the new cuboid to the sum of the surface areas of the three cubes will be ?

Answer: Option D

Let the length of each edge of each cube be a
Then, the cuboid formed by placing 3 cubes adjacently has the dimensions 3a , a and a
Surface area of the cuboid :
$$\begin{array}{rl}& =2\left[3a\times a+a\times a+3a\times a\right]\\ & =2\left[3a^2+a^2+3a^2\right]\\ & =14a^2\end{array}$$
Sum of surface area of 3 cubes :
$$\begin{array}{rl}& =\left(3\times 6a^2\right)\\ & =18a^2\end{array}$$
∴ Required ratio :
$$\begin{array}{rl}& =14a^2:18a^2\\ & =7:9\end{array}$$