A solid cube is cut into two cuboids of equal volumes, Find the ratio of the total surface area of the given cube and that of one of the cuboids.
A) \[3:2\]
B) $2:1$
C) $1:2$
D) $2:3$

Hint:The total surface area of the cube of side $a$ is $6{a^2}$ and the total surface area of the cuboid of length $l$, breadth $b$ and height $h$ is $2(lb + bh + hl)$. When the cube is cut into two cuboids of equal volume the length and breadth of the cuboid will be the same as the length of the cube but height will be half of the side of the cube.

Complete step-by-step answer:
We are given that a solid cube is cut into two cuboids of equal volumes.
We have to find the ratio of the total surface area of the given cube and that of one of the cuboids.
First, we evaluate the total surface area of the cube.
Let the side of the cube be $x$.
Therefore, the total surface area of the cube is $6{x^2}$.
The cube is cut into two cuboids of equal volume.
Now, we find the dimensions of the cuboid.
Let the length $l$, breadth $b$and height $h$ of the cuboid.
When the cube is cut into two cuboids of equal volume the length and breadth of the cuboid will be the same as the length of the cube but height will be half of the side of the cube.
Therefore,
$
  l = x \\
  b = x \\
  h = \dfrac{x}{2} \\
 $
The total surface area of the cuboid is $2(lb + bh + hl)$.
Substitute all the values and evaluate the total surface area.
$
   \Rightarrow 2\left( {x \times x + x \times \dfrac{x}{2} + x \times \dfrac{x}{2}} \right) \\
   = 2\left( {\dfrac{{4{x^2}}}{2}} \right) \\
   = 4{x^2} \\
 $
Evaluate the ratio of the total surface area of the given cube and that of one of the cuboids.
$
   \Rightarrow \dfrac{{6{x^2}}}{{4{x^2}}} = \dfrac{3}{2} \\
   = 3:2 \\
 $
Hence, option (A) is correct.

Note:
While evaluating the dimensions of the cuboid, analyse both the shapes properly and after that write the length, breadth and height of the cuboid. In ratio questions read the question properly in order to evaluate the correct ratio.