If we double one dimension of the cuboid, halve the second dimension of the cuboid, and increase the third dimension by $6 \mathrm{~cm}$, we get a cube that has the same surface area as the original cuboid.

Determine the dimensions of this cuboid.

(M. Petrová)
Let's start from the end, i.e., from the dimensions of the resulting cube, and work backwards to deduce the dimensions of the original cuboid. If we denote the length of the cube's edge in centimeters by $x$, then one dimension of the cuboid is $\frac{x}{2}$ (twice this length is the length of the cube's edge), the second dimension of the cuboid is $2 x$ (half of this length is the length of the cube's edge), and the third dimension is $x-6$ (this length increased by $6 \mathrm{~cm}$ is the length of the cube's edge).

The surface area of the resulting cube is $6 x^{2}$, while the surface area of the original cuboid is

$$
2\left(\frac{x}{2} \cdot 2 x+\frac{x}{2} \cdot(x-6)+2 x \cdot(x-6)\right)=2 x^{2}+x^{2}-6 x+4 x^{2}-24 x=7 x^{2}-30 x .
$$

Since both surface areas are equal, we get the equation:

$$
\begin{aligned}
6 x^{2} & =7 x^{2}-30 x, \\
0 & =x^{2}-30 x=x \cdot(x-30) .
\end{aligned}
$$

Since $x$ cannot be zero ( $x$ is the length of the cube's edge), it must be $x=30(\mathrm{~cm})$.

The dimensions of the original cuboid are therefore:

$$
\frac{30}{2}=15(\mathrm{~cm}), \quad 2 \cdot 30=60(\mathrm{~cm}), \quad 30-6=24(\mathrm{~cm}) .
$$

Scoring suggestion. 1 point for expressing the dimensions of the cuboid in terms of the length of the cube's edge; 1 point each for expressing the surface area of each of the bodies; 2 points for solving the equation (including the case where the solver does not explain the non-zero nature of $x$); 1 point for calculating the dimensions of the original cuboid.