A cube is divided into two cuboids A cube is divided into two cuboids. The surfaces of those cuboids are in the ratio $7: 5$. Calculate the ratio of the volumes.
How can I calculate this?
 A: Without loss of generality we can let the sides of the original cube be $1$.
The larger of the two cuboids has four of its sides equal to say $x$. (The other $8$ sides are $1$.)  Then the smaller cuboid has four of its sides equal to $1-x$.
The surface area of the larger cuboid is $2+4x$. (Two $1\times 1$ faces, and four $x\times 1$ faces). Similarly, the surface area of the smaller cuboid is $2+4(1-x)=6-4x$.
We are told that 
$$\frac{2+4x}{6-4x}=\frac{7}{5}.$$
Solve. We get $x=\frac{2}{3}$, making $1-x=\frac{1}{3}$. So the volumes are in the ratio $2:1$.
A: Suppose the original cube had side $a$. And let dimensions of the resulting cuboids be $a \times a \times b$ and $a \times a \times c$, where $a = b+c$.
The surface areas of the resulting cuboids is $2\left(a \cdot a + a \cdot b + a \cdot b\right) = 2 a(a+2 b)$ and $2a(a+2 c)$, and volumes $a^2 b$ and $a^2 c$ respectively.
It is known that:
$$
  \frac{7}{5} = \frac{a+2b}{a+2c} \quad b+c = a
$$
Solving this gives $b = \frac{2}{3} a$ and $c=\frac{a}{3}$, hence the ratio of volumes is $\frac{a^2 b}{a^2 c} = \frac{b}{c} = 2$.
A: You have a cube with say side length $s$. So the volume is $s^3$. You cut this cube into two cuboids. So one cuboid will have side lengths $s$, $s$, and $s - x$ for some $x$. The other cube will have side lengths $s$, $s$, and $x$. Now


*

*Find expressions that give you the surface area of each of the cuboids. These expressions will be functions of $x$ (and the constant $s$).

*You know that when dividing these two expressions by each other you get $\frac{7}{5}$. So you solve for x.

*Now that you have $x$, you can find the volumes of each of the cuboids. Then you can find their ratio.

