Cuboid perimeters

Answer: $35\text{cm}^3$

Label the widths $a, b, c$

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This perimeter is equal to $a+b+a+b=2a+2b$

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This perimeter is equal to $2a+2c$

Similarly, the third perimeter will be equal to  $2b+2c$

$$2a+2b=12\Rightarrow a+b=6\\

2a+2c=16\Rightarrow a+c=8\\

2b+2c=24\Rightarrow b + c = 12$$ 

Solving by elimination

$\quad\qquad a+b=\ 6\\

\underline{+\ \quad\quad a+c=\ 8\ \ }\\

\quad 2a+b+c=14$

$\quad 2a+b+c=14\\

\underline{-\ \quad\quad b+c=12\ \ }\\

\quad \qquad\quad \ 2a=\ 2$

$\therefore a=1$

So $b=5$, $c=7$, volume $=1\times5\times7=35\text{cm}^3$

Solving by substitution

$a+b=6\Rightarrow b=6-a\\

 a+c=8\Rightarrow c=8-a\\

 \begin{align}b + c = 12\Rightarrow &(6-a)+(8-a)=12\\

\Rightarrow &14-2a=12\\

\Rightarrow &a=1\end{align}$

So $b=5$, $c=7$, volume $=1\times5\times7=35\text{cm}^3$

Adding all of the equations together

$\quad\qquad a+b=\ 6\\

\qquad\quad a+c=\ 8\\

\underline{+\ \quad\quad b+c=12\ \ }\\

\ \ \ 2a+2b+2c=26$

$\therefore a+b+c=13$

$c=13-6=7$

$b=13-8=5$

$a=13-12=1$

So the volume is $1\times5\times7=35\text{cm}^3$.