2410.33 – Battling Percentages

If length, width and height are $l$, $w$ and $h$, then the new dimensions are $l \left(\dfrac{100+p}{100}\right)$, $w \left(\dfrac{100+p}{100}\right)$, and $h \left(\dfrac{100-q}{100}\right)$. Those fractions are imposing, so instead of the percentages $p$ and $q$, let's work with the fractions $x = \dfrac{p}{100}$ and $y = \dfrac{q}{100}$. In these terms the new dimensions are $l (1+x)$, $w (1+x)$, and $h (1-y)$.

The volume of the solid is supposed to be unchanged. That is,

$$l (1+x) w (1+x) h (1-y) = lwh.$$

The original dimensions cancel and we are left with the equation,

$$(1+x)^2(1-y) = 1,$$

in which $y$ is the unknown. Solving is easy enough:

$$\begin{aligned}1-y &= \dfrac{1}{(1+x)^2} \\y &= 1 - \dfrac{1}{(1+x)^2}\end{aligned}$$

Putting back the $p$'s and $q$'s is not so nice:

$$\begin{aligned}q &= 100 y \\&= 100 \left(1 - \dfrac{1}{(1+x)^2}\right) \\[1em]&= 100 \left(1 - \dfrac{1}{\left(1+\dfrac{p}{100}\right)^2}\right) \\[2em]&= 100 \left(1 - \dfrac{100^2}{(100 + p)^2}\right) \\[1em]&= 100 - \dfrac{100^3}{(100 + p)^2}\end{aligned}$$

The answer, believe it or not, is (e).