## 2410.33 – Battling Percentages

In a brick-shaped solid, if the length and width are each $increased$ by $p$ percent, then by what percent $q$ must the height be $decreased$ to maintain the same volume?

1. $100 - \frac{100^2}{(100+p)^2}$
2. $100 - \frac{100}{(100+p)^2}$
3. $100 - \frac{100}{100+p}$
4. $p$
5. $100 - \frac{100^3}{(100+p)^2}$

Solution

If length, width and height are $l$, $w$ and $h$, then the new dimensions are $l \frac{100+p}{100}$, $w \frac{100+p}{100}$, and $h \frac{100-q}{100}$. Those fractions are imposing, so instead of the percentages $p$ and $q$, let's work with the fractions $x = \frac{p}{100}$ and $y = \frac{q}{100}$. 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 &=& \frac{1}{(1+x)^2} \\ y &=& 1 - \frac{1}{(1+x)^2} \end{aligned}

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

\begin{aligned} q &=& 100 y \\ &=& 100 (1 - \frac{1}{(1+x)^2}) \\ &=& 100 (1 - \frac{1}{(1+\frac{p}{100})^2}) \\ &=& 100 (1 - \frac{100^2}{(100 + p)^2}) \\ &=& 100 - \frac{100^3}{(100 + p)^2} \end{aligned}

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