2410.33 – Battling Percentages

In a brick-shaped solid, if the length and width are each increasedincreased by pp percent, then by what percent qq must the height be decreaseddecreased to maintain the same volume?

  1. 100−1002(100+p)2100 - \frac{100^2}{(100+p)^2}
  2. 100−100(100+p)2100 - \frac{100}{(100+p)^2}
  3. 100−100100+p100 - \frac{100}{100+p}
  4. pp
  5. 100−1003(100+p)2100 - \frac{100^3}{(100+p)^2}
Solution

If length, width and height are ll, ww and hh, then the new dimensions are l100+p100l \frac{100+p}{100}, w100+p100w \frac{100+p}{100}, and h100−q100h \frac{100-q}{100}. Those fractions are imposing, so instead of the percentages pp and qq, let's work with the fractions x=p100x = \frac{p}{100} and y=q100y = \frac{q}{100}. In these terms the new dimensions are l(1+x)l (1+x), w(1+x)w (1+x), and h(1−y)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.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,(1+x)^2(1-y) = 1,

in which yy is the unknown. Solving is easy enough:

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

Putting back the pp's and qq' is not so nice:

q=100y=100(1−1(1+x)2)=100(1−1(1+p100)2)=100(1−1002(100+p)2)=100−1003(100+p)2\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).