Question

If z varies directly with x and inversely with the cube of y, when x=8,y=2 and z=3, what is x when y=4 and z=9/32?

Answer 1

$\bf \qquad \qquad \textit{double proportional variation} \\\\ \begin{array}{llll} \textit{\underline{y} varies directly with \underline{x}}\\ \textit{and inversely with \underline{z}} \end{array}\implies y=\cfrac{kx}{z}\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array}\\\\ -------------------------------\\\\ \textit{\underline{z} varies directly with \underline{x} and inversely with the cube of \underline{y}}\qquad z=\cfrac{kx}{y^3}$

$\bf \textit{we also know that } \begin{cases} x=8\\ y=2\\ z=3 \end{cases}\implies 3=\cfrac{k8}{2^3}\implies 3=\cfrac{8k}{8}\implies 3=k \\\\\\ therefore\qquad \boxed{z=\cfrac{3x}{y^3}} \\\\\\ \textit{when y = 4 and }z=\frac{9}{32}\textit{ what is \underline{x}?}\qquad \cfrac{9}{32}=\cfrac{3x}{4^3}\implies \cfrac{9}{32}=\cfrac{3x}{64} \\\\\\ \cfrac{9\cdot 64}{32\cdot 3}=x$