Question

If m varies directly as the square root of y, inversely as p2, and directly as n, what happens to m when y is quadrupled, p is tripled, and n is multiplied by 5?
A. m is divided by 10/9 B. m is multiplied by 9/10 C. It stays the same D. m is multiplied by 10/9

Answer 1

$\bf \begin{array}{cccccclllll} \textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\ \textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\ y&=&{{ k}}&\cdot&x \\ && y={{ k }}x \end{array}\\\\ -----------------------------$

$\bf % inverse proportional variation \begin{array}{llllll} \textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\ \textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\ y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x} \\ &&y=\cfrac{{{ k}}}{x} \end{array}$

now... let's see 

"m varies directly as the square root of y, inversely as p², and directly as n"
$\bf m=\cfrac{k\sqrt{y}\cdot n}{p^2}$

now, let's quadruple "y", multiply "n" by 5 and "p" by 3

$\bf m=\cfrac{k\sqrt{y}\cdot n}{p^2}\qquad \begin{cases} y=4y\\ p=3p\\ n=5n \end{cases}\implies m=\cfrac{k\sqrt{4y}\cdot 5n}{(3p)^2} \\\\\\ m=\cfrac{k\sqrt{2^2y}\cdot 5n}{(3^2p^2)}\implies m=\cfrac{k2\sqrt{y}\cdot 5n}{(9p^2)}\implies m=\cfrac{k\sqrt{y}\cdot n}{p^2}\cdot \cfrac{2\cdot 5}{9} \\\\\\ m=\cfrac{k\sqrt{y}\cdot n}{p^2}\cdot \cfrac{10}{9}\implies \cfrac{9}{10}\cdot m=\cfrac{k\sqrt{y}\cdot n}{p^2}$