Ans: D
Let $x_0$, $y_0$ and $z_0$ be the original values of $x$, $y$ and $z$ respectively. Let $x_1$, $y_1$ and $z_1$ be the new values of $x$, $y$ and $z$. Let $y=\dfrac{kx}{z^2}$, where $k$ is a non-zero constant. Then, we have
Let $x_0$, $y_0$ and $z_0$ be the original values of $x$, $y$ and $z$ respectively. Let $x_1$, $y_1$ and $z_1$ be the new values of $x$, $y$ and $z$. Let $y=\dfrac{kx}{z^2}$, where $k$ is a non-zero constant. Then, we have
$\begin{array}{rcl}
y_0 & = & \dfrac{kx_0}{z_0^2} \\
\end{array} $
and
$\begin{array}{rcl}
y_1 & = & \dfrac{kx_1}{z_1^2} \\
\end{array} $
For $x$ and $z$ being decreased by $20\%$, we have
$\begin{array}{rcl}
x_1 & = & x_0\times (1-20\%) \\
& = & 0.8x_0
\end{array}$
and
$\begin{array}{rcl}
z_1 & = & z_0 \times (1-20\%) \\
& = & 0.8z_0
\end{array}$
Hence, we have
$\begin{array}{rcl}
y_1 & = & \dfrac{k(0.8x_0)}{(0.8z_0)^2} \\
& = & \dfrac{0.8}{0.8^2} \times \dfrac{kx_0}{z_0^2} \\
& = & 1.25 y_0
\end{array}$
Therefore, the percentage change of $y$
$\begin{array}{cl}
= & \dfrac{y_1 – y_0}{y_0} \times 100\% \\
= & \dfrac{1.25y_0 – y_0}{y_0} \times 100\% \\
= & \dfrac{0.25y_0}{y_0} \times 100\% \\
= & 25\%
\end{array}$
Hence, $y$ is increased by $25\%$.
