Ans: A
Let $z=kxy^2$, where $k$ is a non-zero constant. Let $x_0$, $y_0$ and $z_0$ be the original values of $x$, $y$ and $z$ respectively. Let also $x_1$, $y_1$ and $z_1$ be the new values of $x$, $y$ and $z$ respectively. Then we have $z_0 = kx_0y_0^2$, $x_1 = x_0(1-20\%) = 0.8x_0$ and $y_1 = y_0(1+15\%) = 1.15y_0$. And also, we have
Let $z=kxy^2$, where $k$ is a non-zero constant. Let $x_0$, $y_0$ and $z_0$ be the original values of $x$, $y$ and $z$ respectively. Let also $x_1$, $y_1$ and $z_1$ be the new values of $x$, $y$ and $z$ respectively. Then we have $z_0 = kx_0y_0^2$, $x_1 = x_0(1-20\%) = 0.8x_0$ and $y_1 = y_0(1+15\%) = 1.15y_0$. And also, we have
$\begin{array}{rcl}
z_1 & = & kx_1y_1^2 \\
& = & k(0.8x_0)(1.15y_0)^2 \\
& = & 1.058kx_0y_0^2
\end{array}$
Therefore, the percentage changed of $z$
$\begin{array}{cl}
= & \dfrac{z_1-z_0}{z_0} \times 100\% \\
= & \dfrac{1.058kx_0y_0^2 – kx_0y_0^2}{kx_0y_0^2} \times 100\% \\
= & \dfrac{0.058kx_0y_0^2}{kx_0y_0^2} \times 100\% \\
= & 5.8\%
\end{array}$
Hence, $z$ is increased by $5.8\%$.
