Ans: A
Let $z = \dfrac{kx^2}{\sqrt{y}}$, where $k$ is a non-zero constant.
Let $z = \dfrac{kx^2}{\sqrt{y}}$, where $k$ is a non-zero constant.
The new value of $x$
$\begin{array}{cl}
= & x \times (1 – 40\%) \\
= & 0.6x
\end{array}$
The new value of $y$
$\begin{array}{cl}
= & y \times (1 + 44\%) \\
= & 1.44y
\end{array}$
Therefore, the new value of $z$
$\begin{array}{cl}
= & \dfrac{k(0.6x)^2}{\sqrt{1.44y}} \\
= & \dfrac{0.36kx^2}{1.2\sqrt{y}} \\
= & \dfrac{3}{10} \times \dfrac{kx^2}{\sqrt{y}}
\end{array}$
Hence, the percentage change of $z$
$\begin{array}{cl}
= & \dfrac{\frac{3}{10} \times \frac{kx^2}{\sqrt{y}} – \frac{kx^2}{\sqrt{y}}}{\frac{kx^2}{\sqrt{y}}} \times 100\% \\
= & \dfrac{\frac{kx^2}{\sqrt{y}}( \frac{3}{10} \times – 1)}{\frac{kx^2}{\sqrt{y}}} \times 100\% \\
= & -70\%
\end{array}$
Therefore, $z$ is decreased by $70\%$.
