2014-II-10

Ans: C
Let $\theta$ and $r\text{ cm}$ be the angle and the radius of the sector respectively.

The original area of the sector

$\begin{array}{cl}
= & \pi r^2 \times \dfrac{\theta}{360^\circ} \text{ cm}^2
\end{array}$

The new area of the sector

$\begin{array}{cl}
= & \pi [r(1-50\%)]^2 \times \dfrac{\theta(1-x\%)}{360^\circ} \\
= & 0.25 \pi r^2 \times \dfrac{\theta(1-x\%)}{360^\circ} \text{ cm}^2
\end{array}$

Hence, we have

$\begin{array}{rcl}
\pi r^2 \times \dfrac{\theta}{360^\circ} \times (1-90\%) & = & 0.25 \pi r^2 \times \dfrac{\theta(1-x\%)}{360^\circ} \\
0.1 & = & 0.25(1-x\%) \\
1-x\% & = & 0.4 \\
x \% & = & 0.6 \\
x & = & 60
\end{array}$