Ans: D
Let $r$ and $\theta$ be the radius and the angle of the sector respectively.
Let $r$ and $\theta$ be the radius and the angle of the sector respectively.
The original arc length
$\begin{array}{cl}
= & 2\pi r \times \dfrac{\theta}{360^\circ}
\end{array}$
The new arc length
$\begin{array}{cl}
= & 2\pi r(1-60\%) \times \dfrac{\theta(1+k\%)}{360^\circ}
\end{array}$
For the arc length remains unchanged, we have
$\begin{array}{rcl}
2\pi r \times \dfrac{\theta}{360^\circ} & = & 2 \pi r(1-60\%) \times \dfrac{\theta(1+k\%)}{360^\circ} \\
1 & = & 0.4(1+k\%) \\
1+ k\% & = & 2.5 \\
k% & = & 1.5 \\
k\% & = & 150\% \\
k & = & 150
\end{array}$
