A circular spinner is divided into six regions, as shown. Four regions each have a central angle of $x^{\circ}$. The remaining regions have central angles of $20^{\circ}$ and $140^{\circ}$. An arrow is attached to the centre of the circle. The arrow is spun once. What is the probability that the arrow stops on a shaded region?
(A) $\frac{2}{3}$
(B) $\frac{7}{8}$
(D) $\frac{5}{12}$
(E) $\frac{7}{12}$
(C) $\frac{1}{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-040.jpg?height=342&width=336&top_left_y=260&top_left_x=1296)
The six angles around the centre of the spinner add to $360^{\circ}$.

Thus, $140^{\circ}+20^{\circ}+4 x^{\circ}=360^{\circ}$ or $4 x=360-140-20=200$, and so $x=50$.

Therefore, the sum of the central angles of the shaded regions is $140^{\circ}+50^{\circ}+50^{\circ}=240^{\circ}$.

The probability that the spinner lands on a shaded region is the fraction of the entire central angle that is shaded, which equals the sum of the central angles of the shaded regions divided by the total central angle $\left(360^{\circ}\right.$ ), or $\frac{240^{\circ}}{360^{\circ}}=\frac{2}{3}$. (We can ignore the possibility that the spinner lands exactly on one of the dividing lines, since we assume that they are infinitesimally thin.)

ANSWER: (A)