On the spinner shown, $P Q$ passes through centre $O$. If areas labelled $R$ and $S$ are equal, then what percentage of the time will a spin stop on the shaded region?
(A) $50 \%$
(B) $22.5 \%$
(C) $25 \%$
(D) $45 \%$
(E) $12.5 \%$

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-049.jpg?height=342&width=285&top_left_y=1081&top_left_x=1386)
Since $P Q$ passes through centre $O$, then it is a diameter of the circle. Since $\angle Q O T=90^{\circ}$, then $\angle P O T=180^{\circ}-90^{\circ}=90^{\circ}$.

Thus, the area of sector $P O T$ is $\frac{90^{\circ}}{360^{\circ}}=\frac{1}{4}$ or $25 \%$ of the area of the circle.

Since the areas labelled $R$ and $S$ are equal, then each is

$25 \% \div 2=12.5 \%$ of the area of the circle.

Therefore, a spin will stop on the shaded region $12.5 \%$ of the time.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-224.jpg?height=364&width=312&top_left_y=1653&top_left_x=1557)

ANSWER: (E)