In the diagram, $P Q R S$ represents a rectangular piece of paper. The paper is folded along a line $V W$ so that $\angle V W Q=125^{\circ}$. When the folded paper is flattened, points $R$ and $Q$ have moved to points $R^{\prime}$ and $Q^{\prime}$, respectively, and $R^{\prime} V$ crosses $P W$ at $Y$. The measure of $\angle P Y V$ is
(A) $110^{\circ}$
(B) $100^{\circ}$
(C) $95^{\circ}$
(D) $105^{\circ}$
(E) $115^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-046.jpg?height=431&width=529&top_left_y=1606&top_left_x=1210)

## Part C: Each correct answer is worth 8.
Since points $Y, W$ and $Q$ form a straight line segment, then $\angle Y W V=180^{\circ}-\angle V W Q$ and so $\angle Y W V=180^{\circ}-125^{\circ}=55^{\circ}$.

Since $Q^{\prime}$ is the final position of $Q$ after folding, then $\angle Q^{\prime} W V$ is the final position of $\angle Q W V$ after folding, and so $\angle Q^{\prime} W V=\angle Q W V$.

Thus, $\angle Q^{\prime} W V=\angle Q W V=125^{\circ}$ and so $\angle Q^{\prime} W Y=\angle Q^{\prime} W V-\angle Y W V=125^{\circ}-55^{\circ}=70^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-090.jpg?height=431&width=553&top_left_y=1519&top_left_x=840)

Since $Q^{\prime} W$ and $R^{\prime} Y$ are parallel sides of the piece of paper, then $\angle R^{\prime} Y W+\angle Q^{\prime} W Y=180^{\circ}$, and so $\angle R^{\prime} Y W=180^{\circ}-\angle Q^{\prime} W Y=180^{\circ}-70^{\circ}=110^{\circ}$.

Finally, $\angle P Y V$ is opposite $\angle R^{\prime} Y W$ so $\angle P Y V=\angle R^{\prime} Y W=110^{\circ}$.

ANswer: (A)