Stationing of the PVC=117000 and length of the curve is 200m Find the Stationing of the PVI and PUT.

• Stationing of PVC (Point of Vertical Curvature) = $11+000$. This means $11000\, \text{m}$ from the starting reference point.
• Length of the curve ($L$) = $200\, \text{m}$.

For a symmetrical vertical curve, the PVI is located at the midpoint of the curve. Therefore, the distance from PVC to PVI is half the curve length, and the distance from PVI to PVT is also half the curve length.
Distance = $\frac{L}{2} = \frac{200\, \text{m}}{2} = 100\, \text{m}$.

The PVI (Point of Vertical Intersection) is located after the PVC. To find its stationing, add the distance calculated in Step 2 to the PVC stationing.
Stationing of PVI = Stationing of PVC + $\frac{L}{2}$
Stationing of PVI = $11000\, \text{m} + 100\, \text{m}$
Stationing of PVI = $11100\, \text{m}$
In stationing notation, this is $11+100$.

The PVT (Point of Vertical Tangency) is located after the PVI. To find its stationing, add the distance calculated in Step 2 to the PVI stationing.
Stationing of PVT = Stationing of PVI + $\frac{L}{2}$
Stationing of PVT = $11100\, \text{m} + 100\, \text{m}$
Stationing of PVT = $11200\, \text{m}$
In stationing notation, this is $11+200$.

The Stationing of PVI is $\boxed{\text{11+100}}$.
The Stationing of PVT is $\boxed{\text{11+200}}$.