From the grade diagram:
$\dfrac{S_1}{0.042} = \dfrac{260}{0.042 + 0.03}$
$S_1 = 151.67 \, \text{ m}$
$d = S_1 - 130 = 151.67 - 130$
$d = 21.67 \, \text{ m}$
The cross-drainage pipe should be at the lowest point of the curve. Stationing of the lowest point indicated as point A in the figure:
$\text{Sta } A = \text{Sta } PI + d$
$\text{Sta } A = 11488 + 21.67 = 11509.67$
$\text{Sta } A = 11 + 509.67 \, \text{ km}$ Answer for Part 1: [ B ]
Vertical distance between PC and PI:
$a = 130(0.042)$
$a = 5.46 \, \text{ m}$
Vertical distance between PC and the lowest point A:
$A_1 = \frac{1}{2}S_1(0.042) = \frac{1}{2}(151.67)(0.042)$
$A_1 = 3.18 \, \text{ m}$
Elevation of the lowest point A:
$\text{Elev } A = \text{Elev } PI + a - A_1$
$\text{Elev } A = 20.80 + 5.46 - 3.18$
$\text{Elev } A = 23.08 \, \text{ m}$
$\text{Elev of invert } = \text{Elev } A - 0.30 - 0.95$
$\text{Elev of invert } = 23.08 - 0.30 - 0.95$
$\text{Elev of invert } = 21.83 \, \text{ m}$ Answer for Part 2: [ D ]
