118. An ascending gradient of 1 in 100 meets a descending gradient of 1 in 50. The length of summit curve required to provide overtaking sight distance of 500 m will be (A) 938 m (B) 781 m 005 (C) 470 m (D) 170 m

A vertical curve (with EQUAL tangent lengths) is to connect a +1.35% gradient to a -2.25% gradient on a highway designed for a speed of 80 kmph. The K value for the highway is 55 and the minimum required length is to be used. The reduced level and through chainage of the intersection point of the gradients are 78.34 metres and 667.49 metres respectively. i) ii) Calculate the reduced levels and the chainages of the tangent points. Calculate the reduced levels at exact 25 metre multiples of through chainage along the curve. ii) Calculate the through chainage and the reduced level of the highest point on the curve.