A sewer is laid from a manhole A to a manhole B, 250 m away along a gradient of 1 in 125. If the reduced level of the invert at A is 205.75 m and the height of the boning rod is 3 m, the reduced level of the sight rail at B, is

ANS:C - 206.75 m

Invert is the lowest point of sewer and crown is the highest point, therefore, 205.75-2=203.75 is the invert of sewer B. Boning rod is T shaped rod with a vertical portion at invert and a horizontal portion at the crown. So when the boning rod reading is 3 it gives the crown reading of sewer B.

Therefore RL of B =invert RL of B + boning rod height =(203.75+3) = 206.75. The distance between A and B is 250m and B is at a lower level, so in layman terms, for every 125m distance there is a 1m decrease in RL, thus for 250m distance there will be 2m decrease in RL. In mathematical terms, they are simply dividing 250/125 to obtain 2. 205.75 - (250*1/125-3)= 206.75.