SSC CHSL 6 Dec 2015 Shift 1 Paper

Radius of cylinderical roller = 42 cm and height = 120 cm
⇒ Distance covered in 1 revolution by the roller = Curved surface area of the roller = $2\mathit{\pi }\mathit{r}\mathit{h}$
= $2\times \frac{22}{7}\times 42\times 120$
= $44\times 6\times 120=31680\mathit{c}{\mathit{m}}^2$ $=3.168{\mathit{m}}^2$
⇒ Total distance covered in 500 revolutions = $500\times 3.168=1584{\mathit{m}}^2$
Now, cost of levelling the 1 ${\mathit{m}}^2$ ground = Rs. 1.50
∴ Total cost required = $1584\times 1.50=\mathit{R}\mathit{s}.2376$