Question

lown roller is 150 cm and has a diameter of 70 cm to leveling the playground at the rate of ₹ 200 per square ​

Answer 1

The area rolled in one revolution of the roller is equal to the circumference of the roller multiplied by its length=70(pi)*150=10500(pi)sq cm=1.05 sq m. In other words, the surface area of a cylinder of diameter 70 and length 150. 750 revs=750*1.05 sq m=787.5 sq m. The cost of levelling is 787.5*2=$1575.