Question

the diameter of a roller 120 cm long is 84 cm. if it takes 500 complete revolutions to level a a playground determine the cost of levelling it at the rate of 30 paise per sq metre

Answer 1

surface area of on revolution = 2∏rh 
= 2 x 22/ 7 x 60 x  84 
now for 500 revolution = 2 x 22/7 x 60 x 84 x 500
=1584 m2 
now cost leveling   1 m2 = Rs. 0.30
= for 1584 m2 = 0.30 x 1584 =  Rs. 475 . 20

Answer 2

Given,

Diameter of a cylindrical roller = 84 cm

Length of the cylindrical roller = 120 cm

Number of complete revolutions made by the roller to level a playground = 500

The rate of cost of leveling = 30 paise/m^2

To find,

The total cost of leveling the playground.

Solution,

We can simply solve this mathematical problem using the following process:

As per mensuration;

The lateral surface area of a cylinder

= 2π(radius)(length)

Now, according to the question;

The total area of the playground leveled in every single revolution of the cylindrical roller

= the lateral surface area of the cylindrical roller

= 2π(radius of the roller)(length of the roller)

= 2π(diameter of the roller/2)(length of the roller)

= π(diameter of the roller)(length of the roller)

= 22/7(84 cm)(120 cm)

= (22×12×120) cm^2

= 31,680 cm^2

Now, the total area of the playground

= (total area of the playground leveled in every single revolution of the cylindrical roller)×(total number of revolutions made to cover the entire playground)

= 31,680 cm^2 × 500

= 1584×10^4 cm^2

= 1584 m^2

(Since 100 cm = 1 m => 10000 cm^2 = 1 m^2)

Now, according to the question;

The total cost of leveling the playground

= total area of the playground × rate of cost of labeling

= 1584 m^2 × Rs. 0.3

= Rs. 475.2

Hence, the total cost of leveling the playground is equal to Rs. 475.2.