Question

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

Answer 1

Step-by-step explanation:

radius of the roller=d\2=120\2=60cm

length of roller is 84cm

the roller is of cylindrical shape

area of roller=

$2\pi \: r(r + h)$

=2*3.14*60(60+84)

=6.28*60(144)

=54259.2cm^2

=542.592m^2

=542.59m^2

cost of levelling =30 paise per m^2

the roller takes 500 revolution

cost of levelling the playground=500*542.59

=271295 paise

1rupee=100 Paisa

=2712.95 rupees

the cost of levelling the playground is 2712.95 rupees

hope it's correct

Answer 2

diameter=84cm

therefore radius=42cm

height=120cm

area rolled in 1 revolution=c.s.a 

=2*22/7*r*h=2*22/7*42*120

316800 cm^2 = 3.168 m^2

area of field=500*3.168

=1584 m^2

cost=30 p/m^2

therefore, total cost=1584*30

47520 p/m^2