Question

The diameter of a roller 120cm long is 84cm. It it takes 500

complete revolutions to level a playground, find the cost of

levelling it at the rate of 30 paise per square meter.​

Answer 1

Answer:

your answer is ₹52,800,000

Step-by-step explanation:

BECAUSE,,,

we will find the LSA of the roller

obviously it is a cylindrical figure

so LSA=2πrh

•2×22/7×60×84

LSA=31680

now if it takes 500 revolutions to completely roll the ground then the ground's area must be 500 times

then

AREA OF GROUND=31680×500

that is 158400000

now we will find the cost

and the rate is 30 p convert it into ₹ we get

0.3 ₹

divide the area by rate

158400000/0.3

we get the ANSWER

₹52800000

.

.

.

hope it helps

Answer 2

Answer:

d=120 cm

r=d/2

r=120/2 = 60

now,

LSA=2×pie ×r×h

=2×22/7×60×84

=31680 sq. cm

=3.1680sq. m

area of road= 3.1680×500=1584sq. m

cost of levelling = 30×1584=Rs. 47520