Question

The diameter of a road roller, 130 cm long is 91 cm if it takes 450 complete revolutions to level a playground, then find the cost of levelling the ground at 80 paise per square metre.​

Answer 1

Solution :-

We have

Radius of the roller (r) = $\frac{91}{2}$

r = 45.5 cm

Length of the roller (h) = 130 cm

Area covered by ruler in 1 revolution = Curved Surface area of road roller

=》 $2\pi rh$

=》 $2 \times \frac{22}{7} \times 45.5 \times 130$

=》$\frac{44}{7} \times 45.5 \times 13$

=》 37180${cm}^{2}$

Area covered in 450 revolutions

=》 37180×450

=》 16731000${cm}^{2}$

=》 $\frac{16731000}{10000} {m}^{2}$

=》 1673.1 ${m}^{2}$

Cost of levelling the playground = 1673.1× 80p

=》$\frac{1673.1 \times 80}{100} rs$

=》Rs. 1338.48
= Rs.1338.50$(approx)\:\:\:Ans$

Answer 2

Answer:

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