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Ml aggarwal solutions

CHAPTERS

**A road roller has a diameter of 0.7 m and its width is 1.2 m. Find the least number of revolutions that the roller must take in order to level a playground of size 120 m × 44 m.**

Given,

Diameter of a road roller = 0.7 m = 70 cm

So, radius (r) = 70/2 cm = 35 cm = 35/100 m

and width (h) = 1.2 m

Now,

Curved surface area = 2πrh

= (2 × 22/7 × 35/100 × 1.2) m^{2}

= 264/100 m^{2}

Area of playground = 120 m × 44 m

= 120 × 44 m^{2}

= 5280 m^{2}

Hence, the number of revolution made by the road roller = (5280/264) × 100

= 2000 revolutions

Given,

Diameter of a road roller = 0.7 m = 70 cm

So, radius (r) = 70/2 cm = 35 cm = 35/100 m

and width (h) = 1.2 m

Now,

Curved surface area = 2πrh

= (2 × 22/7 × 35/100 × 1.2) m^{2}

= 264/100 m^{2}

Area of playground = 120 m × 44 m

= 120 × 44 m^{2}

= 5280 m^{2}

Hence, the number of revolution made by the road roller = (5280/264) × 100

= 2000 revolutions

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