Question

2. 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 per square metre.​

Answer 1

Given:-

• Diameter of roller is 120 cm.

• Length of roller is 84 cm.

• It takes 500 revolutions to level a playground.

To Find:-

• Area of the playground.

• Cost of leveling the playground at the rate of ₹30/m².

Solution:-

Here, To find area of the play ground we first find Curved surface area of Roller and multiply it with 500 revolutions.

Radius of roller = $\dfrac{d}{2} = \dfrac{120}{2} = 60\ cm$

So, Area of playground = $500 \times (2\pi r h)$

                                       = $500 \times \bigg ( 2 \times \dfrac{22}{7} \times 60 \times 84 \bigg )$

                                       = $500 \times 31,680 \ cm^{2}$

                                       = 1,58,40,000 cm²

Since, 1 m² = 10,000 cm²

So, Area of playground ( in m² ) = $\dfrac{ 1,58,40,000 }{10,000}$

                                                     = 1,584 m²

Hence, Area of playground is 1,584 m².

Now, Cost of leveling the playground = $\boxed{\sf {Rate\ of\ leveling\ \times Area\ of \ playground}}$

                                                               = ₹30/m² × 1,584 m²

                                                               = ₹47,520

Hence, Cost of levelling the playground is ₹47,520.

Answer 2

Answer :-

• Cost of levelling the playground = ₹47,520

Step by step explanation :-

Given :-

1. Diameter of the roller = 84 cm.
2. Length (height) of the roller = 120 cm.
3. Number of revolutions by roller = 500.
4. Rate of levelling per m² = ₹30

To find :-

• Cost of levelling the playground at the rate of ₹30 per m².

Concept :-

• We'll first convert all the given dimensions in m. This is because cost of levelling is given per m² and not cm².

• Then, we'll find out the area of the roller when it completes one revolution. This can be obtained by finding the lateral surface area of roller. Lateral surface area is supposed to be find out because the circles in the roller is not included.

• After finding the area of roller covered after completing one revolution, we'll multiply that value by 500 because it given that it takes 500 revolutions to level the playground.

• After calculating the total area, we'll multiply area by 30 for obtaining the final value.

Solution :-

☆ Converting the dimensions into m :-

1. 84 cm = 0.84 m
2. 120 cm = 1.2 cm

• (Formula used = cm/100)

Radius = Diameter/2

⇒ 1.2/2 = 0.6 cm

Area of roller after covering 1 revolution :-

Lateral Surface area = 2πrh

⠀⠀⠀⠀⠀⠀⠀⠀⠀⇒ 2×(22/7)×0.6×0.84

⠀⠀⠀⠀⠀⠀⠀⠀⠀⇒ 5.24×0.6

⠀⠀⠀⠀⠀⠀⠀⠀⠀⇒ 3.168 m².

Area of the playground :-

Area = Lateral surface area × 500

⠀⠀⠀⠀⠀⠀⠀⠀⠀⇒ 3.168×500

⠀⠀⠀⠀⠀⠀⠀⠀⠀⇒ 1,584 m².

Cost of levelling :-

Cost of levelling = Cost per m² × area of the playground.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⇒ ₹30 × 1,584 m²

⠀⠀⠀⠀⠀⠀⠀⠀⠀⇒ ₹47,520 ✓