The problem asks us to find the volume of a cylindrical litter bin in m³ to 2 decimal places (part a), and then to find the volume of rubbish that can be put in the bin below the level of the hole, also in m³ to 2 decimal places (part b). The dimensions of the cylinder are given in centimeters: diameter = 60 cm and height = 90 cm for the entire bin. The hole is located 70 cm from the base.

1. Problem Description

The problem asks us to find the volume of a cylindrical litter bin in m³ to 2 decimal places (part a), and then to find the volume of rubbish that can be put in the bin below the level of the hole, also in m³ to 2 decimal places (part b). The dimensions of the cylinder are given in centimeters: diameter = 60 cm and height = 90 cm for the entire bin. The hole is located 70 cm from the base.

2. Solution Steps

(a) Find the volume of the litter bin.

The formula for the volume of a cylinder is:

$V = \pi r^2 h$

where $r$ is the radius and $h$ is the height.

The diameter is 60 cm, so the radius is $r = 60/2 = 30$ cm.

The height of the cylinder is $h = 90$ cm.

Therefore, the volume in cm³ is:

$V = \pi (30^2) (90) = \pi (900)(90) = 81000\pi$ cm³.

To convert to m³, we divide by $(100)^3 = 1000000$:

$V = \frac{81000\pi}{1000000} = 0.081\pi$ m³.

$V \approx 0.081 * 3.14159265359 \approx 0.25761059724 \approx 0.26$ m³ (to 2 decimal places).

(b) Find the volume of rubbish below the hole.

The height from the base to the hole is 70 cm.

Using the same radius $r = 30$ cm, we calculate the volume up to the hole.

$V = \pi r^2 h = \pi (30^2) (70) = \pi (900) (70) = 63000\pi$ cm³.

To convert to m³, we divide by $(100)^3 = 1000000$:

$V = \frac{63000\pi}{1000000} = 0.063\pi$ m³.

$V \approx 0.063 * 3.14159265359 \approx 0.1979203371 \approx 0.20$ m³ (to 2 decimal places).

3. Final Answer

(a) The volume of the litter bin is 0.26 m³.

(b) The volume of rubbish that can be put in the bin below the hole is 0.20 m³.