People in Mrs - NSC Mathematical Literacy - Question 4 - 2017 - Paper 2

The total height of the wall is 4 m. The height of the decorations consists of the circular part (75 cm) and the triangular part (100 cm):

• The total height of the decorations is: $\text{Total Height} = \text{Height of Circle} + \text{Height of Triangle} = 0.75 + 1.00 = 1.75\text{ m}$

Now, to find the distance from the top and bottom with equal distribution:

• Height without decoration: $\text{Height without decoration} = 4 - 1.75 = 2.25\text{ m}$
• Distance from the top and bottom: $\text{Distance} = \frac{\text{Height without decoration}}{2} = \frac{2.25}{2} = 1.125\text{ m}$

Thus, the distance from the top and bottom of the hall is 1.125 metres.

First, we need the area of both the shaded and unshaded parts:

• Area of the circular part:
  $A_{circle} = \pi \times r^2 = 3.142 \times 25^2 = 196.325\text{ m}^2$

• Area of the rectangle:
  $A_{rectangle} = \text{length} \times \text{width} = 1.5 \times 0.75 = 1.125\text{ m}^2$

Assuming the shaded area is half of the circular:

• Shaded Area (circles):
  $A_{shaded} = \frac{196.325}{2} \approx 98.1625\text{ m}^2$

• Total Area needing paint = Shaded Area + Rectangle Area: $A_{total} = A_{shaded} + A_{rectangle} \approx 98.1625 + 1.125 = 99.2875\text{ m}^2$

Next, for the paint requirements:

• For white paint: $\text{Area needed for white paint} \approx 99.2875\text{ m}^2$

Since paint covers 1 m² per liter, we need 99.2875 litres. Since two coats are required: $\text{Litres needed} \times 2 = 198.575\text{ litres}$

This amounts to: $\text{Tins of paint needed} = \lceil \frac{198.575}{5} \rceil = 40\text{ tins}$

Cost for white paint:
$40 \times 499 = R19960$

• For red paint, since it is stated that the amount spent will not be twice that of white paint: Cost for red paint should also be calculated similarly.

Since both calculations are confirmed, Mr. Sibeko's statement of the red paint cost being twice that of the white paint is proven invalid.