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 is: 75 cm (the circular part) + 100 cm (the triangular part).

Converting these to metres gives:

• Circular part: 0.75 m
• Triangular part: 1 m

Thus, total height of decorations:

$\text{Total Height} = 0.75 + 1 = 1.75 \text{ m}$

The remaining height from the wall is:

$\text{Remaining Height} = 4 - 1.75 = 2.25 \text{ m}$

Since the decorations are equidistant from the top and bottom, the space above and below the decorations is:

$\text{Space from Top and Bottom} = \frac{2.25}{2} = 1.125 \text{ m}$

First, calculate the area of the shaded part and the white part. Using the area formulas:

• Area of Rectangle:

$\text{Area}_{rect} = length \times width = 1.5 \times 0.75 = 1.125 \text{ m}²$

• Area of Triangles: Each triangle has a base of 0.75 m (half of 1.5 m) and height of 0.75 m:

$\text{Area}_{triangle} = \frac{1}{2} \times base \times height = \frac{1}{2} \times 0.75 \times 0.75 = 0.28125 \text{ m}²$

Since there are two triangles:

$\text{Total Area}_{triangles} = 2 \times 0.28125 = 0.5625 \text{ m}²$

Total Area of Decorations:

$\text{Total Area} = \text{Area}_{rect} + \text{Total Area}_{triangles} = 1.125 + 0.5625 = 1.6875 \text{ m}²$

Now, for 15 decorations:

$\text{Total Area}_{15} = 1.6875 \times 15 = 25.3125 \text{ m}²$

Calculating Paint Needed:

• Red Paint:

$\text{Litres}_{red} = 25.3125 \text{ m}² \times 2 = 50.625 \text{ litres}$

• Cans: As 1 can covers 5 litres:

$\text{Cans}_{red} = \frac{50.625}{5} = 10.125 \rightarrow 11 \text{ cans}$

Cost for Red Paint:

$\text{Cost}_{red} = 11 \times 505 = R5555$

• White Paint: Similar calculation gives:

$\text{Litres}_{white} = 50.625 \text{ litres}$

$\text{Cans}_{white} = 11 \text{ cans}$

Cost for White Paint:

$\text{Cost}_{white} = 11 \times 499 = R5489$

Conclusion: The amount spent for red paint (R5555) is not twice that of white paint (R5489), hence Mr. Sibeko's statement is not valid.