3.1 The Big Five Marathon is an annual event in South Africa - NSC Mathematical Literacy - Question 3 - 2018 - Paper 2

The general direction a runner is heading when passing the 20 km mark is Southeast or East of South.

The runner was correct as he was at a height of 1,400 m at the 5 km mark and then at 1,708 m at the 10 km mark. There is an increase in height of 308 m, confirming his claim of running at form 1 level due to the acute increase in altitude.

Cut-off times are established to ensure that runners complete the marathon within a specific time frame. This helps manage safety, medical support, and resource allocation effectively during the event.

To verify the claim:

1. Calculate the speed required for the half-marathon runner to beat the cut-off time: $\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{16.5\text{ km}}{5 \text{h}} = 3.3\text{ km/h}.$

2. Compare this with the full marathon runner's speed. Assuming he runs at: $\text{Speed}_{full} = (3.3 + 2.7) \text{ km/h} = 6.0 \text{ km/h}.$

This shows that the full marathon runner would indeed be faster and can beat the half-marathon runner's time based on the calculated speeds.

First, calculate the radius from the diameter: $\text{Radius} = \frac{31.2 \text{ cm}}{2} = 15.6 \text{ cm}.$

Next, using the formula for the volume of a cylinder: $\text{Volume} = \pi \times (15.6)^{2} \times \text{height} = 20,000 \text{ cm}^{3}.$

Rearranging for height: $\text{height} = \frac{20,000}{\pi \times (15.6)^{2}} \approx 25.1 \text{ cm}.$

Thus, the maximum height of water is approximately 25.1 cm.

The area of the rectangular pallet is calculated as: $\text{Area} = 100\text{ cm} \times 120\text{ cm} = 12,000 \text{cm}^2.$

Next, determine the area occupied by the bucket. The radius of the bucket is: $\text{Radius} = 15.6 \text{ cm}.$ Thus, Area of 11 buckets: $\text{Area}_{bucket} = 11 \times (\pi \times (15.6)^{2}) = 11 \times 764.63712 \approx 8411.00832\text{ cm}^2.$

Finally, calculate unused area: $\text{Unused Area} = 12,000 - 8,411.00832 \approx 3,588.99168\text{ cm}^2.$

Based on the earlier calculated dimensions of the bucket arrangement, length C can be deduced as follows: The pallet length for 12 buckets arranged in 3 rows of 4 buckets each would be: $C = 4 \times (31.2 \text{ cm}) = 124.8 \text{ cm}.$

The original length is 120 cm. The new length required is 124.8 cm. The percentage increase is calculated as: $\text{Percentage Increase} = \frac{\text{New Length} - \text{Original Length}}{\text{Original Length}} \times 100 = \frac{124.8 - 120}{120} \times 100 \approx 4\%.$