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

When passing the 20 km mark, the general direction a marathon runner is heading is to the South East.

When a runner states they are at their highest level or altitude, it signifies they are at the peak of their performance. This situation is often experienced when they are at the highest elevation point in the marathon, which is the highest point in the route, thus coinciding with the highest altitude of the race.

Cut-off times for a marathon are instituted primarily to ensure the safety and health of all participants. They help race organizers manage the event more effectively, allowing for timely medical support and ensuring that properties are closed off when necessary for the completion of the race. Additionally, this maintains an orderly race environment.

To verify the claim, we first calculate the required speeds:

1. Speed required for half-marathon:

   Given distance $= 16.5 ext{ km}$ and time $= 5 ext{ hours}$:

   $Speed_{half-marathon} = \frac{Distance}{Time} = \frac{16.5}{5} = 3.3 ext{ km/h}$

2. Speed required for full marathon to beat 5.25 hours:

   With distance $= 31.5 ext{ km}$ and time $= 5.25 ext{ hours}$:

   $Speed_{full-marathon} = \frac{31.5}{5.25} = 6 ext{ km/h}$

3. Difference in speed:

   $Difference = Speed_{full-marathon} - Speed_{half-marathon} = 6 - 3.3 = 2.7 ext{ km/h}$

Thus, the runner's assertion is CORRECT.

The volume of a cylinder is given by:

$Volume = \pi \times (radius)^2 \times height$

1. Calculate the radius:

   Given outside diameter $= 31.2 ext{ cm}$, the inner diameter is:

   Inner diameter = 31.2 cm - 2 × 0.2 cm = 30.8 cm

   Therefore, the radius = 30.8 / 2 = 15.4 cm.

2. Set volume to be 20 liters (20000 cm³):

   Equating volume:

   $20000 = 3.142 \times (15.4)^2 \times height$

   This simplifies to:

   $height = \frac{20000}{3.142 \times 237.16} = 26.84 ext{ cm}$

1. Area of the rectangular floor:

   The length is $120 ext{ cm}$ and the width is $100 ext{ cm}$:

   $Area_{rectangular} = length \times width = 120 \times 100 = 12000 ext{ cm}^2$

2. Area of one bucket (radius = 15.4 cm):

   $Area_{bucket} = \pi \times (15.4)^2 \approx 749.13 ext{ cm}^2$

3. Area used by 11 buckets:

   $Total\,Area_{buckets} = 11 \times Area_{bucket} = 11 \times 749.13 \approx 8240.43 ext{ cm}^2$

4. Unused area calculation:

   $Unused\,Area = Area_{rectangular} - Total\,Area_{buckets} = 12000 - 8240.43 = 3759.57 ext{ cm}^2$

Given the total length of the pallet is 120 cm and the diameter of the buckets is 31.2 cm:

To calculate length C, we subtract the total diameter occupied by 4 buckets:

$C = 120 - (4 imes 31.2) = 120 - 124.8 = -4.8 ext{ cm}$

Since this is a negative value, it implies an adjustment in pallet dimensions is necessary.

To accommodate 12 buckets arranged in three rows of four each:

1. Calculate the needed space for 12 buckets:

   $Length_{new} = 12 \times 31.2 = 374.4 ext{ cm}$

2. Percentage Increase:

   Percentage\,Increase = \frac{Length_{new} - Length_{original}}{Length_{original}} \times 100 = \frac{374.4 - 120}{120} \times 100 = 212 ext{%}

Thus, the pallet length should be increased by 212%.