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Soccer fields need to be watered regularly. Windmills can be used to extract water from underground to store in a cylindrical tank. This water is used to wet the soccer field. On ANNEXURE C, in the ANSWER BOOK, the diagram of the windmill and the water storage tank is shown.

Mathematics

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3.2.1

  1. Determine the number of blades.

    The problem refers to "choosing ONE of the blades" but does not provide Annexure C, which would show the diagram of the windmill. We will assume a common number of blades for a windmill, which is 3.

  2. Calculate the probability.

    The probability of randomly choosing one specific blade out of 3 blades is 13\frac{1}{3}.

    P(choosing one blade)=13P(\text{choosing one blade}) = \frac{1}{3}
  3. Round the probability to three decimal places.

    13=0.3333...≈0.333\frac{1}{3} = 0.3333... \approx \mathbf{0.333}

    The probability of randomly choosing one of the blades is 0.333\boxed{\mathbf{0.333}}.

    3.2.2

  4. Identify the radius and calculate the diameter.

    The radius of the circle formed by the rotating blades is 1.2 m1.2 \text{ m}.
    The diameter is twice the radius.

    Diameter=2×Radius=2×1.2 m=2.4 m\text{Diameter} = 2 \times \text{Radius} = 2 \times 1.2 \text{ m} = 2.4 \text{ m}
  5. Calculate the total distance (circumference) using the given formula.

    The distance a point on the tip of blade A will make after one rotation is the circumference of the circle.

    Circumference=3.142×Diameter\text{Circumference} = 3.142 \times \text{Diameter} Circumference=3.142×2.4 m\text{Circumference} = 3.142 \times 2.4 \text{ m} Circumference=7.5408 m\text{Circumference} = \mathbf{7.5408 \text{ m}}

    The total distance is 7.5408 m\boxed{\mathbf{7.5408 \text{ m}}}.

    3.2.3

  6. Analyze the relationship between wind speed and water flow.

    When the wind blows faster, the windmill rotates faster. This faster rotation directly leads to the pump moving more quickly, which in turn increases the speed at which water flows through the pipe.

  7. Choose the correct word.

    The relationship is a direct proportion.
    The correct word is direct\boxed{\text{direct}}.

    3.2.4

    a) Show, with calculations, how 4.123875 m34.123875 \text{ m}^3 was calculated.

  8. Identify the dimensions of the tank and the required water level increase.

    The inner diameter of the cylindrical tank is 5 m5 \text{ m}.
    The radius of the tank is 5 m2=2.5 m\frac{5 \text{ m}}{2} = 2.5 \text{ m}.
    The required increase in water level is 21 cm21 \text{ cm}.

  9. Convert the height increase to meters.

    21 cm=0.21 m21 \text{ cm} = 0.21 \text{ m}
  10. Calculate the volume of water needed using the given formula.

    The volume of a cylinder is given by 3.142×(radius)2×height3.142 \times (\text{radius})^2 \times \text{height}.

    Volume=3.142×(2.5 m)2×0.21 m\text{Volume} = 3.142 \times (2.5 \text{ m})^2 \times 0.21 \text{ m} Volume=3.142×6.25 m2×0.21 m\text{Volume} = 3.142 \times 6.25 \text{ m}^2 \times 0.21 \text{ m} Volume=3.142×1.3125 m3\text{Volume} = 3.142 \times 1.3125 \text{ m}^3 Volume=4.123875 m3\text{Volume} = \mathbf{4.123875 \text{ m}^3}

    The calculation shows that 4.123875 m3\boxed{\mathbf{4.123875 \text{ m}^3}} was correctly calculated.

    b) Determine the time it will take for the water flowing into the tank to increase the water level by 21 cm. Give your answer in hours and minutes, rounded to the nearest minute.

  11. Convert the volume of water needed from cubic meters to litres.

    We need 4.123875 m34.123875 \text{ m}^3 of water.
    Given: 1000ℓ=1 m31000 \ell = 1 \text{ m}^3.

    Volume in litres=4.123875 m3×1000ℓm3=4123.875ℓ\text{Volume in litres} = 4.123875 \text{ m}^3 \times 1000 \frac{\ell}{\text{m}^3} = 4123.875 \ell
  12. Convert the volume from litres to gallons.

    Given: 1 gallon=3.78541178ℓ1 \text{ gallon} = 3.78541178 \ell.

    Volume in gallons=4123.875ℓ3.78541178ℓgallon≈1089.40 gallons\text{Volume in gallons} = \frac{4123.875 \ell}{3.78541178 \frac{\ell}{\text{gallon}}} \approx 1089.40 \text{ gallons}
  13. Calculate the time in minutes.

    The speed of water flowing through the discharge pipe is 5 gallons per minute5 \text{ gallons per minute}.

    Time in minutes=Volume in gallonsFlow rate\text{Time in minutes} = \frac{\text{Volume in gallons}}{\text{Flow rate}} Time in minutes=1089.40 gallons5gallonsminute≈217.88 minutes\text{Time in minutes} = \frac{1089.40 \text{ gallons}}{5 \frac{\text{gallons}}{\text{minute}}} \approx 217.88 \text{ minutes}
  14. Convert the total minutes to hours and minutes, rounded to the nearest minute.

    Hours=217.88 minutes60minuteshour≈3.631 hours\text{Hours} = \frac{217.88 \text{ minutes}}{60 \frac{\text{minutes}}{\text{hour}}} \approx 3.631 \text{ hours}

    This is 3 full hours.

    Remaining minutes=(3.631−3) hours×60minuteshour\text{Remaining minutes} = (3.631 - 3) \text{ hours} \times 60 \frac{\text{minutes}}{\text{hour}} Remaining minutes=0.631 hours×60minuteshour≈37.86 minutes\text{Remaining minutes} = 0.631 \text{ hours} \times 60 \frac{\text{minutes}}{\text{hour}} \approx 37.86 \text{ minutes}

    Rounding to the nearest minute, this is 38 minutes38 \text{ minutes}.
    The total time is 3 hours and 38 minutes\boxed{\mathbf{3 \text{ hours and } 38 \text{ minutes}}}.

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