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Determine how many times larger the area of the pool cover is compared to the area of the pool. Show all the work.

Mathematics

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1.9.2
The dimensions of the pool (length and breadth) and the pool cover are not provided in the question. Therefore, it is not possible to determine how many times larger the area of the pool cover is compared to the area of the pool.

1.10.1
The length and breadth of the pool are not provided in the question.

  1. Calculate the average depth of the pool.

    Average depth=Shallow end depth+Deep end depth2=1 m+2 m2=1.5 m\text{Average depth} = \frac{\text{Shallow end depth} + \text{Deep end depth}}{2} = \frac{1 \text{ m} + 2 \text{ m}}{2} = 1.5 \text{ m}
  2. Express the total capacity (volume) of the pool using variables for length and breadth.

    Let LL be the length and BB be the breadth of the pool.

    Volume=Length×Breadth×Average depth\text{Volume} = \text{Length} \times \text{Breadth} \times \text{Average depth} Volume=L×B×1.5 m3\text{Volume} = L \times B \times 1.5 \text{ m}^3

    To convert to litres, use 1 m3=1000 litres1 \text{ m}^3 = 1000 \text{ litres}.

    Volume=1.5×L×B×1000 litres=1500LB litres\text{Volume} = 1.5 \times L \times B \times 1000 \text{ litres} = \boxed{1500LB \text{ litres}}

    The total capacity of the pool is 1500LB1500LB litres, where LL and BB are the length and breadth of the pool in meters, respectively.

    1.10.2
    The total volume of the pool (from 1.10.1) is required to answer this question, but the length and breadth of the pool are not provided.

  3. Determine the amount of chlorine in litres per 1000 litres of pool water.

    Given: 1.5 teaspoons per 1000 litres of pool water.
    Given: 1 teaspoon = 5 ml.

    1.5 teaspoons=1.5×5 ml=7.5 ml1.5 \text{ teaspoons} = 1.5 \times 5 \text{ ml} = 7.5 \text{ ml}

    Convert ml to litres:

    7.5 ml=7.51000 litres=0.0075 litres7.5 \text{ ml} = \frac{7.5}{1000} \text{ litres} = 0.0075 \text{ litres}

    So, 0.0075 litres of chlorine are needed for every 1000 litres of pool water.

  4. Express the total litres of chlorine required using the pool's volume.

    Let VpoolV_{pool} be the volume of the pool in litres.

    Litres of chlorine=Vpool1000 litres×0.0075 litres\text{Litres of chlorine} = \frac{V_{pool}}{1000 \text{ litres}} \times 0.0075 \text{ litres}

    Substituting Vpool=1500LBV_{pool} = 1500LB litres from 1.10.1:

    Litres of chlorine=1500LB1000×0.0075=1.5LB×0.0075=0.01125LB litres\text{Litres of chlorine} = \frac{1500LB}{1000} \times 0.0075 = 1.5LB \times 0.0075 = \boxed{0.01125LB \text{ litres}}

    The number of litres of chlorine required is 0.01125LB0.01125LB litres, where LL and BB are the length and breadth of the pool in meters.

    1.11.1

  5. Identify the given filling rate and conversion factor.

    Given: Filling rate = 9 gallons per minute.
    Given: 1 gallon = 3.78541 litres.

  6. Convert the filling rate from gallons per minute to litres per minute.

    Litres per minute=9 gallons/minute×3.78541 litres/gallon\text{Litres per minute} = 9 \text{ gallons/minute} \times 3.78541 \text{ litres/gallon} Litres per minute=34.06869 litres/minute\text{Litres per minute} = 34.06869 \text{ litres/minute} Litres per minute≈34.07 litres/minute\text{Litres per minute} \approx \boxed{34.07 \text{ litres/minute}}

    1.11.2
    The total volume of the pool (from 1.10.1) is required to answer this question, but the length and breadth of the pool are not provided.

  7. State the formula for time taken.

    Time=Total Volume of PoolFilling Rate\text{Time} = \frac{\text{Total Volume of Pool}}{\text{Filling Rate}}
  8. Substitute the expressions for total volume and filling rate.

    From 1.10.1, Vpool=1500LBV_{pool} = 1500LB litres.
    From 1.11.1, Filling Rate = 34.06869 litres/minute.

    Time in minutes=1500LB litres34.06869 litres/minute\text{Time in minutes} = \frac{1500LB \text{ litres}}{34.06869 \text{ litres/minute}} Time in minutes≈44.028LB minutes\text{Time in minutes} \approx 44.028LB \text{ minutes}
  9. Convert the time from minutes to days.

    Time in days=Time in minutes60 minutes/hour×24 hours/day\text{Time in days} = \frac{\text{Time in minutes}}{60 \text{ minutes/hour} \times 24 \text{ hours/day}} Time in days=44.028LB1440 days\text{Time in days} = \frac{44.028LB}{1440} \text{ days} Time in days≈0.030575LB days\text{Time in days} \approx \boxed{0.030575LB \text{ days}}

    The time taken to fill the pool is approximately 0.030575LB0.030575LB days, where LL and BB are the length and breadth of the pool in meters. The final answer would need to be rounded up to the nearest day once numerical values for LL and BB are known.

    1.12
    No question was asked for section 1.12. The provided information describes the dimensions of a fire pit.

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