Math Problem Statement

A community of 20,000 population had a water consumption rate of 24 MLD in 2010. In 2020, the population was 36,000 and the water consumption rate was 36 MLD. Assume that the annual percentage rate of population growth remains equal over time, but the per capita water consumption rate stabilizes at 600 L/d. What is the projected water demand (MLD) in 2030?

Solution

To solve this problem, we can follow these steps:

  1. Determine the population growth rate:

    • In 2010, the population was 20,000.
    • In 2020, the population was 36,000.

    The formula for the annual growth rate rr (expressed as a decimal) is given by:

    Pt=P0×(1+r)tP_{t} = P_{0} \times (1 + r)^t where:

    • PtP_{t} is the population at time tt (36,000 in 2020),
    • P0P_{0} is the population at the initial time (20,000 in 2010),
    • tt is the time difference (10 years from 2010 to 2020),
    • rr is the annual growth rate (what we need to find).

    Substituting the values:

    36,000=20,000×(1+r)1036,000 = 20,000 \times (1 + r)^{10}

    Solving for rr:

    1.8=(1+r)101.8 = (1 + r)^{10}

    Taking the 10th root of both sides:

    1+r=1.81101.06061 + r = 1.8^{\frac{1}{10}} \approx 1.0606

    So,

    r=1.06061=0.0606or6.06%per year.r = 1.0606 - 1 = 0.0606 \quad \text{or} \quad 6.06\% \, \text{per year}.

  2. Estimate the population in 2030: Now, using the same growth rate, we can project the population in 2030 (20 years after 2010):

    P2030=20,000×(1.0606)20P_{2030} = 20,000 \times (1.0606)^{20}

    Calculating:

    P203020,000×3.21564,300P_{2030} \approx 20,000 \times 3.215 \approx 64,300

  3. Estimate the per capita water demand: The problem states that the per capita water consumption stabilizes at 600 liters per day (L/d). To find the total water demand in 2030, we multiply the population by the per capita water consumption:

    Total water demand=64,300people×600L/d1,000,000MLD\text{Total water demand} = \frac{64,300 \, \text{people} \times 600 \, \text{L/d}}{1,000,000} \, \text{MLD}

    Total water demand=38,580,000L/d1,000,000=38.58MLD\text{Total water demand} = \frac{38,580,000 \, \text{L/d}}{1,000,000} = 38.58 \, \text{MLD}

Thus, the projected water demand in 2030 is approximately 38.58 MLD.


Would you like further details or have any questions about this calculation?

Here are 5 related questions to explore:

  1. How does the formula for population growth change if the growth rate is not constant?
  2. How would the water demand change if the per capita consumption increased to 700 L/d instead?
  3. What if the population growth rate slowed down after 2020? How would that affect the demand?
  4. How does stabilizing the per capita consumption affect water resource management?
  5. What are other factors, besides population growth, that could influence future water demand?

Tip: Always verify whether growth rates or trends in a problem are constant, as real-world changes often involve variable rates.

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Math Problem Analysis

Mathematical Concepts

Population Growth
Exponential Growth
Water Demand Calculation

Formulas

P_t = P_0 × (1 + r)^t
Water Demand = (Population × Per Capita Water Consumption) / 1,000,000

Theorems

Exponential Growth Model

Suitable Grade Level

Grades 10-12