Math Problem Statement

The population of Ratnanagar was 40,000 two years ago, and the current population is 44,100. Find the population growth rate. Two years ago, the population of Jajarkot district was 120,000, and now it is 122,412. If the crude death rate of the district was 5%, what was the crude birth rate? The current number of students in a school is 1,000. If 5 students carry 1 student every year, find the number of students after 2 years.

Solution

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Question 11

The population of Ratnanagar was 40,000 two years ago, and the current population is 44,100. Find the population growth rate.

Solution

We use the population growth formula:
P=P0(1+r)tP = P_0 (1 + r)^t
Where:

  • P=44,100P = 44,100 (current population)
  • P0=40,000P_0 = 40,000 (initial population)
  • t=2t = 2 (years)
  • rr is the growth rate we need to find.

Rearranging:
r=(PP0)1t1r = \left(\frac{P}{P_0}\right)^{\frac{1}{t}} - 1

Substituting values:
r=(44,10040,000)121r = \left(\frac{44,100}{40,000}\right)^{\frac{1}{2}} - 1 r=(1.1025)0.51=1.051=0.05r = (1.1025)^{0.5} - 1 = 1.05 - 1 = 0.05

The growth rate is 5%.


Question 12

Two years ago, the population of Jajarkot district was 120,000, and now it is 122,412. If the crude death rate of the district was 5%, what was the crude birth rate?

Solution

The net growth rate is calculated as:
r=(PP0)1t1r = \left(\frac{P}{P_0}\right)^{\frac{1}{t}} - 1

Using the given data:

  • P=122,412P = 122,412, P0=120,000P_0 = 120,000, t=2t = 2.
    r=(122,412120,000)121r = \left(\frac{122,412}{120,000}\right)^{\frac{1}{2}} - 1 r=(1.0201)0.51=1.011=0.01or 1%.r = (1.0201)^{0.5} - 1 = 1.01 - 1 = 0.01 \, \text{or } 1\%.

The net growth rate is 1% per year.

The birth rate is given by:
Birth rate=Net growth rate+Death rate\text{Birth rate} = \text{Net growth rate} + \text{Death rate} Birth rate=1%+5%=6%.\text{Birth rate} = 1\% + 5\% = 6\%.

The crude birth rate is 6%.


Question 13

The current number of students in a school is 1,000. If 5 students carry 1 student every year, find the number of students after 2 years.

Solution

Let the annual growth rate of students be rr:
r = \frac{5}{1000} = 0.005 \, \text{(or 0.5% per year)}.

Using the formula:
P=P0(1+r)tP = P_0 (1 + r)^t Where:

  • P0=1,000P_0 = 1,000, r=0.005r = 0.005, and t=2t = 2.
    P=1000(1+0.005)2P = 1000 (1 + 0.005)^2 P=1000(1.005)2=1000×1.010025=1010.025P = 1000 (1.005)^2 = 1000 \times 1.010025 = 1010.025

The number of students after 2 years is approximately 1,010.


Summary of Answers

  1. Population growth rate for Ratnanagar: 5%
  2. Crude birth rate for Jajarkot: 6%
  3. Number of students after 2 years: 1,010

Let me know if you want further clarifications or a deeper explanation for any part!


Related Questions

  1. What would be the population of Ratnanagar after 5 years if the growth rate stays the same?
  2. How can we calculate the crude death rate if only net growth and birth rates are provided?
  3. What would be the student population after 5 years if the same growth rate is applied?
  4. What assumptions are involved in using the population growth formula?
  5. How does doubling time relate to the growth rate in these scenarios?

Tip

Always use precise values for intermediate steps to avoid rounding errors in population growth calculations.

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Math Problem Analysis

Mathematical Concepts

Population growth
Crude birth rate
Exponential growth
Percentage calculations

Formulas

P = P0 (1 + r)^t
r = (P / P0)^(1/t) - 1
Birth rate = Net growth rate + Death rate

Theorems

Exponential growth formula

Suitable Grade Level

Grades 8-10