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Question

Initially, the population of a town was 80,000 and increases at the rate of R%. If the population becomes 1,06,480 after three years, then the value of R is:

The correct answer is
10

Population Growth Rate Calculation

This problem involves calculating the annual percentage rate at which a town's population grew over a period of three years. We are given the initial population, the final population, and the time duration. This type of problem is typically solved using the concept of compound growth, similar to how compound interest works.

Understanding Population Growth Formula

The formula for population growth, assuming a constant annual growth rate, is:

$ P_t = P_0 \left(1 + \frac{R}{100}\right)^t $

Where:

  • $P_t$ is the population after t years (the final population).
  • $P_0$ is the initial population.
  • $R$ is the annual growth rate in percent (%).
  • $t$ is the number of years.

Applying the Formula to the Problem

Let's identify the values given in the question:

  • Initial Population ($P_0$) = 80,000
  • Final Population ($P_t$) = 1,06,480
  • Time Period ($t$) = 3 years
  • Annual Growth Rate = $R\%$

Now, we substitute these values into the population growth formula:

$ 1,06,480 = 80,000 \left(1 + \frac{R}{100}\right)^3 $

Step-by-Step Calculation of Rate R

Our goal is to find the value of R. We can rearrange the equation to solve for it.

Step 1: Isolate the Growth Factor Term

Divide both sides of the equation by the initial population ($P_0$):

$ \frac{1,06,480}{80,000} = \left(1 + \frac{R}{100}\right)^3 $

Simplify the fraction:

$ \frac{106480}{80000} = \frac{10648}{8000} $

We can simplify this fraction further by dividing both the numerator and the denominator by common factors. Dividing both by 8:

$ \frac{10648 \div 8}{8000 \div 8} = \frac{1331}{1000} $

So the equation becomes:

$ \frac{1331}{1000} = \left(1 + \frac{R}{100}\right)^3 $

Step 2: Find the Cube Root

To eliminate the exponent '3', we take the cube root of both sides of the equation:

$ \sqrt[3]{\frac{1331}{1000}} = \sqrt[3]{\left(1 + \frac{R}{100}\right)^3} $

We know that $11^3 = 1331$ and $10^3 = 1000$. Therefore:

$ \frac{11}{10} = 1 + \frac{R}{100} $

Converting the fraction to a decimal:

$ 1.1 = 1 + \frac{R}{100} $

Step 3: Solve for R

Now, isolate the term $\frac{R}{100}$ by subtracting 1 from both sides:

$ 1.1 - 1 = \frac{R}{100} $

$ 0.1 = \frac{R}{100} $

Finally, multiply both sides by 100 to find the value of R:

$ R = 0.1 \times 100 $

$ R = 10 $

Conclusion

The value of R, representing the annual percentage growth rate of the town's population, is 10.

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Important Questions from Percentage (Notes)

  1. Anubhav spent 14% of his income on electricity bills, 28% on rent and 18% on shopping. If $\frac{1}{4}$ th of the remaining amount is ₹5120, how much did he spend on electricity bills ?
  2. The total population of a town is 50,000. The number of males and females increases by 10% and 15% respectively and consequently the population of the town becomes 56,000. What was the number of males in the town?
  3. Gokul donates $17\%$ of his savings to old age home, $17\%$ of the savings to orphanage and $17\%$ of his savings to foundations for medical help. The remaining amount of Rs.$9800$ of savings is been deposited in bank. Find the salary of Gokul, if $50\%$ of the salary is his savings amount. (In Rs.)
  4. Which of the following multipliers will increase a given number by 25.3 % ?
  5. If A earns \(33\frac{1}{3}%\) more than B, then how much percent does B earn less than A?

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