Compound Interest

Case Study : In 2015, the population of a town was 20000. During 2016 and 2017, it increases by 4% and 5% every year respectively. In 2018, due to an epidemic and migration to cities, the population decreases at 5%. Based on the above information answer the following questions:

Q1: The population of the town in 2016 was:

Step 1: Identify the baseline population from 2015 and the growth rate for 2016.
Population in 2015 \( (P) = 20000 \)
Growth Rate for 2016 \( (R_1) = 4\% \)
Step 2: Compute the population at the end of 2016 using the uniform growth relation.
Formula: \( A_{2016} = P \times \left(1 + \frac{R_1}{100}\right) \)
\( A_{2016} = 20000 \times \left(1 + \frac{4}{100}\right) \)
\( A_{2016} = 20000 \times \frac{104}{100} \)
\( A_{2016} = 200 \times 104 \)
\( A_{2016} = 20800 \)
Answer:a. 20800

Q2: The difference in the population of the town at the end of the years 2017 and 2015 was:

Step 1: Identify the baseline population from 2016 and the growth rate for 2017.
Population in 2016 \( (A_{2016}) = 20800 \)
Growth Rate for 2017 \( (R_2) = 5\% \)
Step 2: Calculate the population at the end of 2017.
Formula: \( A_{2017} = A_{2016} \times \left(1 + \frac{R_2}{100}\right) \)
\( A_{2017} = 20800 \times \left(1 + \frac{5}{100}\right) \)
\( A_{2017} = 20800 \times \frac{105}{100} \)
\( A_{2017} = 208 \times 105 \)
\( A_{2017} = 21840 \)
Step 3: Compute the difference between the population in 2017 and 2015.
\( \text{Difference} = A_{2017} – P_{2015} \)
\( \text{Difference} = 21840 – 20000 \)
\( \text{Difference} = 1840 \)
Answer:d. 1840

Q3: The population of the town at the end of the year 2018 was:

Step 1: Identify the baseline population from 2017 and the net rate of decline for 2018.
Population in 2017 \( (A_{2017}) = 21840 \)
Decline Rate for 2018 \( (R_3) = 5\% \)
Step 2: Compute the final population at the end of 2018 using the depreciation relation.
Formula: \( A_{2018} = A_{2017} \times \left(1 – \frac{R_3}{100}\right) \)
\( A_{2018} = 21840 \times \left(1 – \frac{5}{100}\right) \)
\( A_{2018} = 21840 \times \frac{95}{100} \)
\( A_{2018} = 218.4 \times 95 \)
\( A_{2018} = 20748 \)
Answer:c. 20748

Q4: Which of the following expressions gives the population at the end of the year 2018?

Step 1: Construct the net chain multiplying factor based on two sequential increases and one sequential decrease profile.
\( \text{Expression} = P \times \left(1 + \frac{4}{100}\right) \times \left(1 + \frac{5}{100}\right) \times \left(1 – \frac{5}{100}\right) \)
Step 2: Simplify each bracket to match the structural options.
\( \text{Expression} = 20000 \times \frac{104}{100} \times \frac{105}{100} \times \frac{95}{100} \)
Answer:b. 20000 \(\frac{104}{100}\ \frac{105}{100}\ \frac{95}{100}\)

Q5: If the population of the town at the end of 2019 was 21163, then during 2019, the population increases at the rate of:

Step 1: Identify baseline population from 2018 and final population of 2019.
Population in 2018 \( (P_{2018}) = 20748 \)
Population in 2019 \( (A_{2019}) = 21163 \)
Let the growth rate during 2019 be \( R_4\% \).
Step 2: Set up the growth equation profile.
\( A_{2019} = P_{2018} \times \left(1 + \frac{R_4}{100}\right) \)
\( 21163 = 20748 \times \left(1 + \frac{R_4}{100}\right) \)
Step 3: Isolate the fraction components to solve for \( R_4 \).
\( 1 + \frac{R_4}{100} = \frac{21163}{20748} \)
\( \frac{R_4}{100} = \frac{21163}{20748} – 1 \)
\( \frac{R_4}{100} = \frac{21163 – 20748}{20748} \)
\( \frac{R_4}{100} = \frac{415}{20748} \)
\( R_4 = \frac{41500}{20748} \)
\( R_4 = 2\% \)
Answer:a. 2%