Compound Interest — Case-Study Based Questions

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:

1. The population of the town in 2016 was:
   (a) 20800
   (b) 20500
   (c) 20460
   (d) 20300

2. The difference in the population of the town at the end of the years 2017 and 2015 was:
   (a) 1500
   (b) 1700
   (c) 1800
   (d) 1840

3. The population of the town at the end of the year 2018 was:
   (a) 20100
   (b) 20500
   (c) 20748
   (d) 20850

4. Which of the following expressions gives the population at the end of the year 2018?
   
   (a) $20000 \times \dfrac{104}{100} \times \dfrac{105}{100} \times \dfrac{105}{100}$
   
   (b) $20000 \times \dfrac{104}{100} \times \dfrac{105}{100} \times \dfrac{95}{100}$
   
   (c) $20000 \times \dfrac{104}{100} \times \dfrac{95}{100}\times \dfrac{95}{100}$
   
   (d) $20000 \times \dfrac{96}{100} \times \dfrac{95}{100} \times \dfrac{95}{100}$

5. If the population of the town at the end of 2019 was 21,163, then during 2019, the population increases at the rate of:
   (a) 2%
   (b) 3%
   (c) 3.5% (d) 4%

Answer

1. Given,

Population (in 2015) = 20000

r = 4% p.a.

n = 1 year

By formula,

Population in 2016 = $P \times \Big(1 + \dfrac{r}{100}\Big)^n$

Substituting the values in formula,

$\text{Population in 2016 } = 20000 \times \Big(1 + \dfrac{4}{100}\Big)^1 \\[1em] =20000 \times \Big(\dfrac{100 + 4}{100}\Big)^1 \\[1em] =20000 \times \Big(\dfrac{104}{100}\Big)^1 \\[1em] =20000 \times (1.04)\\[1em] =20800.$

Hence, option (a) is correct option.

2. Given,

Population (in 2016) = 20800

r = 5% p.a.

n = 1 year

By formula,

Population in 2017 = $P \times \Big(1 + \dfrac{r}{100}\Big)^n$

Substituting the values in formula,

$\text{Population in 2017} = 20800 \times \Big(1 + \dfrac{5}{100}\Big)^1 \\[1em] =20800 \times \Big(\dfrac{100 + 5}{100}\Big)^1 \\[1em] =20800 \times \Big(\dfrac{105}{100}\Big)^1 \\[1em] =20800 \times (1.05)\\[1em] =21840.$

The difference in population at the end of the years 2017 and 2015 = 21840 - 20000 = 1840.

Hence, option (d) is correct option.

3. Given,

Population (in 2017) = 21840

r (decrease) = 5% p.a.

n = 1 year

By formula,

Population in 2018 = $P \times \Big(1 - \dfrac{r}{100}\Big)^n$

Substituting the values in formula,

$\text{Population in 2018}=21840 \times \Big(1 - \dfrac{5}{100}\Big)^1 \\[1em] =21840 \times \Big(\dfrac{100 - 5}{100}\Big)^1 \\[1em] =21840 \times \Big(\dfrac{95}{100}\Big)^1 \\[1em] =21840 \times (0.95)\\[1em] =20748.$

Hence, option (c) is correct option.

4. Given,

P = ₹ 20,000

r₁ = 4%

r₂ = 5%

r₃ = 5% (decrease)

By formula,

Population at the end = $P \times \Big(1 + \dfrac{r_1}{100}\Big) \times \Big(1 + \dfrac{r_2}{100}\Big) \times \Big(1 - \dfrac{r_3}{100}\Big)$

Substituting the values in formula,

$\Rightarrow \text{Population at the end of year 2018} = 20000 \times \Big(1 + \dfrac{4}{100}\Big) \times \Big(1 + \dfrac{5}{100}\Big) \times \Big(1 - \dfrac{5}{100}\Big) \\[1em] = 20000 \times \Big(\dfrac{100 + 4}{100}\Big) \times \Big(\dfrac{100 + 5}{100}\Big) \times \Big(\dfrac{100 - 5}{100}\Big) \\[1em] = 20000 \times \Big(\dfrac{104}{100}\Big) \times \Big(\dfrac{105}{100}\Big) \times \Big(\dfrac{95}{100}\Big)$

Hence, option (b) is correct option.

5. Given,

A (population in 2019) = 21,163

P (population in 2018) = 20,748

n = 1 year

By formula,

A = $P \times \Big(1 + \dfrac{r}{100}\Big)^n$

Substituting the values in formula,

$\Rightarrow 21163 = 20748 \times \Big(1 + \dfrac{R}{100}\Big)^1 \\[1em] \Rightarrow \dfrac{21163}{20748} = \Big(1 + \dfrac{R}{100}\Big)^1 \\[1em] \Rightarrow 1.02 = \Big(1 + \dfrac{R}{100}\Big)^1 \\[1em] \Rightarrow 1.02 - 1 = (\dfrac{R}{100}) \\[1em] \Rightarrow 0.02 = (\dfrac{R}{100}) \\[1em] \Rightarrow 0.02 \times 100 = R \\[1em] \Rightarrow R = 2$

Hence, option (a) is correct option.