Uniform Rate of Growth

We will discuss here how to apply the principle of compound interest in the problems of uniform rate of growth or appreciation.

The word growth can be used in several ways:

(i) The growth of industries in the country

(ii) The rapid growth of plants or inflation, etc.

If the rate of growth occurs at the same rate, we call it as uniform increase or growth

When the growth of industries or, production in any particular industry is takes into consideration:

Then formula Q = P(1 + $\frac{r}{100}$)$^{n}$ can be used as:

Production after n years = Initial (original) production (1 + $\frac{r}{100}$)$^{n}$ where rate of growth in production is r%.

In the similar manner, the formula Q = P(1 + $\frac{r}{100}$)$^{n}$ can be used for the growth of plants, growth of inflation, etc.

If the present value P of a quantity increase at the rate of r% per unit of time then the value Q of the quantity after n units of time is given by

Q = P(1 + $\frac{r}{100}$)$^{n}$ and growth = Q - P = P{(1 + $\frac{r}{100}$)$^{n}$ - 1}

(i) If the present population of a town = P, rate of growth of population = r % p.a. then the population of the town after n years is Q, where

Q = P(1 + $\frac{r}{100}$)$^{n}$ and growth of population = Q - P = P{(1 + $\frac{r}{100}$)$^{n}$ - 1}

(ii) If the present price of a house = P, rate of appreciation in price of the house = r % p.a. then the price of the house after n years is Q, where

Q = P(1 + $\frac{r}{100}$)$^{n}$ and appreciation in price = Q - P = P{(1 + $\frac{r}{100}$)$^{n}$ - 1}

Increase in population, increase in number of students in academic institutions, increase in production in the fields of agriculture and industry are examples of uniform increase or growth.

Solved examples on the principle of compound interest in the uniform rate of growth (appreciation):

1. The population of a village increases by 10% every year. If the present population is 6000, what will be the population of the village after 3 years?

Solution:

The present population P = 6000,

Rate (r) = 10

Unit of time being year (n) = 3

Q = P(1 + $\frac{r}{100}$)$^{n}$

⟹ Q = 6000(1 + $\frac{10}{100}$)$^{3}$

⟹ Q = 6000(1 + $\frac{1}{10}$)$^{3}$

⟹ Q = 6000($\frac{11}{10}$)$^{3}$

⟹ Q = 6000 × ($\frac{11}{10}$) × ($\frac{11}{10}$) × ($\frac{11}{10}$)

⟹ Q = 7986

Therefore, the population of the village will be 7986 after 3 years.

2. The present population of Berlin is 2000000. If the rate of increase of population of Berlin at the end of a year is 2% of the population at the beginning of the year, find the population of Berlin after 3 years?

Solution:

Population of Berlin after 3 years

Q = P(1 + $\frac{r}{100}$)$^{n}$

⟹ Q = 200000(1 + $\frac{2}{100}$)$^{3}$

⟹ Q= 200000(1 + $\frac{1}{50}$)$^{3}$

⟹ Q= 200000($\frac{51}{50}$)$^{3}$

⟹ Q= 200000($\frac{51}{50}$) × ($\frac{51}{50}$) × ($\frac{51}{50}$)

⟹ Q = 2122416

Therefore, the population of Berlin after 3 years = 2122416

3. A man buys a plot of land for $ 150000. If the value of the land appreciates by 12% every year then find the profit that the man will make by selling the plot after 2 years.

Solution:

The present price of the land, P = $ 150000, r = 12 and n = 2

Q = P(1 + $\frac{r}{100}$)$^{n}$

⟹ Q = $ 150000(1 + $\frac{12}{100}$)$^{2}$

⟹ Q = $ 150000(1 + $\frac{3}{25}$)$^{2}$

⟹ Q = $ 150000($\frac{28}{25}$)$^{2}$

⟹ Q = $ 150000 × ($\frac{28}{25}$) × ($\frac{28}{25}$)

⟹ Q = $ 188160

Therefore, the required profit = Q – P = $ 188160 - $ 150000 = $ 38160

● Compound Interest

Compound Interest

Compound Interest with Growing Principal