Compound Interest — Exercise 2.3

The present population of a town is 200000. Its population increases by 10% in the first year and 15% in the second year. Find the population of the town at the end of two years.

Answer

By formula,

V = $V_0\Big(1 + \dfrac{r_1}{100}\Big)\Big(1 + \dfrac{r_2}{100}\Big)$

Putting values in formula we get,

$V = 200000\Big(1 + \dfrac{10}{100}\Big)\Big(1 + \dfrac{15}{100}\Big) \\[1em] = 200000 \times \dfrac{110}{100} \times \dfrac{115}{100} \\[1em] = 20 \times 110 \times 115 \\[1em] = 253000.$

Hence, the population of the town at the end of two years is 253000.

The present population of a town is 15625. If the population increases at the rate of 4% every year, what will be the increase in the population in next 3 years?

Answer

By formula,

V = $V_0\Big(1 + \dfrac{r}{100}\Big)^n$

Putting values in formula we get,

$V = 15625\Big(1 + \dfrac{4}{100}\Big)^3 \\[1em] = 15625 \times \Big(\dfrac{104}{100}\Big)^3 \\[1em] = 15625 \times \Big(\dfrac{26}{25}\Big)^3 \\[1em] = 15625 \times \dfrac{26}{25} \times \dfrac{26}{25} \times \dfrac{26}{25} \\[1em] = \dfrac{15625 \times 17576}{15625} \\[1em] = 17576.$

Increase in population = 17576 - 15625 = 1951.

Hence, the increase in population in next 3 years = 1951.

The population of a city increases each year by 4% of what it had been at the beginning of each year. If its present population is 6760000, find :

(i) its population 2 years hence

(ii) its population 2 years ago.

Answer

(i) By formula,

V = $V_0\Big(1 + \dfrac{r}{100}\Big)^n$

Putting values in formula we get,

$V = 6760000\Big(1 + \dfrac{4}{100}\Big)^2 \\[1em] = 6760000 \times \Big(\dfrac{104}{100}\Big)^2 \\[1em] = 6760000 \times \Big(\dfrac{26}{25}\Big)^2 \\[1em] = 6760000 \times \dfrac{26}{25} \times \dfrac{26}{25} \\[1em] = \dfrac{6760000 \times 676}{625} \\[1em] = 10816 \times 676 \\[1em] = 7311616.$

Hence, the population after 2 years = 7311616.

(ii) Let population 2 years ago be P and present population will be the final population.

By formula,

V = $V_0\Big(1 + \dfrac{r}{100}\Big)^n$

Putting values in formula we get,

$\Rightarrow 6760000 = P \times \Big(1 + \dfrac{4}{100}\Big)^2 \\[1em] \Rightarrow 6760000 = P \times \Big(\dfrac{104}{100}\Big)^2 \\[1em] \Rightarrow 6760000 = P \times \Big(\dfrac{26}{25}\Big)^2 \\[1em] \Rightarrow 6760000 = P \times \dfrac{26}{25} \times \dfrac{26}{25} \\[1em] \Rightarrow 6760000 = P \times \dfrac{676}{625} \\[1em] \Rightarrow P = \dfrac{6760000 \times 625}{676} \\[1em] \Rightarrow P = 625 \times 10000 \\[1em] \Rightarrow P = 6250000.$

Hence, the population 2 years ago was 6250000.

The cost of a refrigerator is ₹9000. Its value depreciates at the rate of 5% every year. Find the total depreciation in its value at the end of 2 years.

Answer

Depreciation formula,

V = $V_0\Big(1 - \dfrac{r}{100}\Big)^n$

Putting values in formula we get,

$V = ₹9000\Big(1 - \dfrac{5}{100}\Big)^2 \\[1em] = ₹9000 \times \Big(\dfrac{95}{100}\Big)^2 \\[1em] = ₹9000 \times \Big(\dfrac{19}{20}\Big)^2 \\[1em] = ₹9000 \times \dfrac{19}{20} \times \dfrac{19}{20} \\[1em] = ₹\dfrac{9000 \times 361}{400} \\[1em] = ₹8122.5$

Depreciation = ₹9000 - ₹8122.5 = ₹877.5

Hence, the total depreciation in its value at the end of 2 years = ₹877.5

Dinesh purchased a scooter for ₹24000. The value of scooter is depreciating at the rate of 5% per annum. Calculate its value after 3 years.

Answer

Depreciation formula,

V = $V_0\Big(1 - \dfrac{r}{100}\Big)^n$

Putting values in formula we get,

$V = ₹24000\Big(1 - \dfrac{5}{100}\Big)^3 \\[1em] = ₹24000 \times \Big(\dfrac{95}{100}\Big)^3 \\[1em] = ₹24000 \times \Big(\dfrac{19}{20}\Big)^3 \\[1em] = ₹24000 \times \dfrac{19}{20} \times \dfrac{19}{20} \times \dfrac{19}{20} \\[1em] = ₹\dfrac{24000 \times 6859}{8000} \\[1em] = ₹(3 \times 6859) \\[1em] = ₹20577.$

Hence, the value of scooter after 3 years = ₹20577.

A farmer increases his output of wheat in his farm every year by 8%. This year he produced 2187 quintals of wheat. What was the yearly produce of wheat two years ago?

Answer

Let yearly produce 2 years ago be P. Then present production will be the final production.

By formula,

V = $V_0\Big(1 + \dfrac{r}{100}\Big)^n$

Putting values in formula we get,

$\Rightarrow 2187 = P \times \Big(1 + \dfrac{8}{100}\Big)^2 \\[1em] \Rightarrow 2187 = P \times \Big(\dfrac{108}{100}\Big)^2 \\[1em] \Rightarrow 2187 = P \times \Big(\dfrac{27}{25}\Big)^2 \\[1em] \Rightarrow 2187 = P \times \dfrac{27}{25} \times \dfrac{27}{25} \\[1em] \Rightarrow 2187 = P \times \dfrac{729}{625} \\[1em] \Rightarrow P = \dfrac{2187 \times 625}{729} \\[1em] \Rightarrow P = 3 \times 625 \\[1em] \Rightarrow P = 1875.$

Hence, the yearly produce of wheat two years ago was 1875 quintals.

The value of a property decreases every year at the rate of 5%. If its present value is ₹411540, what was its value three years ago?

Answer

Depreciation formula,

V = $V_0\Big(1 - \dfrac{r}{100}\Big)^n$

Let value three years ago be V₀.

Putting values in formula we get,

$\Rightarrow ₹411540 = V_0\Big(1 - \dfrac{5}{100}\Big)^3 \\[1em] \Rightarrow ₹411540 = V_0\Big(\dfrac{95}{100}\Big)^3 \\[1em] \Rightarrow ₹411540 = V_0\Big(\dfrac{19}{20}\Big)^3 \\[1em] \Rightarrow ₹411540 = V_0 \times \dfrac{19}{20} \times \dfrac{19}{20} \times \dfrac{19}{20} \\[1em] \Rightarrow V_0 = ₹\dfrac{411540 \times 20 \times 20 \times 20}{19 \times 19 \times 19} \\[1em] \Rightarrow V_0 = ₹\dfrac{411540 \times 8000}{6859} \\[1em] \Rightarrow V_0 = ₹(60 \times 8000) \\[1em] \Rightarrow V_0 = ₹480000.$

Hence, the value of property 3 years ago was ₹480000.

Ahmed purchased an old scooter for ₹16000. If the cost of the scooter after 2 years depreciates to ₹14440, find the rate of depreciation.

Answer

Let rate of depreciation be r%.

Depreciation formula,

V = $V_0\Big(1 - \dfrac{r}{100}\Big)^n$

Putting values in formula we get,

$\Rightarrow 14440 = 16000\Big(1 - \dfrac{r}{100}\Big)^2 \\[1em] \dfrac{14440}{16000} = \Big(1 - \dfrac{r}{100}\Big)^2 \\[1em] \dfrac{1444}{1600} = \Big(1 - \dfrac{r}{100}\Big)^2 \\[1em] \Big(\dfrac{38}{40}\Big)^2 = \Big(1 - \dfrac{r}{100}\Big)^2 \\[1em] 1 - \dfrac{r}{100} = \dfrac{38}{40} \\[1em] \dfrac{r}{100} = 1 - \dfrac{38}{40} \\[1em] \dfrac{r}{100} = \dfrac{2}{40} \\[1em] r = \dfrac{200}{40} \\[1em] r = 5$

Hence, the rate of depreciation = 5%.

A factory increased its production of cars from 80000 in the year 2011-2012 to 92610 in 2014-2015. Find the annual rate of growth of production of cars.

Answer

Let rate of growth be r% per annum.

Growth formula,

V = $V_0\Big(1 + \dfrac{r}{100}\Big)^n$

Putting values in formula we get,

$\Rightarrow 92610 = 80000\Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \dfrac{92610}{80000} = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \dfrac{9261}{8000} = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Big(\dfrac{21}{20}\Big)^3 = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] 1 + \dfrac{r}{100} = \dfrac{21}{20} \\[1em] \dfrac{r}{100} = \dfrac{21}{20} - 1 \\[1em] \dfrac{r}{100} = \dfrac{21 - 20}{20} \\[1em] \dfrac{r}{100} = \dfrac{1}{20} \\[1em] r = \dfrac{100}{20} \\[1em] r = 5$

Hence, the annual rate of growth of production of cars is 5%.

The value of a machine worth ₹500000 is depreciating at the rate of 10% every year. In how many years will its value be reduced to ₹364500?

Answer

Let value be depreciated from ₹500000 to ₹364500 in n years.

Depreciation formula,

V = $V_0\Big(1 - \dfrac{r}{100}\Big)^n$

Putting values in formula we get,

$\Rightarrow 364500 = 500000\Big(1 - \dfrac{10}{100}\Big)^n \\[1em] \Rightarrow \dfrac{364500}{500000} = \Big(\dfrac{100 - 10}{100}\Big)^n \\[1em] \Rightarrow \dfrac{729}{1000} = \Big(\dfrac{90}{100}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{9}{10}\Big)^3 = \Big(\dfrac{9}{10}\Big)^n \\[1em] \therefore n = 3.$

Hence, the value of machine will be depreciated from ₹500000 to ₹364500 in 3 years.

Mahindra set up a factory by investing ₹2500000. During the first two years, his profits were 5% and 10% respectively. If each year the profit was on previous year's capital, calculate his total profit.

Answer

When rate of interest are different then total amount is given by,

$A = P\Big(1 + \dfrac{r_1}{100}\Big)\Big(1 + \dfrac{r_2}{100}\Big)$

Substituting values we get,

$A = ₹2500000\Big(1 + \dfrac{5}{100}\Big)\Big(1 + \dfrac{10}{100}\Big) \\[1em] = ₹2500000\Big(1 + \dfrac{1}{20}\Big)\Big(1 + \dfrac{1}{10}\Big) \\[1em] = ₹2500000 \times \dfrac{21}{20} \times \dfrac{11}{10} \\[1em] = \dfrac{₹2500000 \times 21 \times 11}{20 \times 10} \\[1em] = ₹12500 \times 21 \times 11 \\[1em] = ₹2887500.$

Profit = Amount - Principal = ₹2887500 - ₹2500000 = ₹387500

Hence, profit = ₹387500.

The value of a property is increasing at the rate of 25% every year. By what percent will the value of the property increase after 3 years?

Answer

Let initial value of property be V₀.

By growth formula,

$V = V_0\Big(1 + \dfrac{r}{100}\Big)^n.$

Substituting values we get,

$V = V_0\Big(1 + \dfrac{25}{100}\Big)^3 \\[1em] = V_0\Big(1 + \dfrac{1}{4}\Big)^3 \\[1em] = V_0\Big(\dfrac{5}{4}\Big)^3 \\[1em] = V_0 \times \dfrac{125}{64} \\[1em] = \dfrac{125V_0}{64}.$

Change in value (C) = Final value - Initial value.

$\therefore C = \dfrac{125V_0}{64} - V_0 \\[1em] = \dfrac{125V_0 - 64V_0}{64} \\[1em] = \dfrac{61V_0}{64}.$

Percentage increase = $\dfrac{\text{Change in value}}{\text{Original value}} \times 100.$

Substituting values we get,

$\text{Percentage increase} = \dfrac{\dfrac{61V_0}{64}}{V_0} \times 100 \\[1em] = \dfrac{61V_0}{64V_0} \times 100 \\[1em] = \dfrac{6100}{64} \\[1em] = \dfrac{1525}{16} \\[1em] = 95\dfrac{5}{16}$

Hence, after 3 years the value of property will increase by $95\dfrac{5}{16}$

Mr. Durani bought a plot of land for ₹180000 and a car for ₹320000 at the same time. The value of the plot of land grows uniformly at the rate of 30% p.a., while the value of the car depreciates by 20% in the first year and by 15% p.a. thereafter. If he sells the plot of land as well as the car after 3 years, what will be his profit or loss?

Answer

Since, value of land grows uniformly at the rate of 30% p.a. hence, by growth formula,

$V = V_0\Big(1 + \dfrac{r}{100}\Big)^n$.

Substituting values we get value of land after 3 years,

$V = ₹180000\Big(1 + \dfrac{30}{100}\Big)^3 \\[1em] = ₹180000\Big(\dfrac{130}{100}\Big)^3 \\[1em] = ₹180000\Big(\dfrac{13}{10}\Big)^3 \\[1em] = ₹180000 \times \dfrac{13}{10} \times \dfrac{13}{10} \times \dfrac{13}{10} \\[1em] = ₹180000 \times \dfrac{2197}{1000} \\[1em] = ₹180 \times 2197 \\[1em] = ₹395460.$

Given, the value of the car depreciates by 20% in the first year and by 15% p.a. thereafter,

Hence, value of car after 3 years = $V = V_0\Big(1 - \dfrac{r_1}{100}\Big)\Big(1 - \dfrac{r_2}{100}\Big)\Big(1 - \dfrac{r_3}{100}\Big)$

Substituting values, we get value of car after 3 years,

$V = ₹320000\Big(1 - \dfrac{20}{100}\Big)\Big(1 - \dfrac{15}{100}\Big)\Big(1 - \dfrac{15}{100}\Big) \\[1em] = ₹320000 \times \dfrac{80}{100} \times \dfrac{85}{100} \times \dfrac{85}{100} \\[1em] = ₹(32 \times 4 \times 17 \times 85) \\[1em] = ₹184960.$

Present value of land and car = ₹180000 + ₹320000 = ₹500000.

Total value of land and car after 3 years = ₹395460 + ₹184960 = ₹580420.

Profit = Amount - Initial value = ₹580420 - ₹500000 = ₹80420.

After 3 years profit of Mr. Durani would be ₹80420.