Compound Interest [Applications] — Exercise 3(D)

When the present population (P) of a certain locality increases by r% per year, the population in n years will be :

1. $P\Big(1 + \dfrac{r}{100}\Big)^n$ - P

2. $P\Big(1 + \dfrac{r}{100}\Big)^n$ + P

3. $P\Big(1 + \dfrac{r}{100}\Big)^n$

4. $P\Big(1 + \dfrac{r}{100n}\Big)$

Answer

For growth :

Value after n years = Present value $\times \Big(1 + \dfrac{r}{100}\Big)^n = P \times \Big(1 + \dfrac{r}{100}\Big)^n.$

Hence, Option 3 is the correct option.

The population of a town decreases by 10% in a particular year and then increases by 15% in the next year. The population at the end of two years is :

1. $\Big(1 + \dfrac{10}{100}\Big)^3\Big(1 + \dfrac{15}{100}\Big)$ times

2. $\Big(1 - \dfrac{10}{100}\Big)\Big(1 + \dfrac{15}{100}\Big)$ times

3. $\Big(1 - \dfrac{10}{100}\Big)\Big(1 - \dfrac{15}{100}\Big)$ times

4. $\Big(1 + \dfrac{10}{100}\Big)\Big(1 - \dfrac{15}{100}\Big)$ times

Answer

Given,

The population of a town decreases by 10% in a particular year and then increases by 15% in the next year.

Let present population be P.

$\text{Population after 2 years} = P \Big(1 - \dfrac{10}{100}\Big)\Big(1 + \dfrac{15}{100}\Big).$

Hence, Option 2 is the correct option.

When the cost of a machine decreases by r% per year; the cost of machine in 3 years is :

1. $\Big(1 + \dfrac{r}{100}\Big)^3$ times

2. $\Big(1 - \dfrac{r}{100}\Big)^3$ times

3. $\Big(1 - \dfrac{r}{100}\Big)^2$ times

4. $\Big(1 + \dfrac{r}{100}\Big)^2$ times

Answer

For depreciation :

Value after n years = Present value $\times \Big(1 - \dfrac{r}{100}\Big)^n$

So,

Value of machine after 3 years = Present value $\times \Big(1 - \dfrac{r}{100}\Big)^3$

Hence, Option 2 is the correct option.

On a certain sum the rate of C.I. is x% per annum for the first two years and y% per annum for the next three years. Then the amount after 5 years is :

1. $\Big(1 + \dfrac{x}{100}\Big)\Big(1 + \dfrac{y}{100}\Big)$ times

2. $\Big(1 + \dfrac{x}{100}\Big)^3\Big(1 + \dfrac{y}{100}\Big)^2$ times

3. $\Big(1 + \dfrac{x}{100}\Big)^2\Big(1 + \dfrac{y}{100}\Big)^3$ times

4. $\Big(1 + \dfrac{x}{100}\Big)^2\Big(1 - \dfrac{y}{100}\Big)^3$ times

Answer

By formula,

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

Given,

On a certain sum the rate of C.I. is x% per annum for the first two years and y% per annum for the next three years.

$\text{Amount after 5 years} = P\Big(1 + \dfrac{x}{100}\Big)^2\Big(1 + \dfrac{y}{100}\Big)^3.$

Hence, Option 3 is the correct option.

The cost (₹ x) of a machine increases by 20% in the first two years and then decreases by 25% in the next two years. Then the cost of machine becomes :

1. ₹ $x \times \dfrac{120}{100} \times \dfrac{75}{100}$

2. ₹ $x \times \Big(\dfrac{120}{100}\Big)^2 \times \dfrac{75}{100}$

3. ₹ $x \times \Big(\dfrac{120}{100}\Big)^2 \times \Big(\dfrac{75}{100}\Big)^2$

4. ₹ $x \times \Big(\dfrac{80}{100}\Big)^2 \times \Big(\dfrac{75}{100}\Big)^2$

Answer

According to question :

The cost (₹ x) of a machine increases by 20% in the first two years and then decreases by 25% in the next two years.

$\Rightarrow x \times \Big(1 + \dfrac{20}{100}\Big)^2 \times \Big(1 - \dfrac{25}{100}\Big)^2 \\[1em] \Rightarrow x \times \Big(\dfrac{120}{100}\Big)^2 \times \Big(\dfrac{75}{100}\Big)^2.$

Hence, Option 3 is the correct option.

The cost of a machine is supposed to depreciate each year by 12% of its value at the beginning of the year. If the machine is valued at ₹ 44,000 at the beginning of 2008, find its value :

(i) at the end of 2009.

(ii) at the beginning of 2007.

Answer

(i) For depreciation :

Value after n years = Present value $\times \Big(1 - \dfrac{r}{100}\Big)^n$

The cost of machine depreciates at the beginning of the next year, or we can say that at end of each year.

So, Value at end of 2009 = Value at beginning of 2010.

$\text{Value at beginning of 2010 }= 44000 \times \Big(1 - \dfrac{12}{100}\Big)^2 \\[1em] = 44000 \times \Big(\dfrac{88}{100}\Big)^2 \\[1em] = 44000 \times \Big(\dfrac{22}{25}\Big)^2 \\[1em] = 44000 \times \dfrac{484}{625} \\[1em] = ₹ 34073.60$

Hence, value at the end of 2009 = ₹ 34073.60

(ii) Let value at beginning of 2007 be ₹ x.

After one year it becomes ₹ 44000.

For depreciation :

Value after n years = Present value $\times \Big(1 - \dfrac{r}{100}\Big)^n$

Substituting values we get :

$\Rightarrow 44000 = x \times \Big(1 - \dfrac{12}{100}\Big)^1 \\[1em] \Rightarrow 44000 = x \times \Big(\dfrac{88}{100}\Big) \\[1em] \Rightarrow 44000 = \dfrac{88x}{100} \\[1em] \Rightarrow x = \dfrac{44000 \times 100}{88} \\[1em] \Rightarrow x = ₹ 50000.$

Hence, value at beginning of 2007 = ₹ 50000.

The value of an article decreased for two years at the rate of 10% per year and then in the third year it increased by 10%. Find the original value of the article, if its value at the end of 3 years is ₹ 40,095.

Answer

Let original value be ₹ x.

Given, it decreases by 10% for two years. So,

For depreciation :

Value after n years = Present value $\times \Big(1 - \dfrac{r}{100}\Big)^n$

$\text{Value after 2 years} = x \times \Big(1 - \dfrac{10}{100}\Big)^2 \\[1em] = x \times \Big(\dfrac{90}{100}\Big)^2\\[1em] = x \times \Big(\dfrac{9}{10}\Big)^2 \\[1em] = x \times \dfrac{81}{100} \\[1em] = \dfrac{81x}{100}.$

Given,

In the third year value is increased by 10%.

For growth :

Value after n year = Present value $\times \Big(1 + \dfrac{r}{100}\Big)^n$

$\text{Value after 1 year} = \dfrac{81x}{100} \times \Big(1 + \dfrac{10}{100}\Big)^1 \\[1em] = \dfrac{81x}{100} \times \dfrac{110}{100} \\[1em] = \dfrac{891x}{1000}.$

Given,

At the end of 3 years value is ₹ 40095.

$\therefore \dfrac{891x}{1000} = 40095 \\[1em] \Rightarrow x = \dfrac{40095 \times 1000}{891} \\[1em] \Rightarrow x = ₹ 45000.$

Hence, original value of article = ₹ 45000.

According to a census taken towards the end of the year 2009, the population of a rural town was found to be 64000. The census authority also found that the population of this particular town had a growth of 5% per annum. In how many years after 2009 did the population of this town reach 74088 ?

Answer

Let in n years population reaches from 64000 to 74088.

For growth :

Value after n year = Present value $\times \Big(1 + \dfrac{r}{100}\Big)^n$

Substituting values we get :

$\Rightarrow 74088 = 64000 \times \Big(1 + \dfrac{5}{100}\Big)^n \\[1em] \Rightarrow \dfrac{74088}{64000} = \Big(\dfrac{105}{100}\Big)^n \\[1em] \Rightarrow \dfrac{9261}{8000} = \Big(\dfrac{21}{20}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{21}{20}\Big)^3 = \Big(\dfrac{21}{20}\Big)^n \\[1em] \Rightarrow n = 3.$

In 3 years the population will reach 74088.

The population of a town decreased by 12% during 1998 and then increased by 8% during 1999. Find the population of the town, at the beginning of 1998, if at the end of 1999 its population was 285120.

Answer

Let population in the beginning of 1998 be x.

Given,

The population of a town decreased by 12% during 1998 and then increased by 8% during 1999.

Population at the end of 1999 = 285120.

$\therefore 285120 = x\Big(1 - \dfrac{12}{100}\Big)\Big(1 + \dfrac{8}{100}\Big) \\[1em] \Rightarrow 285120 = x \times \dfrac{88}{100} \times \dfrac{108}{100} \\[1em] \Rightarrow 285120 = \dfrac{9504x}{10000} \\[1em] \Rightarrow x = \dfrac{285120 \times 10000}{9504} \\[1em] \Rightarrow x = 300000.$

Hence, population of town at beginning of 1998 = 300000.

A sum of money, invested at compound interest, amounts to ₹ 16500 in 1 year and to ₹ 19965 in 3 years. Find the rate per cent and the original sum of money invested.

Answer

Let original sum of money invested be ₹ x and rate of percent be r%.

By formula,

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

Given,

The sum of money, invested at compound interest, amounts to ₹ 16500 in 1 year.

$\Rightarrow A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] \Rightarrow 16500 = x \times \Big(1 + \dfrac{r}{100}\Big)^1 \\[1em] \Rightarrow 16500 = x\Big(1 + \dfrac{r}{100}\Big) ......(1)$

The sum of money, invested at compound interest, amounts to ₹ 19965 in 3 years.

$\Rightarrow A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] \Rightarrow 19965 = x \times \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow 19965 = x\Big(1 + \dfrac{r}{100}\Big)^3 ......(2)$

Dividing equation (2) by (1), we get :

$\Rightarrow \dfrac{19965}{16500} = \dfrac{x\Big(1 + \dfrac{r}{100}\Big)^3}{x\Big(1 + \dfrac{r}{100}\Big)} \\[1em] \Rightarrow \dfrac{121}{100} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{11}{10}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big) = \dfrac{11}{10} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{11}{10} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{1}{10} \\[1em] \Rightarrow r = \dfrac{100}{10} = 10$

Substituting value of r in equation (1), we get :

$\Rightarrow 16500 = x\Big(1 + \dfrac{10}{100}\Big) \\[1em] \Rightarrow 16500 = x \times \Big(\dfrac{110}{100}\Big) \\[1em] \Rightarrow x = \dfrac{16500 \times 100}{110} \\[1em] \Rightarrow x = ₹ 15000.$

Hence, rate percent = 10% and sum invested = ₹ 15000.

The difference between C.I. and S.I. on ₹ 7500 for two years is ₹ 12 at the same rate of interest per annum. Find the rate of interest.

Answer

Given,

P = ₹ 7500

Time = 2 years

Let rate of interest be r%.

By formula,

S.I. = $\dfrac{P \times R \times T}{100}$

Substituting values we get :

$S.I. = \dfrac{7500 \times r \times 2}{100} \\[1em] = 150r.$

By formula,

C.I. = A - P

$C.I. = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = 7500 \times \Big(1 + \dfrac{r}{100}\Big)^2 - 7500 \\[1em] = 7500 \times \Big(\dfrac{100 + r}{100}\Big)^2 - 7500 \\[1em] = 7500 \times \Big(\dfrac{10000 + r^2 + 200r}{10000}\Big) - 7500 \\[1em] = \dfrac{3}{4} \times (10000 + r^2 + 200r) - 7500 \\[1em] = 7500 + \dfrac{3r^2}{4} + 150r - 7500 \\[1em] = \dfrac{3r^2}{4} + 150r$

The difference between C.I. and S.I. on ₹ 7500 for two years is ₹ 12.

∴ C.I. - S.I. = ₹ 12

$\therefore \dfrac{3r^2}{4} + 150r - 150r = 12 \\[1em] \Rightarrow \dfrac{3r^2}{4} = 12 \\[1em] \Rightarrow r^2 = \dfrac{12 \times 4}{3} \\[1em] \Rightarrow r^2 = 16 \\[1em] \Rightarrow r = \sqrt{16} = 4$

Hence, rate of interest = 4%.

A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in 10 years. Find in how many years will the money become twenty-seven times of itself at the same rate of interest p.a.

Answer

Let rate of interest be r% and sum of money be ₹ P.

By formula,

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

Given,

₹ P becomes three times of itself in 10 years.

$\therefore 3P = P\Big(1 + \dfrac{r}{100}\Big)^{10} \\[1em] \Rightarrow \dfrac{3P}{P} = \Big(1 + \dfrac{r}{100}\Big)^{10} \\[1em] \Rightarrow 3 = \Big(1 + \dfrac{r}{100}\Big)^{10} .....(1)$

Let in n years money will become 27 times.

$\Rightarrow P\Big(1 + \dfrac{r}{100}\Big)^n = 27P \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = \dfrac{27P}{P} \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = 27 \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = 3^3 \\[1em]$

From equation (1)

$\Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = \Big[\Big(1 + \dfrac{r}{100}\Big)^{10}\Big]^3 \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n =\Big(1 + \dfrac{r}{100}\Big)^{30} \\[1em] \Rightarrow n = 30 \text{ years}.$

Hence, in 30 years money will becomes 27 times of itself.

Mr. Sharma borrowed a certain sum of money at 10% per annum compounded annually. If by paying ₹ 19360 at the end of second year and ₹ 31944 at the end of the third year he clears the debt; find the sum borrowed by him.

Answer

Let sum borrowed be ₹ x.

By formula,

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

At the end of 2 years :

$A = x \times \Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] = x \times \Big(\dfrac{110}{100}\Big)^2 \\[1em] = x \times \Big(\dfrac{11}{10}\Big)^2 \\[1em] = \dfrac{121x}{100}.$

Given,

He pays back ₹ 19360 at the end of second year. So,

Principal for third year = ₹ $\Big(\dfrac{121x}{100} - 19360\Big)$

Amount after 3 year :

$A = \Big(\dfrac{121x}{100} - 19360\Big) \times \Big(1 + \dfrac{10}{100}\Big)^1 \\[1em] = \Big(\dfrac{121x}{100} - 19360\Big) \times \dfrac{110}{100} \\[1em] = \dfrac{1331x}{1000} - 21296.$

Given,

On giving ₹ 31944 at the end of the third year he clears the debt.

$\therefore \dfrac{1331x}{1000} - 21296 = 31944 \\[1em] \Rightarrow \dfrac{1331x}{1000} = 31944 + 21296 \\[1em] \Rightarrow \dfrac{1331x}{1000} = 53240 \\[1em] \Rightarrow x = \dfrac{53240 \times 1000}{1331} \\[1em] \Rightarrow x = ₹ 40000.$

Hence, sum borrowed = ₹ 40000.

The difference between compound interest for a year payable half-yearly and simple interest on a certain sum of money lent out at 10% for a year is ₹ 15. Find the sum of money lent out.

Answer

Let sum of money lent out be ₹ x.

Calculating C.I. payable half-yearly :

P = ₹ x

Rate of interest = 10%

Time = 1 year

C.I. = A - P

$C.I. = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2} - P \\[1em] = x \times \Big(1 + \dfrac{10}{200}\Big)^{1 \times 2} - x \\[1em] = x \times \Big(\dfrac{210}{200}\Big)^2 - x \\[1em] = \dfrac{441x}{400} - x \\[1em] = \dfrac{441x - 400x}{400} \\[1em] = ₹ \dfrac{41x}{400}.$

Calculating S.I. :

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{x \times 10 \times 1}{100} \\[1em] = ₹ \dfrac{x}{10}.$

Given,

Difference between compound interest for a year payable half-yearly and simple interest on ₹ x lent out at 10% for a year is ₹ 15.

$\therefore \dfrac{41x}{400} - \dfrac{x}{10} = 15 \\[1em] \Rightarrow \dfrac{41x - 40x}{400} = 15 \\[1em] \Rightarrow \dfrac{x}{400} = 15 \\[1em] \Rightarrow x = 15 \times 400 = ₹ 6000.$

Hence, sum of money lent out = ₹ 6000.

The ages of Pramod and Rohit are 16 years and 18 years respectively. In what ratio must they invest money at 5% p.a. compounded yearly so that both get the same sum on attaining the age of 25 years ?

Answer

Let Pramod invest ₹ x and Rohit invest ₹ y.

By formula,

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

For Pramod :

P = ₹ x

r = 5%

n = 9 years (As it will take 9 years for Pramod to reach 25 years of age)

Substituting values we get :

$\text{Amount received by Pramod} = x\Big(1 + \dfrac{5}{100}\Big)^9 \\[1em] = x \times \Big(\dfrac{105}{100}\Big)^9 \\[1em] = x \times \Big(\dfrac{21}{20}\Big)^9.$

For Rohit :

P = ₹ y

r = 5%

n = 7 years (As it will take 7 years for Pramod to reach 25 years of age)

Substituting values we get :

$\text{Amount received by Rohit} = y\Big(1 + \dfrac{5}{100}\Big)^7 \\[1em] = y \times \Big(\dfrac{105}{100}\Big)^7 \\[1em] = y \times \Big(\dfrac{21}{20}\Big)^7.$

Since,

Amount received by both are equal.

$\Rightarrow x \times \Big(\dfrac{21}{20}\Big)^9 = y \times \Big(\dfrac{21}{20}\Big)^7 \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{\Big(\dfrac{21}{20}\Big)^7}{\Big(\dfrac{21}{20}\Big)^9} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{1}{\Big(\dfrac{21}{20}\Big)^2} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{20^2}{21^2} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{400}{441}.$

Hence, ratio in which sum must be invested = 400 : 441.