Compound Interest [Applications] — Exercise 3(B)

On ₹ 6000, the difference between C.I. and S.I. in 2 years and at 10% compound interest, compounded per year, is :

1. ₹ 120

2. ₹ 600

3. ₹ 60

4. ₹ 180

Answer

For S.I. :

Principal = ₹ 6000

Rate = 10%

Time = 2 years

S.I. = $\dfrac{6000 \times 10 \times 2}{100}$ = ₹ 1200.

For C.I. :

C.I. = A - P

$= P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = 6000 \times \Big(1 + \dfrac{10}{100}\Big)^2 - 6000 \\[1em] = 6000 \times \Big(\dfrac{110}{100}\Big)^2 - 6000 \\[1em] = 6000 \times \Big(\dfrac{11}{10}\Big)^2 - 6000 \\[1em] = 6000 \times \dfrac{121}{100} - 6000 \\[1em] = 7260 - 6000 \\[1em] = ₹ 1260.$

Difference between C.I. and S.I. = ₹ 1260 - ₹ 1200 = ₹ 60.

Hence, Option 3 is the correct option.

₹ 10000 amounts to ₹ 12500 in one year. The rate of interest per year is :

1. 15%

2. 12.5%

3. 20%

4. 25%

Answer

Let rate of interest be r%.

Given,

P = ₹ 10000

A = ₹ 12500

n = 1 year

By formula,

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

Substituting values we get :

$\Rightarrow 12500 = 10000 \times \Big(1 + \dfrac{r}{100}\Big)^1 \\[1em] \Rightarrow \dfrac{12500}{10000} = \Big(1 + \dfrac{r}{100}\Big) \\[1em] \Rightarrow \dfrac{5}{4} = 1 + \dfrac{r}{100}\\[1em] \Rightarrow \dfrac{5}{4} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{5 - 4}{4} = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{1}{4} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{100}{4} = 25$

Hence, Option 4 is the correct option.

The C.I. on ₹ 16000 in 2 years at the rate of 20% per annum is :

1. ₹ 19360

2. ₹ 7040

3. ₹ 23040

4. ₹ 22400

Answer

Given,

P = ₹ 16000

n = 2 years

r = 20%

By formula,

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

C.I. = A - P

$C.I. = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = 16000 \times \Big(1 + \dfrac{20}{100}\Big)^2 - 16000 \\[1em] = 16000 \times \Big(\dfrac{120}{100}\Big)^2 - 16000 \\[1em] = 16000 \times \Big(\dfrac{6}{5}\Big)^2 - 16000 \\[1em] = 16000 \times \dfrac{36}{25} - 16000 \\[1em] = 23040 - 16000 \\[1em] = ₹ 7040.$

Hence, Option 2 is the correct option.

Simple interest, at the same rate for the same period as given above in part (c) is :

1. ₹ 7040

2. ₹ 6400

3. ₹ 3200

4. ₹ 1280

Answer

By formula,

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

Substituting values we get :

$S.I. = \dfrac{16000 \times 20 \times 2}{100} \\[1em] = ₹ 6400.$

Hence, Option 2 is the correct option.

The difference between C.I. and S.I. in 2 years as given above for parts (c) and (d) is :

1. ₹ 640

2. ₹ 3840

3. ₹ 1920

4. ₹ 1280

Answer

From part (c), we get :

C.I. = ₹ 7040

From part (d), we get :

S.I. = ₹ 6400

C.I. - S.I. = ₹ 7040 - ₹ 6400 = ₹ 640.

Hence, Option 1 is the correct option.

The difference between simple interest and compound interest on a certain sum is ₹ 54.40 for 2 years at 8 percent per annum. Find the sum.

Answer

Given,

n = 2 years

r = 8%

Let sum of money be ₹ P.

C.I. = A - P

$= P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = P\Big(1 + \dfrac{8}{100}\Big)^2 - P \\[1em] = P \times \Big(\dfrac{108}{100}\Big)^2 - P \\[1em] = P \times \Big(\dfrac{27}{25}\Big)^2 - P\\[1em] = P \times \dfrac{729}{625} - P \\[1em] = \dfrac{729P}{625} - P \\[1em] = \dfrac{729P - 625P}{625} \\[1em] = \dfrac{104P}{625}.$

By formula,

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{P \times 8 \times 2}{100} \\[1em] = \dfrac{4P}{25}.$

Given,

Difference between S.I. and C.I. = ₹ 54.40

$\Rightarrow \dfrac{104P}{625} - \dfrac{4P}{25} = 54.40 \\[1em] \Rightarrow \dfrac{104P - 100P}{625} = 54.40 \\[1em] \Rightarrow \dfrac{4P}{625} = 54.40 \\[1em] \Rightarrow P = \dfrac{54.40 \times 625}{4} \\[1em] \Rightarrow P = ₹ 8500.$

Hence, sum = ₹ 8500.

Pramod and Anand each lent the same sum of money for 2 years at 5% at simple interest and compound interest respectively. Anand received ₹ 15 more than Pramod. Find the amount of money lent by each and the interest received.

Answer

Let sum of money be ₹ x.

For Pramod :

P = ₹ x

Time = 2 years

Rate = 5%

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

Substituting values we get :

$S.I. = \dfrac{x \times 5 \times 2}{100} \\[1em] = \dfrac{x}{10}.$

For Anand :

P = ₹ x

Time (n) = 2 years

Rate (r) = 5%

C.I. = A - P

$C.I. = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = x \times \Big(1 + \dfrac{5}{100}\Big)^2 - x \\[1em] = x \times \Big(\dfrac{105}{100}\Big)^2 - x \\[1em] = x \times \Big(\dfrac{21}{20}\Big)^2 - x \\[1em] = x \times \dfrac{441}{400} - x \\[1em] = \dfrac{441x}{400} - x \\[1em] = \dfrac{441x - 400x}{400} \\[1em] = \dfrac{41x}{400}.$

Given,

Anand received ₹ 15 more than Pramod.

∴ C.I. - S.I. = ₹ 15

$\Rightarrow \dfrac{41x}{400} - \dfrac{x}{10} = 15 \\[1em] \Rightarrow \dfrac{41x - 40x}{400} = 15 \\[1em] \Rightarrow \dfrac{x}{400} = 15 \\[1em] \Rightarrow x = 400 \times 15 = ₹ 6000. \\[1em] S.I. = \dfrac{x}{10} = \dfrac{6000}{10} = ₹ 600. \\[1em] C.I. = \dfrac{41x}{400} = \dfrac{41 \times 6000}{400} = ₹ 615.$

Hence, sum lent by each = ₹ 6000 and interest received by Pramod = ₹ 600 and Anand = ₹ 615.

Simple interest on a sum of money for 2 years at 4% is ₹ 450. Find the compound interest on the same sum and at the same rate for 2 years.

Answer

Let the sum be ₹ x.

Given,

Simple interest on the sum of money for 2 years at 4% is ₹ 450.

By formula,

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

Substituting values we get :

$\Rightarrow 450 = \dfrac{x \times 4 \times 2}{100} \\[1em] \Rightarrow x = \dfrac{450 \times 100}{8} \\[1em] \Rightarrow x = 225 \times 25 \\[1em] \Rightarrow x = ₹5625.$

By formula,

C.I. = A - P

$= P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = 5625 \times \Big(1 + \dfrac{4}{100}\Big)^2 - 5625 \\[1em] = 5625 \times \Big(\dfrac{104}{100}\Big)^2 - 5625 \\[1em] = 5625 \times \Big(\dfrac{26}{25}\Big)^2 - 5625 \\[1em] = 5625 \times \dfrac{676}{625} - 5625 \\[1em] = 6084 - 5625 \\[1em] = ₹ 459.$

Hence, the compound interest = ₹ 459.

Compound interest on a certain sum of money at 5% per annum for two years is ₹ 246. Calculate simple interest on the same sum for 3 years at 6% per annum.

Answer

Let sum of money be ₹ x.

By formula,

$\Rightarrow C.I. = A - P \\[1em] \Rightarrow 246 = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] \Rightarrow 246 = x \times \Big(1 + \dfrac{5}{100}\Big)^2 - x \\[1em] \Rightarrow 246 = x \times \Big(\dfrac{105}{100}\Big)^2 - x \\[1em] \Rightarrow 246 = x \times \Big(\dfrac{21}{20}\Big)^2 - x \\[1em] \Rightarrow 246 = x \times \dfrac{441}{400} - x \\[1em] \Rightarrow 246 = \dfrac{441x}{400} - x \\[1em] \Rightarrow 246 = \dfrac{441x - 400x}{400} \\[1em] \Rightarrow 246 = \dfrac{41x}{400} \\[1em] \Rightarrow x = \dfrac{246 \times 400}{41} \\[1em] \Rightarrow x = ₹ 2400.$

For S.I. :

P = x = ₹ 2400

R = 6%

T = 3 years

By formula,

$\Rightarrow S.I. = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow S.I. = \dfrac{x \times 6 \times 3}{100} \\[1em] \Rightarrow S.I. = \dfrac{2400 \times 6 \times 3}{100} \\[1em] \Rightarrow S.I. = ₹ 432.$

Hence, simple interest = ₹ 432.

A sum of money, invested at compounded interest, amounts to ₹ 19360 in 2 years and to ₹ 23425.60 in 4 years. Find the rate percent and the original sum of money.

Answer

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

By formula,

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

Given,

Sum amounts to ₹ 19360 in two years.

$\therefore P\Big(1 + \dfrac{r}{100}\Big)^2 = 19360$ ......(1)

Given,

Sum amounts to ₹ 23425.60 in four years.

$\therefore P\Big(1 + \dfrac{r}{100}\Big)^4 = 23425.60$ ......(2)

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

$\Rightarrow \dfrac{P\Big(1 + \dfrac{r}{100}\Big)^4}{P\Big(1 + \dfrac{r}{100}\Big)^2} = \dfrac{23425.60}{19360} \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^2 = \dfrac{146.41}{121} \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^2 = \Big(\dfrac{12.1}{11}\Big)^2 \\[1em] \Rightarrow 1 + \dfrac{r}{100} = \dfrac{12.1}{11} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{12.1}{11} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{12.1 - 11}{11} \\[1em] \Rightarrow r = \dfrac{1.1}{11} \times 100 \\[1em] \Rightarrow r = 10$

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

$\Rightarrow P\Big(1 + \dfrac{r}{100}\Big)^2 = 19360 \\[1em] \Rightarrow P\Big(1 + \dfrac{10}{100}\Big)^2 = 19360 \\[1em] \Rightarrow P\Big(\dfrac{110}{100}\Big)^2 = 19360 \\[1em] \Rightarrow P \times \Big(\dfrac{11}{10}\Big)^2 = 19360 \\[1em] \Rightarrow P \times \dfrac{121}{100} = 19360 \\[1em] \Rightarrow P = \dfrac{19360 \times 100}{121} \\[1em] \Rightarrow P = 160 \times 100 \\[1em] \Rightarrow P = ₹ 16000.$

Hence, sum of money = ₹ 16000 and rate of interest = 10%.

A sum of money let out at C.I. at a certain rate per annum becomes three times of itself in 8 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 8 years.

$\therefore 3P = P\Big(1 + \dfrac{r}{100}\Big)^8 \\[1em] \Rightarrow \dfrac{3P}{P} = \Big(1 + \dfrac{r}{100}\Big)^8 \\[1em] \Rightarrow 3 = \Big(1 + \dfrac{r}{100}\Big)^8 .....(1)$

Let in n years money becomes 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)

$\Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = \Big[\Big(1 + \dfrac{r}{100}\Big)^8\Big]^3 \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n =\Big(1 + \dfrac{r}{100}\Big)^{24} \\[1em] \Rightarrow n = 24 \text{ years}.$

Hence, in 24 years money will becomes 27 times of itself.

On what sum of money will compound interest (payable annually) for 2 years be the same as simple interest on ₹ 9430 for 10 years, both at the rate of 5 percent per annum ?

Answer

For S.I. :

P = ₹ 9430

T = 10 years

R = 5%

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{9430 \times 5 \times 10}{100} \\[1em] = ₹ 4715.$

Since, C.I. = S.I. = ₹ 4715

Let sum on which C.I. = ₹ 4715 for 2 years at 5% be ₹x.

$\Rightarrow C.I. = A - P \\[1em] \Rightarrow 4715 = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] \Rightarrow 4715 = x \times \Big(1 + \dfrac{5}{100}\Big)^2 - x \\[1em] \Rightarrow 4715 = x \times \Big(\dfrac{105}{100}\Big)^2 - x \\[1em] \Rightarrow 4715 = x \times \Big(\dfrac{21}{20}\Big)^2 - x \\[1em] \Rightarrow 4715 = \dfrac{441x}{400} - x \\[1em] \Rightarrow 4715 = \dfrac{441x - 400x}{400} \\[1em] \Rightarrow 4715 = \dfrac{41x}{400} \\[1em] \Rightarrow x = \dfrac{4715 \times 400}{41} \\[1em] \Rightarrow x = 115 \times 400 \\[1em] \Rightarrow x = ₹ 46000.$

Hence, sum of money = ₹ 46000.

Simple interest on a certain sum of money for 4 years at 4% per annum exceeds the compound interest on the same sum for 3 years at 5 percent per annum by ₹ 228. Find the sum.

Answer

Let sum of money be ₹ x.

For S.I. :

P = ₹ x

Time (T) = 4 years

Rate of interest (R) = 4%

By formula,

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

Substituting values we get :

$\Rightarrow S.I. = \dfrac{x \times 4 \times 4}{100} \\[1em] = \dfrac{4x}{25}.$

For C.I. :

P = ₹ x

Rate of interest (r) = 5%

Time (n) = 3 years

By formula,

$\Rightarrow C.I. = A - P \\[1em] \Rightarrow C.I. = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = x \times \Big(1 + \dfrac{5}{100}\Big)^3 - x \\[1em] = x \times \Big(\dfrac{105}{100}\Big)^3 - x \\[1em] = x \times \Big(\dfrac{21}{20}\Big)^3 - x \\[1em] = \dfrac{9261x}{8000} - x \\[1em] = \dfrac{9261x - 8000x}{8000} \\[1em] = \dfrac{1261x}{8000}.$

Given,

Simple interest on ₹ x for 4 years at 4% per annum exceeds the compound interest on the same sum for 3 years at 5 percent per annum by ₹ 228.

$\therefore \dfrac{4x}{25} - \dfrac{1261x}{8000} = 228 \\[1em] \Rightarrow \dfrac{1280x - 1261x}{8000} = 228 \\[1em] \Rightarrow \dfrac{19x}{8000} = 228 \\[1em] \Rightarrow x = \dfrac{228 \times 8000}{19} \\[1em] \Rightarrow x = ₹ 96000.$

Hence, sum of money = ₹ 96000.

A certain sum of money amounts to ₹ 23400 in 3 years at 10% per annum simple interest. Find the amount of the same sum in 2 years at 10% p.a. compound interest.

Answer

Let sum of money be ₹ x.

Given,

It amounts to ₹ 23400 in 3 years at 10% per annum simple interest.

By formula,

$\Rightarrow A = P + S.I. \\[1em] \Rightarrow 23400 = x + \dfrac{x \times 10 \times 3}{100} \\[1em] \Rightarrow 23400 = x + \dfrac{3x}{10} \\[1em] \Rightarrow \dfrac{13x}{10} = 23400 \\[1em] \Rightarrow x = \dfrac{23400 \times 10}{13} \\[1em] \Rightarrow x = ₹ 18000.$

In case of compound interest :

By formula,

$A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = 18000 \times \Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] = 18000 \times \Big(\dfrac{110}{100}\Big)^2 \\[1em] = 18000 \times \Big(\dfrac{11}{10}\Big)^2 \\[1em] = 18000 \times \Big(\dfrac{121}{100}\Big) \\[1em] = ₹ 21780.$

Hence, amount on same sum in 2 years at 10% p.a. compound interest = ₹ 21780.

Mohit borrowed a certain sum at 5% per annum compound interest and cleared this loan by paying ₹ 12600 at the end of the first year and ₹ 17640 at the end of the second year. Find the sum borrowed.

Answer

Let Mohit borrowed ₹ x and ₹ y.

By formula,

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

For ₹ 12600,

P = ₹ x

A = ₹ 12600

r = 5%

n = 1 year

Substituting values in formula we get :

$\Rightarrow 12600 = x \times \Big(1 + \dfrac{5}{100}\Big)^1 \\[1em] \Rightarrow 12600 = x \times \dfrac{105}{100} \\[1em] \Rightarrow x = \dfrac{12600 \times 100}{105} \\[1em] \Rightarrow x = ₹ 12000.$

For ₹ 17640,

P = ₹ y

A = ₹ 17640

r = 5%

n = 2 year

Substituting values in formula we get :

$\Rightarrow 17640 = y \times \Big(1 + \dfrac{5}{100}\Big)^2 \\[1em] \Rightarrow 17640 = y \times \Big(\dfrac{105}{100}\Big)^2 \\[1em] \Rightarrow 17640 = y \times \Big(\dfrac{21}{20}\Big)^2 \\[1em] \Rightarrow y = \dfrac{17640 \times 20^2}{21^2} \\[1em] \Rightarrow y = ₹ 16000.$

Total money borrowed = ₹ (x + y) = ₹ (12000 + 16000) = ₹ 28000.

Hence, sum borrowed = ₹ 28000.