Compound Interest — Exercise 2(A)

Calculate the amount and the compound interest on ₹ 25,000 for 2 years at 8% per annum, compounded annually.

Answer

For first year :

P = ₹ 25,000

T = 1 year

R = 8%

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

$= \dfrac{25000 \times 8 \times 1}{100}$ = ₹ 2,000.

Amount = P + I = ₹ 25,000 + ₹ 2,000 = ₹ 27,000.

For second year :

P = ₹ 27,000

T = 1 year

R = 8%

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

$= \dfrac{27000 \times 8 \times 1}{100}$ = ₹ 2160.

Amount = P + I = ₹ 27,000 + ₹ 2,160 = ₹ 29,160.

Compound interest = Final amount - Initial principal

= ₹ 29,160 - ₹ 25,000 = ₹ 4,160.

Hence, compound interest = ₹ 4,160 and amount = ₹ 29,160.

Rohit borrows ₹ 62,500 from Arun for 2 years at 10% per annum, simple interest. He immediately lends out this sum to Kunal at 10% per annum for the same period, compounded annually. Calculate Rohit's profit in the transaction at the end of two years.

Answer

For Rohit,

P = ₹ 62,500

T = 2 year

R = 10% per annum simple interest

Interest Rohit pays to Arun:

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

$= \dfrac{62500 \times 10 \times 2}{100}$ = ₹ 12,500.

For Kunal,

For first year :

P = ₹ 62,500

T = 1 year

R = 10% per annum compounded annually

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

$= \dfrac{62500 \times 10 \times 1}{100}$ = ₹ 6,250.

Amount = P + I = ₹ 62,500 + ₹ 6,250 = ₹ 68,750.

For second year :

P = ₹ 68,750

T = 1 year

R = 10% per annum compounded annually

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

$= \dfrac{68750 \times 10 \times 1}{100}$ = ₹ 6,875.

Amount = P + I = ₹ 68,750 + ₹ 6,875 = ₹ 75,625.

Compound interest = Final amount - Initial principal

= ₹ 75,625 - ₹ 62,500 = ₹ 13,125.

∴ Interest Kunal pays to Rohit = ₹ 13,125

Rohit's profit = Compound interest received from Kunal - Simple interest paid to Arun = ₹ 13,125 - ₹ 12,500 = ₹ 625.

Hence, Rohit's profit in the transaction at the end of two years = ₹ 625.

A man invests ₹ 10,000 for 3 years at a certain rate of interest, compounded annually. At the end of one year, it amounts to ₹ 11,200. Calculate :

(i) the rate of interest per annum;

(ii) the interest accrued in the second year;

(iii) the amount at the end of the third year.

Answer

(i) Given,

P = ₹ 10,000

T = 3 year

Amount at the end of first year = ₹ 11,200

Interest in the first year = Amount - Principal

= ₹ 11,200 - ₹ 10,000 = ₹ 1,200.

So, for 1 year interest equals to ₹ 1,200 on ₹ 10,000. Let rate of interest be R%. Substituting values we get :

$\Rightarrow I = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow 1200 = \dfrac{10000 \times R \times 1}{100} \\[1em] \Rightarrow 1200 = 100 \times R \\[1em] \Rightarrow R = \dfrac{1200}{100} \\[1em] \Rightarrow R = 12$

Hence, the rate of interest per annum = 12% p.a.

(ii) Given,

For second year :

P = ₹ 11,200

T = 1 year

R = 12%

Interest accrued in the second year,

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

$= \dfrac{11200 \times 12 \times 1}{100}$

= ₹ 1,344.

Hence, the interest accrued in the second year = ₹ 1,344.

(iii) For third year,

P = ₹ 11,200 + ₹ 1,344 = ₹ 12,544

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

$= \dfrac{12544 \times 12 \times 1}{100}$

= ₹ 1,505.28

Amount at the end of the third year = P + I = ₹ 12,544 + ₹ 1,505.28 = ₹ 14,049.28

Hence, the amount at the end of the third year = ₹ 14,049.28.

Sudhakar borrows ₹ 22,500 at 10% per annum, compounded annually. If he repays ₹ 11,250 at the end of first year and ₹ 12,550 at the end of the second year, find the amount of loan outstanding against him at the end of the third year.

Answer

For first year :

P = ₹ 22,500

T = 1 year

R = 10 %

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

$= \dfrac{22500 \times 10 \times 1}{100}$ = ₹ 2,250.

Amount = P + I = ₹ 22,500 + ₹ 2,250 = ₹ 24,750.

Amount payed at end of first year = ₹ 11,250.

Amount left at beginning of second year = ₹ 24,750 - ₹ 11,250 = ₹ 13,500.

For second year :

P = ₹ 13,500

R = 10%

T = 1 year

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

$= \dfrac{13500 \times 10 \times 1}{100}$ = ₹ 1,350.

Amount = P + I = ₹ 13,500 + ₹ 1,350 = ₹ 14,850.

Amount payed at end of second year = ₹ 12,550.

Amount left at beginning of third year = ₹ 14,850 - ₹ 12,550 = ₹ 2,300

For third year :

P = ₹ 2,300

R = 10%

T = 1 year

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

$= \dfrac{2300 \times 10 \times 1}{100}$ = ₹ 230.

Amount due at the end of third year = P + I = ₹ 2,300 + ₹ 230 = ₹ 2,530.

Hence, the amount outstanding at the end of the third year = ₹ 2,530.

A man borrows ₹ 15,000 at 12% per annum, compounded annually. If he repays ₹ 4,400 at end of each year, find the amount outstanding against him at the beginning of third year.

Answer

For first year :

P = ₹ 15,000

T = 1 year

R = 12%

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

$= \dfrac{15000 \times 12 \times 1}{100}$ = ₹ 1,800.

Amount = P + I = ₹ 15,000 + ₹ 1,800 = ₹ 16,800.

Amount payed at end of first year = ₹ 4,400.

Amount left at beginning of second year = ₹ 16,800 - ₹ 4,400 = ₹ 12,400.

For second year :

P = ₹ 12,400

R = 12%

T = 1 year

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

$= \dfrac{12400 \times 12 \times 1}{100}$ = ₹ 1,488.

Amount = P + I = ₹ 12,400 + ₹ 1,488 = ₹ 13,888.

Amount payed at end of second year = ₹ 4,400.

Amount left at beginning of third year = ₹ 13,888 - ₹ 4,400 = ₹ 9,488.

Hence, amount left at beginning of third year = ₹ 9,488.

Mr. Ravi borrows ₹ 16,000 for 2 years. The rate of interest for the two successive years are 10% and 12% respectively. If he repays ₹ 5,600 at the end of first year, find the amount outstanding at the end of the second year.

Answer

For first year :

P = ₹ 16,000

T = 1 year

R = 10 %

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

$= \dfrac{16000 \times 10 \times 1}{100}$ = ₹ 1,600.

Amount = P + I = ₹ 16,000 + ₹ 1,600 = ₹ 17,600.

Amount payed at end of first year = ₹ 5,600.

Amount left at beginning of second year = ₹ 17,600 - ₹ 5,600 = ₹ 12,000.

For second year :

P = ₹ 12,000

R = 12%

T = 1 year

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

$= \dfrac{12000 \times 12 \times 1}{100}$ = ₹ 1,440.

Amount = P + I = ₹ 12,000 + ₹ 1,440 = ₹ 13,440.

Hence, amount outstanding at end of second year = ₹ 13,440.

Calculate the amount of ₹ 30,000 at the end of 2 years 4 months, compounded annually at 10% per annum.

Answer

For first year :

P = ₹ 30,000

T = 1 year

R = 10%

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

$= \dfrac{30000 \times 10 \times 1}{100}$ = ₹ 3,000.

Amount = P + I = ₹ 30,000 + ₹ 3,000 = ₹ 33,000.

For second year :

P = ₹ 33,000

T = 1 year

R = 10%

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

$= \dfrac{33000 \times 10 \times 1}{100}$ = ₹ 3,300

Amount = P + I = ₹ 33,000 + ₹ 3,300 = ₹ 36,300

For next 4 months :

P = ₹ 36,300

T = 4 months = $\dfrac{4}{12}$ year = $\dfrac{1}{3}$ year

R = 10%

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

$= \dfrac{36300 \times 10 \times \dfrac{1}{3}}{100}$

$= \dfrac{363000}{300}$ = ₹ 1,210.

Amount = P + I = ₹ 36,300 + ₹ 1,210 = ₹ 37,510.

Hence, final amount = ₹ 37,510.

Calculate the amount of ₹ 31,250 at the end of $2\dfrac{1}{2}$ years, compounded annually at 8% per annum.

Answer

For first year :

P = ₹ 31,250

T = 1 year

R = 8%

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

$= \dfrac{31250 \times 8 \times 1}{100}$ = ₹ 2,500.

Amount = P + I = ₹ 31,250 + ₹ 2,500 = ₹ 33,750.

For second year :

P = ₹ 33,750

T = 1 year

R = 8%

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

$= \dfrac{33750 \times 8 \times 1}{100}$ = ₹ 2,700.

Amount = P + I = ₹ 33,750 + ₹ 2,700 = ₹ 36,450.

For next $\dfrac{1}{2}$ year :

P = ₹ 36,450

T = $\dfrac{1}{2}$ year

R = 8%

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

$= \dfrac{36450 \times 8 \times \dfrac{1}{2}}{100}$

$= \dfrac{291600}{200}$ = ₹ 1,458.

Amount = P + I = ₹ 36,450 + ₹ 1,458 = ₹ 37,908.

Hence, final amount = ₹ 37,908.

Calculate the amount and the compound interest on ₹ 15,000 for 2 years compounded annually, the rates of interest for successive years being 8% and 9% per annum respectively.

Answer

For first year :

P = ₹ 15,000

T = 1 year

R = 8%

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

$= \dfrac{15000 \times 8 \times 1}{100}$ = ₹ 1,200.

Amount = P + I = ₹ 15,000 + ₹ 1,200 = ₹ 16,200.

For second year :

P = ₹ 16,200

T = 1 year

R = 9%

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

$= \dfrac{16200 \times 9 \times 1}{100}$ = ₹ 1,458.

Amount = P + I = ₹ 16,200 + ₹ 1,458 = ₹ 17,658.

Compound interest = Final amount - Initial principal

= ₹ 17,658 - ₹ 15,000 = ₹ 2,658.

Hence, final amount = ₹ 17,658 and compound interest = ₹ 2,658.

Calculate the amount and the compound interest on ₹ 25,000 for 3 years compounded annually, the rates of interest for successive years being 8%, 9% and 10% respectively.

Answer

For first year :

P = ₹ 25,000

T = 1 year

R = 8%

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

$= \dfrac{25000 \times 8 \times 1}{100}$ = ₹ 2,000.

Amount = P + I = ₹ 25,000 + ₹ 2,000 = ₹ 27,000.

For second year :

P = ₹ 27,000

T = 1 year

R = 9%

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

$= \dfrac{27000 \times 9 \times 1}{100}$ = ₹ 2,430.

Amount = P + I = ₹ 27,000 + ₹ 2,430 = ₹ 29,430.

For third year :

P = ₹ 29,430

T = 1 year

R = 10%

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

$= \dfrac{29430 \times 10 \times 1}{100}$ = ₹ 2,943.

Amount = P + I = ₹ 29,430 + ₹ 2,943 = ₹ 32,373.

Compound interest = Final amount - Initial principal

= ₹ 32,373 - ₹ 25,000 = ₹ 7,373.

Hence, final amount = ₹ 32,373 and compound interest = ₹ 7,373.

Peter invested ₹ 2,40,000 for 2 years at 10% per annum compounded annually. If 20% of the accrued interest at the end of each year is deducted as income tax, find the amount he received at the end of 2 years.

Answer

For first year :

P = ₹ 2,40,000

T = 1 year

R = 10%

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

$= \dfrac{240000 \times 10 \times 1}{100}$ = ₹ 24,000.

Income tax deducted = 20% of Interest

= $\dfrac{20}{100} \times 24000$ = ₹ 4,800

Interest after deduction = ₹ 24,000 - ₹ 4,800 = ₹ 19,200.

Amount = P + I = ₹ 2,40,000 + ₹ 19,200 = ₹ 2,59,200.

For second year :

P = ₹ 2,59,200

T = 1 year

R = 10%

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

$= \dfrac{259200 \times 10 \times 1}{100}$ = ₹ 25,920.

Income tax deducted = 20% of Interest

= $\dfrac{20}{100} \times 25920$ = ₹ 5,184.

Interest after deduction = ₹ 25,920 - ₹ 5,184 = ₹ 20,736.

Amount = P + I = ₹ 2,59,200 + ₹ 20,736 = ₹ 2,79,936.

Hence, final amount received at the end of 2 years = ₹ 2,79,936.

Find the amount and the compound interest on ₹ 10,000 for 1 year at 12% per annum, compounded half-yearly.

Answer

Given,

Rate = 12%

Half yearly rate (R) = $\dfrac{\text{Rate}}{2} = \dfrac{12}{2}$ = 6%

For first half year :

P = ₹ 10,000

T = 1 half year

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

$= \dfrac{10000 \times 6 \times 1}{100}$ = ₹ 600.

Amount = P + I = ₹ 10,000 + ₹ 600 = ₹ 10,600.

For second half year :

P = ₹ 10,600

T = 1 half year

Half yearly rate = 6%

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

$= \dfrac{10600 \times 6 \times 1}{100}$ = ₹ 636.

Amount = P + I = ₹ 10,600 + ₹ 636 = ₹ 11,236.

Compound interest = Final amount - Initial principal

= ₹ 11,236 - ₹ 10,000 = ₹ 1,236.

Hence, final amount = ₹ 11,236 and compound interest = ₹ 1,236.

Find the amount and the compound interest on ₹ 64,000 for $1\dfrac{1}{2}$ year at 15% per annum, compounded half-yearly.

Answer

Given,

Rate = 15%

Half yearly rate (R) = $\dfrac{\text{Rate}}{2} = \dfrac{15}{2}$ % = 7.5%

Time = $1\dfrac{1}{2}$ year = $\dfrac{3}{2} \times 2$ = 3 half-year.

For first half year :

P = ₹ 64,000

T = 1 half year

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

$= \dfrac{64,000 \times 7.5 \times 1}{100}$ = ₹ 4,800

Amount = P + I = ₹ 64,000 + ₹ 4,800 = ₹ 68,800

For second half year :

P = ₹ 68,800

Half yearly rate (R) = 7.5%

T = 1 half year

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

$= \dfrac{68800 \times 7.5 \times 1}{100}$ = ₹ 5,160.

Amount = P + I = ₹ 68,800 + ₹ 5,160 = ₹ 73,960.

For third half year :

P = ₹ 73,960

Half yearly rate (R) = 7.5%

T = 1 year

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

$= \dfrac{73960 \times 7.5 \times 1}{100}$ = ₹ 5,547.

Amount = P + I = ₹ 73,960 + ₹ 5,547 = ₹ 79,507.

Compound interest = Final amount - Initial principal

= ₹ 79,507 - ₹ 64,000 = ₹ 15,507.

Hence, final amount = ₹ 79,507 and compound interest = ₹ 15,507.

The simple interest on a sum of money for 2 years at 10% p.a. is ₹ 1,700. Find:

(i) the sum of money,

(ii) the compound interest on this sum for 1 year, payable half yearly at the same rate.

Answer

(i) Given,

The simple interest on a sum of money for 2 years at 10% p.a. is ₹ 1700.

I = ₹ 1,700

T = 2 year

R = 10%

Let sum of money be ₹ P.

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

Substituting values we get :

$\Rightarrow 1700 = \dfrac{P \times 10 \times 2}{100} \\[1em] \Rightarrow 1700 = \dfrac{P \times 20}{100}\\[1em] \Rightarrow 1700 = \dfrac{P}{5} \\[1em] \Rightarrow P = 1700 \times 5 = 8,500.$

Hence, the sum of money = ₹ 8,500

(ii) Given,

For first half year :

P = ₹ 8,500

R = 10%

Half yearly rate = $\dfrac{Rate}{2} = \dfrac{10}{2}$ = 5%

T = 1 half year

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

$= \dfrac{8500 \times 5 \times 1}{100} = ₹ 425$

Amount = P + I = ₹ 8,500 + ₹ 425 = ₹ 8,925.

For second half year :

P = ₹ 8,925

Half yearly rate = 5%

T = 1 half year

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

$= \dfrac{8925 \times 5 \times 1}{100} = ₹ 446.25$

Amount = P + I = ₹ 8,925 + ₹ 446.25 = ₹ 9,371.25

Compound interest = Final Amount - Initial Pincipal

= ₹ 9,371.25 - ₹ 8,500

= ₹ 871.25

Hence, compound interest = ₹ 871.25