Compound Interest [Basic Concepts] — Exercise 2(C)

₹ 300 and ₹ 360 are the compound interest for two consecutive years. The rate of interest is :

1. 1.2%

2. 12%

3. 120%

4. 20%

Answer

Difference between C.I. of two successive years = ₹ 360 - ₹ 300 = ₹ 60

∴ ₹ 60 is the interest of one year on ₹ 300.

By formula,

Rate of interest = $\dfrac{100 \times I}{P \times T} = \dfrac{100 \times 60}{300 \times 1}$ = 20%.

Hence, Option 4 is the correct option.

A certain sum of money amounts to ₹ 5000 at the end of 5th year and to ₹ 6000 at the end of 6th year. The rate of interest is :

1. 120%

2. 20%

3. 1.2%

4. 12%

Answer

Difference between amounts of two successive years = ₹ 6000 - ₹ 5000 = ₹ 1000

∴ ₹ 1000 is the interest of one year on ₹ 5000.

By formula,

Rate of interest = $\dfrac{100 \times I}{P \times T} = \dfrac{100 \times 1000}{5000 \times 1}$ = 20%.

Hence, Option 2 is the correct option.

At the end of 2020, the compound interest amounted to ₹ 3850 at 10% C.I. The C.I. on the same sum and at same rate amounted at the end of 2019 was :

1. ₹ 3500

2. ₹ 4235

3. ₹ 3181

4. ₹ 3182

Answer

Let C.I. in 2019 be ₹ x.

Difference in C.I. of successive years = ₹ 3850 - ₹ x.

So,

∴ ₹ (3850 - x) is the interest of one year on ₹ x.

By formula,

Rate of interest = $\dfrac{100 \times I}{P \times T}$

Substituting values we get :

$\Rightarrow 10 = \dfrac{100 \times (3850 - x)}{x \times 1} \\[1em] \Rightarrow 10x = 100(3850 - x) \\[1em] \Rightarrow x = 10(3850 - x) \\[1em] \Rightarrow x = 38500 - 10x \\[1em] \Rightarrow 11x = 38500 \\[1em] \Rightarrow x = \dfrac{38500}{11} \\[1em] \Rightarrow x = 3500.$

Hence, Option 1 is the correct option.

A sum of money, lent out at C.I. amounts to ₹ 4500 in 6 years. If rate of C.I. is 12%, the same money will amount in 7 years to rupees :

1. 540

2. 5040

3. 5400

4. 4725

Answer

Let money amounts to ₹ x.

Difference between amounts of two successive years = ₹ x - ₹ 4500

∴ ₹ (x - 4500) is the interest of one year on ₹ 4500.

By formula,

Rate of interest = $\dfrac{100 \times I}{P \times T}$

Substituting values we get :

$\Rightarrow 12 = \dfrac{100 \times (x - 4500)}{4500 \times 1} \\[1em] \Rightarrow 12 \times 45 = x - 4500 \\[1em] \Rightarrow 540 = x - 4500 \\[1em] \Rightarrow x = 540 + 4500 \\[1em] \Rightarrow x = 5040.$

Hence, Option 2 is the correct option.

For two consecutive years, a sum lent out at C.I. earns ₹ 600 and ₹ 690 respectively. The rate of C.I. is :

1. 12%

2. 18%

3. 15%

4. 5%

Answer

Difference between C.I. of two successive years = ₹ 690 - ₹ 600 = ₹ 90

∴ ₹ 90 is the interest of one year on ₹ 600.

By formula,

Rate of interest = $\dfrac{100 \times I}{P \times T} = \dfrac{100 \times 90}{600 \times 1}$ = 15%.

Hence, Option 3 is the correct option.

For two consecutive years, a sum of money lent out at C.I. amounts to ₹ 2400 and ₹ 2760 respectively. The rate of interest is :

1. 5%

2. 15%

3. 18%

4. 10%

Answer

Difference between amounts of two successive years = ₹ 2760 - ₹ 2400 = ₹ 360

∴ ₹ 360 is the interest of one year on ₹ 2400.

By formula,

Rate of interest = $\dfrac{100 \times I}{P \times T} = \dfrac{100 \times 360}{2400 \times 1}$ = 15%.

Hence, Option 2 is the correct option.

A certain sum amounts to ₹ 5292 in two years and ₹ 5556.60 in three years, interest being compounded annually. Find :

(i) the rate of interest

(ii) the original sum.

Answer

(i) Given,

Amount in two years = ₹ 5292

Amount in three years = ₹ 5556.60

Difference between the amounts of two successive years

= ₹ 5556.60 - ₹ 5292 = ₹ 264.60

∴ ₹ 264.60 is the interest of one year on ₹ 5292.

By formula,

Rate of interest = $\dfrac{I \times 100}{P \times T} = \dfrac{264.60 \times 100}{5292 \times 1} = \dfrac{100}{20}$ = 5%.

Hence, rate of interest = 5%.

(ii) Let original sum be ₹ x.

For 1st year :

P = ₹ x

R = 5%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{x \times 5 \times 1}{100} = \dfrac{x}{20}$.

Amount = P + I = $x + \dfrac{x}{20} = \dfrac{21x}{20}$.

For second year :

P = ₹ $\dfrac{21x}{20}$

R = 5%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{\dfrac{21x}{20} \times 5 \times 1}{100} = \dfrac{21x}{400}$.

Amount = P + I = $\dfrac{21x}{20} + \dfrac{21x}{400} = \dfrac{420x + 21x}{400} = \dfrac{441x}{400}$.

Given,

Amount after 2 years = ₹ 5292

$\therefore \dfrac{441x}{400} = 5292 \\[1em] \Rightarrow x = \dfrac{5292 \times 400}{441} = 4800.$

Hence, original sum = ₹ 4800.

Mohit invests ₹ 8000 for 3 years at a certain rate of interest, compounded annually. At the end of one year, it amounts to ₹ 9440. Calculate :

(i) the rate of interest per annum.

(ii) the amount at the end of the second year.

(iii) the interest accrued in the third year.

Answer

(i) Given,

Mohit invests ₹ 8000 (P)

Amount at end of one year = ₹ 9440

Interest = Amount - P = ₹ 9440 - ₹ 8000 = ₹ 1440.

∴ ₹ 1440 is the interest of one year on ₹ 8000.

By formula,

Rate of interest = $\dfrac{I \times 100}{P \times T} = \dfrac{1440 \times 100}{8000 \times 1} = \dfrac{144}{8}$ = 18%.

Hence, rate of interest = 18%.

(ii) For second year :

P = ₹ 9440

R = 18%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{9440 \times 18 \times 1}{100} = \dfrac{169920}{100}$ = ₹ 1699.20

Amount = P + I = ₹ 9440 + ₹ 1699.20 = ₹ 11,139.20

Hence, amount at the end of second year = ₹ 11,139.20

(iii) For third year :

P = ₹ 11,139.20

R = 18%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{11139.20 \times 18 \times 1}{100} = \dfrac{200505.60}{100}$ = ₹ 2,005.06

Hence, interest accrued in third year = ₹ 2,005.06

The compound interest, calculated yearly, on a certain sum of money for the second year is ₹ 1089 and for the third year it is ₹ 1197.90. Calculate the rate of interest and the sum of money.

Answer

Difference between C.I. of two successive years = ₹ 1197.90 - ₹ 1089 = ₹ 108.9

∴ ₹ 108.9 is the interest of one year on ₹ 1089.

By formula,

Rate of interest = $\dfrac{100 \times I}{P \times T} = \dfrac{100 \times 108.9}{1089 \times 1}$ = 10%.

Let sum of money be ₹ x.

For first year :

P = ₹ x

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{x \times 10 \times 1}{100} = \dfrac{x}{10}$.

A = P + I = $x + \dfrac{x}{10} = \dfrac{11x}{10}$

For second year :

P = ₹ $\dfrac{11x}{10}$

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{\dfrac{11x}{10} \times 10 \times 1}{100} = \dfrac{11x}{100}$.

Given,

C.I. for 2nd year = ₹ 1089

$\therefore \dfrac{11x}{100} = 1089 \\[1em] \Rightarrow x = \dfrac{1089 \times 100}{11} = 9900.$

Hence, rate of interest = 10% and sum of money = ₹ 9900.

A sum is invested at compound interest compounded yearly. If the interest for two successive years be ₹ 5700 and ₹ 7410, calculate the rate of interest.

Answer

Difference between C.I. of two successive years = ₹ 7410 - ₹ 5700 = ₹ 1710

∴ ₹ 1710 is the interest of one year on ₹ 5700.

By formula,

Rate of interest = $\dfrac{100 \times I}{P \times T} = \dfrac{100 \times 1710}{5700 \times 1}$ = 30%.

Hence, the rate of interest = 30%.

The cost of a machine depreciated by ₹ 4000 during the first year and by ₹ 3600 during the second year. Calculate :

(i) the rate of depreciation.

(ii) the original cost of the machine.

(iii) its cost at the end of third year.

Answer

(i) Difference between depreciation in value between the first and second years is ₹ 4,000 - ₹ 3,600 = ₹ 400

So, the depreciation of one year on ₹ 4,000 = ₹ 400

By formula,

Rate of depreciation = $\dfrac{100 \times I}{P \times T} = \dfrac{100 \times 400}{4000 \times 1}$ = 10%.

Hence, rate of depreciation = 10%.

(ii) Let cost of machine be ₹ x.

Given,

Depreciation in first year = ₹ 4000

Depreciation % = 10%

$\therefore \dfrac{x \times 1\times 10}{100} = 4000$

x = 40000.

Hence, original cost of machine = ₹ 40000.

(iii) Value of machine at beginning of third year = Original value - Depreciation in first and second years

= ₹ 40000 - (₹ 4000 + ₹ 3600)

= ₹ 40000 - ₹ 7600

= ₹ 32400.

For third year :

P = ₹ 32400

T = 1 year

Depreciation % = 10%

Depreciation = $\dfrac{P \times R \times T}{100} = \dfrac{32400 \times 10 \times 1}{100}$ = ₹3240.

Value at the end of third year = ₹ 32400 - ₹ 3240 = ₹ 29160.

Hence, value of machine at the end of third year = ₹ 29160.

Ramesh invests ₹ 12800 for three years at the rate of 10% per annum compound interest. Find :

(i) the sum due to Ramesh at the end of the first year.

(ii) the interest he earns for the second year.

(iii) the total amount due to him at the end of third year.

Answer

(i) For first year :

P = ₹ 12800

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{12800 \times 10 \times 1}{100}$ = ₹ 1280.

Amount = P + I = ₹ 12800 + ₹ 1280 = ₹ 14080.

Hence, sum due at the end of first year = ₹ 14080.

(ii) For second year :

P = ₹ 14080

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{14080 \times 10 \times 1}{100}$ = ₹ 1408.

Hence, interest for second year = ₹ 1408.

(iii) Amount at end of second year = P + I = ₹ 14080 + ₹ 1408 = ₹ 15488.

For third year :

P = ₹ 15488

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{15488 \times 10 \times 1}{100}$ = ₹ 1548.80

Amount = P + I = ₹ 15488 + ₹ 1548.80 = ₹ 17036.80

Hence, amount due at end of third year = ₹ 17036.80

A certain sum of money is put at compound interest, compounded half-yearly. If the interest for two successive half-years are ₹ 650 and ₹ 760.50; find the rate of interest.

Answer

Difference between C.I. of two successive half-years = ₹ 760.50 - ₹ 650 = ₹ 110.50

∴ ₹ 110.50 is the interest of $\dfrac{1}{2}$ year on ₹ 650.

By formula,

Rate of interest = $\dfrac{100 \times I}{P \times T} = \dfrac{100 \times 110.50}{650 \times \dfrac{1}{2}} = \dfrac{22100}{650}$ = 34%.

Hence, the rate of interest = 34%.

Geeta borrowed ₹ 15000 for 18 months at a certain rate of interest compounded semi-annually. If at the end of six months it amounted to ₹ 15600; calculate :

(i) the rate of interest per annum.

(ii) the total amount of money that Geeta must pay at the end of 18 months in order to clear the account.

Answer

(i) Difference between C.I. of two successive half-years = ₹ 15600 - ₹ 15000 = ₹ 600

∴ ₹ 600 is the interest of $\dfrac{1}{2}$ year on ₹ 15000.

By formula,

Rate of interest = $\dfrac{100 \times I}{P \times T} = \dfrac{100 \times 600}{15000 \times \dfrac{1}{2}} = \dfrac{120}{15}$ = 8%

Hence, the rate of interest = 8%.

(ii) For 2nd half-year :

P = ₹ 15600

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

R = 8%

I = $\dfrac{P \times R \times T}{100} = \dfrac{15600 \times 8 \times \dfrac{1}{2}}{100}$ = ₹ 624.

Amount = P + I = ₹ 15600 + ₹ 624 = ₹ 16224.

For 3rd half-year :

P = ₹ 16224

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

R = 8%

I = $\dfrac{P \times R \times T}{100} = \dfrac{16224 \times 8 \times \dfrac{1}{2}}{100}$ = ₹ 648.96

Amount = P + I = ₹ 16224 + ₹ 648.96 = ₹ 16872.96

Hence, amount needed to pay at the end of 18 months = ₹ 16872.96

₹ 8000 is lent out at 7% compound interest for 2 years. At the end of the first year ₹ 3560 are returned. Calculate :

(i) the interest paid for the second year.

(ii) the total interest paid in two years

(iii) the total amount of money paid in two years to clear the debt.

Answer

(i) For first year :

P = ₹ 8000

R = 7%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{8000 \times 7 \times 1}{100}$ = ₹ 560.

Amount = P + I = ₹ 8000 + ₹ 560 = ₹ 8560.

Amount paid back at end of first year = ₹ 3560

Amount left = ₹ 8560 - ₹ 3560 = ₹ 5000.

For second year :

P = ₹ 5000

R = 7%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{5000 \times 7 \times 1}{100}$ = ₹ 350.

Amount = P + I = ₹ 5000 + ₹ 350 = ₹ 5350

Hence interest paid in second year = ₹ 350.

(ii) Total interest paid in two years = ₹ 350 + ₹ 560 = ₹ 910.

Hence, total interest paid in two years = ₹ 910.

(iii) Amount of money paid in two years to clear the debt = Amount at end of 2nd year + Money paid back at end of first year

= ₹ 5350 + ₹ 3560 = ₹ 8910.

Hence, total amount of money paid in two years to clear the debt = ₹ 8910.

Find the sum invested at 10% compounded annually, on which the interest for the third year, exceeds the interest of the first year by ₹ 252.

Answer

Let sum of money be ₹ x.

For first year :

P = ₹ x

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{x \times 10 \times 1}{100} = \dfrac{x}{10}$

Amount = P + I = $x + \dfrac{x}{10} = \dfrac{11x}{10}$

For second year :

P = ₹ $\dfrac{11x}{10}$

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{\dfrac{11x}{10} \times 10 \times 1}{100} = \dfrac{11x}{100}$

Amount = P + I = $\dfrac{11x}{10} + \dfrac{11x}{100} = \dfrac{110x + 11x}{100} = \dfrac{121x}{100}$.

For third year :

P = ₹ $\dfrac{121x}{100}$

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{\dfrac{121x}{100} \times 10 \times 1}{100} = \dfrac{121x}{1000}$.

Given,

Interest for the third year exceeds the interest of the first year by ₹ 252.

$\therefore \dfrac{121x}{1000} - \dfrac{x}{10} = 252 \\[1em] \Rightarrow \dfrac{121x - 100x}{1000} = 252 \\[1em] \Rightarrow \dfrac{21x}{1000} = 252 \\[1em] \Rightarrow x = \dfrac{252 \times 1000}{21} = 12000.$

Hence, sum = ₹ 12000.

A man borrows ₹ 10000 at 10% compound interest compounded yearly. At the end of each year, he pays back 30% of the sum borrowed. How much money is left unpaid just after the second year ?

Answer

30% of sum borrowed = $\dfrac{30}{100} \times 10000$ = ₹ 3000.

So, at the end of each year ₹ 3000 is returned back.

For first year :

P = ₹ 10000

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{1000} = \dfrac{10000 \times 10 \times 1}{100}$ = ₹ 1000

Amount = P + I = ₹ 10000 + ₹ 1000 = ₹ 11000.

Amount left to pay at end of first year = ₹ 11000 - ₹ 3000 = ₹ 8000.

For second year :

P = ₹ 8000

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{8000 \times 10 \times 1}{100}$ = ₹ 800

Amount = P + I = ₹ 8000 + ₹ 800 = ₹ 8800.

Amount left to pay at end of second year = ₹ 8800 - ₹ 3000 = ₹ 5800.

Hence, amount left to pay after second year = ₹ 5800.

A man borrows ₹ 10000 at 10% compound interest compounded yearly. At the end of each year, he pays back 20% of the amount for that year. How much money is left unpaid just after the second year ?

Answer

For first year :

P = ₹ 10000

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{10000 \times 10 \times 1}{100}$ = ₹ 1000

Amount = P + I = ₹ 10000 + ₹ 1000 = ₹ 11000.

Amount paid back = 20% of the amount for that year

= $\dfrac{20}{100} \times 11000$ = ₹ 2200

Amount left to pay at end of first year = ₹ 11000 - ₹ 2200 = ₹ 8800.

For second year :

P = ₹ 8800

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{8800 \times 10 \times 1}{100}$ = ₹ 880

Amount = P + I = ₹ 8800 + ₹ 880 = ₹ 9680.

Amount paid back = 20% of the amount for that year

= $\dfrac{20}{100} \times 9680$ = ₹ 1936

Amount left to pay at end of second year = ₹ 9680 - ₹ 1936 = ₹ 7744.

Hence, amount left to pay after second year = ₹ 7744.