Compound Interest — Exercise 2.1

Find the amount and the compound interest on ₹8000 at 5% per annum for 2 years.

Answer

Principal for first year = ₹8000.

Interest for the first year = ₹ $\dfrac{8000 \times 5 \times 1}{100}$ = ₹400.

Amount after one year = ₹8000 + ₹400 = ₹8400.

Principal for the second year = ₹8400.

Interest for the second year = ₹ $\dfrac{8400 \times 5 \times 1}{100}$ = ₹420.

Amount after 2 years = ₹8400 + ₹420 = ₹8820.

Compound interest for 2 years = Final amount - Principal = ₹8820 - ₹8000 = ₹820.

Hence, the amount and compound interest on ₹8000 at 5% per annum after 2 years is ₹8820 and ₹820 respectively.

A man invested ₹46875 at 4% per annum compound interest for 3 years. Calculate :

(i) the interest for the first year.

(ii) the amount standing to his credit at the end of the second year.

(iii) the interest for the third year.

Answer

(i) Principal for first year = ₹46875.

Interest for the first year = ₹ $\dfrac{46875 \times 4 \times 1}{100} = \dfrac{187500}{100}$ = ₹1875.

Hence, the interest for the first year = ₹1875.

(ii) Amount after one year = ₹46875 + ₹1875 = ₹48750.

Principal for the second year = ₹48750.

Interest for the second year = ₹ $\dfrac{48750 \times 4 \times 1}{100} = \dfrac{195000}{100}$ = ₹1950.

Amount after 2 years = ₹48750 + ₹1950 = ₹50700.

Hence, the amount standing to his credit at the end of the second year is ₹50700.

(iii) Principal for third year = ₹50700.

Interest for the third year = ₹ $\dfrac{50700 \times 4 \times 1}{100} = \dfrac{202800}{100}$ = ₹2028.

Hence, the interest for the third year = ₹2028.

Calculate the compound interest for the second year on ₹8000 invested for 3 years at 10% p.a.

Also find the sum due at the end of third year.

Answer

Principal for first year = ₹8000.

Interest for the first year = ₹ $\dfrac{8000 \times 10 \times 1}{100}$ = ₹800.

Amount after one year = ₹8000 + ₹800 = ₹8800.

Principal for the second year = ₹8800.

Interest for the second year = ₹ $\dfrac{8800 \times 10 \times 1}{100}$ = ₹880.

Amount after 2 years = ₹8800 + ₹880 = ₹9680.

Principal for the third year = ₹9680.

Interest for the third year = ₹ $\dfrac{9680 \times 10 \times 1}{100}$ = ₹968.

Amount after 3 years = ₹9680 + ₹968 = ₹10648.

Hence, the compound interest for the second year on ₹8000 invested for 3 years at 10% p.a. is ₹880 and the sum due at the end of third year is ₹10648.

Ramesh invested ₹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 three years.

Answer

(i) Principal for first year = ₹12800.

Interest for the first year = ₹ $\dfrac{12800 \times 10 \times 1}{100} = \dfrac{128000}{100}$ = ₹1280.

Amount after first year = ₹12800 + ₹1280 = ₹14080

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

(ii) Principal for the second year = ₹14080.

Interest for the second year = ₹ $\dfrac{14080 \times 10 \times 1}{100} = \dfrac{140800}{100}$ = ₹1408.

Hence, the interest Ramesh earns for second year = ₹1408.

(iii) Amount after 2 years = ₹14080 + ₹1408 = ₹15488.

Interest for the third year = ₹ $\dfrac{15488 \times 10 \times 1}{100} = \dfrac{154880}{100}$ = ₹1548.80

Amount after 3 years = ₹15488 + ₹1548.80 = ₹17036.80

Hence, the total amount due to Ramesh at the end of third year is ₹17036.80

The simple interest on a sum of money for 2 years at 12% per annum is ₹1380. Find :

(i) the sum of money.

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

Answer

Given, S.I. = ₹1380, rate = 12% p.a. and time = 2 years.

(i) Let the sum of money be P, then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow 1380 = \dfrac{P \times 12 \times 2}{100} \\[1em] \Rightarrow 138000 = 24P \\[1em] \Rightarrow P = \dfrac{138000}{24} \\[1em] \Rightarrow P = 5750.$

Hence, the sum of money is ₹5750.

(ii) Since, the rate of interest is 12% per annum, therefore, the rate of interest half-yearly = 6%.

Principal for first half-year = ₹5750.

Interest for first half-year = ₹ $\dfrac{5750 \times 6 \times 1}{100} = \dfrac{34500}{100}$ = ₹345.

∴ Amount after first half-year = ₹5750 + ₹345 = ₹6095.

Principal for the second half-year = ₹6095.

Interest for the 2nd half-year = ₹ $\dfrac{6095 \times 6 \times 1}{100} = \dfrac{36570}{100}$ = ₹365.70

∴ Compound interest on the above sum for one year payable half-yearly = ₹345 + ₹365.70 = ₹710.70

Hence, the compound interest on ₹5750 for one year payable half-yearly at the same rate is ₹710.70

A person invests ₹10000 for two years at a certain rate of interest, compounded annually. At the end of one year this sum amounts to ₹11200. Calculate :

(i) the rate of interest per annum.

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

Answer

(i) Let the rate of interest be R.

Given, amount at the end of first year = ₹11200.

Interest = Amount - Principal = ₹11200 - ₹10000 = ₹1200.

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

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

Hence, the rate of interest is 12% per annum.

(ii) Amount after first year = ₹11200.

Interest for second year = $\dfrac{11200 \times 12 \times 1}{100} = \dfrac{134400}{100}$ = ₹1344.

Amount at the end of second year = ₹11200 + ₹1344 = ₹12544.

Hence, the amount at the end of second year = ₹12544.

Mr. Lalit invested ₹5000 at a certain rate of interest, compounded annually for two years. At the end of first year it amounts to ₹5325. Calculate :

(i) the rate of interest.

(ii) the amount at the end of second year, to the nearest rupee.

Answer

(i) Let the rate of interest be R.

Given, amount at the end of first year = ₹5325.

Interest = Amount - Principal = ₹5325 - ₹5000 = ₹325.

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

$\Rightarrow 325 = \dfrac{5000 \times R \times 1}{100} \\[1em] \Rightarrow 325 = 50R \\[1em] \Rightarrow R = \dfrac{325}{50} \\[1em] \Rightarrow R = 6.5$

Hence, the rate of interest is 6.5% per annum.

(ii) Amount after first year = ₹5325.

Interest for second year = $\dfrac{5325 \times 6.5 \times 1}{100} = \dfrac{34612.5}{100}$ = ₹346.125.

Amount at the end of second year = ₹5325 + ₹346.125 = ₹5671.125.

Hence, the amount at the end of second year to the nearest rupee = ₹5671.

A man invests ₹5000 for three years at a certain rate of interest, compounded annually. At the end of one year it amounts to ₹5600. 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) Let the rate of interest be R.

Given, amount at the end of first year = ₹5600.

Interest = Amount - Principal = ₹5600 - ₹5000 = ₹600.

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

$\Rightarrow 600 = \dfrac{5000 \times R \times 1}{100} \\[1em] \Rightarrow 600 = 50R \\[1em] \Rightarrow R = \dfrac{600}{50} \\[1em] \Rightarrow R = 12$

Hence, the rate of interest is 12% per annum.

(ii) Amount after first year = ₹5600.

Principal for second year = ₹5600.

Interest for second year = $\dfrac{5600 \times 12 \times 1}{100} = \dfrac{67200}{100}$ = ₹672.

Hence, the interest accrued in second year = ₹672.

(iii) Amount after second year = ₹5600 + ₹672 = ₹6272

Principal for third year = ₹6272.

Interest for third year = $\dfrac{6272 \times 12 \times 1}{100} = \dfrac{75264}{100}$ = ₹752.64

Amount after third year = ₹6272 + ₹752.64 = ₹7024.64

Hence, the amount at the end of third year = ₹7024.64

Find the amount and the compound interest on ₹2000 at 10% p.a. for $2\dfrac{1}{2}$ years, compounded annually.

Answer

Principal for first year = ₹2000.

Interest for the first year = ₹ $\dfrac{2000 \times 10 \times 1}{100}$ = ₹200.

Amount after one year = ₹2000 + ₹200 = ₹2200.

Principal for the second year = ₹2200.

Interest for the second year = ₹ $\dfrac{2200 \times 10 \times 1}{100}$ = ₹220

Amount after 2 years = ₹2200 + ₹220 = ₹2420.

Principal for next $\dfrac{1}{2}$ year = ₹2420.

Interest for the next $\dfrac{1}{2}$ year = ₹ $\dfrac{2420 \times 10 \times \dfrac{1}{2}}{100} = \dfrac{12100}{100}$ = ₹121.

Amount after $2\dfrac{1}{2}$ years = ₹2420 + ₹121 = ₹2541.

Compound interest for $2\dfrac{1}{2}$ years = Final amount - Principal = ₹2541 - ₹2000 = ₹541.

Hence, the amount and compound interest on ₹2000 at 10% per annum after $2\dfrac{1}{2}$ years is ₹2541 and ₹541 respectively.

Find the amount and the compound interest on ₹50000 for $1\dfrac{1}{2}$ years at 8% per annum, the interest being compounded semi-annually.

Answer

Since, the rate of interest is 8% per annum, therefore, the rate of interest half-yearly = $\dfrac{1}{2}$ of 8% = 4%.

Principal for first half-year = ₹50000.

Interest for first half-year = $\dfrac{50000 \times 4 \times 1}{100} = \dfrac{200000}{100}$ = ₹2000.

Amount after first half-year = ₹50000 + ₹2000 = ₹52000.

Principal for second half-year = ₹52000.

Interest for the second half-year = $\dfrac{52000 \times 4 \times 1}{100} = \dfrac{208000}{100}$ = ₹2080.

Amount after one year = ₹52000 + ₹2080 = ₹54080.

Principal for third half-year = ₹54080.

Interest for the third half-year = $\dfrac{54080 \times 4 \times 1}{100} = \dfrac{216320}{100}$ = ₹2163.20

Amount after $1\dfrac{1}{2}$ year = ₹54080 + ₹2163.20 = ₹56243.20

Compound interest for $1\dfrac{1}{2}$ year = Final amount - principal = ₹56243.20 - ₹50000 = ₹6243.20

Hence, the amount and the compound interest on ₹50000 for $1\dfrac{1}{2}$ years at 8% per annum, the interest being compounded semi-annually are ₹56243.20 and ₹6243.20 respectively.

Calculate the amount and the compound interest on ₹5000 in 2 years when the rate of interest for successive years is 6% and 8% respectively.

Answer

Principal for first year = ₹5000.

Interest for the first year = ₹ $\dfrac{5000 \times 6 \times 1}{100} = \dfrac{30000}{100}$ = ₹300.

Amount after one year = ₹5000 + ₹300 = ₹5300.

Principal for the second year = ₹5300.

Interest for the second year = ₹ $\dfrac{5300 \times 8 \times 1}{100} = \dfrac{42400}{100}$ = ₹424.

Amount after 2 years = ₹5300 + ₹424 = ₹5724.

Compound interest = Final amount - Principal = ₹5724 - ₹5000 = ₹724.

Hence, the amount and the compound interest are ₹5724 and ₹724 respectively.

A sum of ₹9600 is invested for 3 years at 10% per annum at compound interest.

(i) What is the sum due at the end of the first year?

(ii) What is the sum due at the end of the second year?

(iii) Find the compound interest earned in 2 years.

(iv) Find the difference between the answers in (ii) and (i) and find the interest on this sum for one year.

(v) Hence, write down the compound interest for the third year.

Answer

(i) Principal for first year = ₹9600.

Interest for the first year = ₹ $\dfrac{9600 \times 10 \times 1}{100} = \dfrac{96000}{100}$ = ₹960.

Amount after one year = ₹9600 + ₹960 = ₹10560.

Hence, the amount due at the end of first year = ₹10560.

(ii) Principal for second year = ₹10560.

Interest for the second year = ₹ $\dfrac{10560 \times 10 \times 1}{100} = \dfrac{105600}{100}$ = ₹1056.

Amount after 2 years = ₹10560 + ₹1056 = ₹11616.

Hence, the amount due at the end of second year = ₹11616.

(iii) Compound interest = Final amount - Principal = ₹11616 - ₹9600 = ₹2016.

Hence, the compound interest earned in 2 years = ₹2016.

(iv) Difference between (ii) and (i) = ₹11616 - ₹10560 = ₹1056.

Interest on the above sum for 1 year = $\dfrac{1056 \times 10 \times 1}{100} = \dfrac{10560}{100}$ = ₹105.60

Hence, the difference = ₹1056 and interest earned on it = ₹105.60

(v) Principal for third year = ₹11616.

Interest for the third year = ₹ $\dfrac{11616 \times 10 \times 1}{100} = \dfrac{116160}{100}$ = ₹1161.60

Hence, the compound interest for third year = ₹1161.60

The simple interest on a certain sum of money for 2 years at 10% per annum is ₹1600. Find the amount due and the compound interest on this sum of money at the same rate after 3 years, interest being reckoned annually.

Answer

Let the sum of money be ₹x.

Given, simple interest = ₹1600.

$\therefore 1600 = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow 1600 = \dfrac{x \times 10 \times 2}{100} \\[1em] \Rightarrow 1600 = \dfrac{20x}{100} \\[1em] \Rightarrow 160000 = 20x \\[1em] \Rightarrow x = \dfrac{160000}{20} \\[1em] \Rightarrow x = ₹8000.$

Principal for first year = ₹8000.

Interest for the first year = ₹ $\dfrac{8000 \times 10 \times 1}{100} = \dfrac{80000}{100}$ = ₹800.

Amount after one year = ₹8000 + ₹800 = ₹8800.

Principal for the second year = ₹8800.

Interest for the second year = ₹ $\dfrac{8800 \times 10 \times 1}{100} = \dfrac{88000}{100}$ = ₹880.

Amount after 2 years = ₹8800 + ₹880 = ₹9680.

Principal for the third year = ₹9680.

Interest for the third year = ₹ $\dfrac{9680 \times 10 \times 1}{100} = \dfrac{96800}{100}$ = ₹968

Amount after 3 years = ₹9680 + ₹968 = ₹10648

Compound interest = Final amount - Principal = ₹10648 - ₹8000 = ₹2648.

Hence, the amount and compound interest due = ₹10648 and ₹2648 respectively.

Vikram borrowed ₹20000 from a bank at 10% per annum simple interest. He lent it to his friend Venkat at the same rate but compounded annually. Find his gain after $2\dfrac{1}{2}$ years.

Answer

Simple interest = $\dfrac{P \times R \times T}{100}.$

∴ S.I. = $\dfrac{20000 \times 10 \times 2.5}{100} = \dfrac{500000}{100}$ = ₹5000.

Calculating, compound interest :

Principal for first year = ₹20000.

Interest for first year = $\dfrac{20000 \times 10 \times 1}{100} = \dfrac{200000}{100}$ = ₹2000.

Amount after first year = ₹20000 + ₹2000 = ₹22000.

Principal for second year = ₹22000.

Interest for second year = $\dfrac{22000 \times 10 \times 1}{100} = \dfrac{220000}{100}$ = ₹2200.

Amount after second year = ₹22000 + ₹2200 = ₹24200.

Principal for next $\dfrac{1}{2}$ year = ₹24200.

Interest for next $\dfrac{1}{2}$ year = $\dfrac{24200 \times 10 \times \dfrac{1}{2}}{100} = \dfrac{242000}{200}$ = ₹1210.

Amount after $2\dfrac{1}{2}$ year = ₹24200 + ₹1210 = ₹25410.

Compound interest = Final amount - principal = ₹25410 - ₹20000 = ₹5410.

Difference between C.I. and S.I. = ₹5410 - ₹5000 = ₹410.

Hence, Venkat gained ₹410 after $2\dfrac{1}{2}$ years.

A man borrows ₹6000 at 5% compound interest. If he repays ₹1200 at the end of each year, find the amount outstanding at the beginning of the third year.

Answer

Principal for first year = ₹6000, rate = 5%.

Interest for first year = $\dfrac{6000 \times 5 \times 1}{100} = \dfrac{30000}{100}$ = ₹300.

Amount after first year = ₹6000 + ₹300 = ₹6300.

Money refunded at the end of first year = ₹1200.

Principal for second year = ₹6300 - ₹1200 = ₹5100.

Interest for second year = $\dfrac{5100 \times 5 \times 1}{100} = \dfrac{25500}{100}$ = ₹255.

Amount after second year = ₹5100 + ₹255 = ₹5355.

Money refunded at the end of second year = ₹1200.

Principal for third year = ₹5355 - ₹1200 = ₹4155.

Hence, the amount outstanding at the beginning of third year = ₹4155.

Mr. Raina deposits ₹ 1,600 in a bank every year in the beginning of the year, at 5% per annum compound interest. Calculate the amount due to him at the end of 2 years. Also find his gain in two years.

Answer

For 1st year,

P = ₹ 1,600

R = 5%

T = 1 year

Using formula,

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

Substituting the values, we get

$\Rightarrow I = \dfrac{1,600 \times 5 \times 1}{100}\\[1em] = 16 \times 5 \\[1em] = 80.$

A = P + I = ₹ 1,600 + ₹ 80 = ₹ 1,680

For 2nd year,

P = ₹ 1,680 + ₹ 1,600 (Amount deposited at beginning of every year) = ₹3,280

R = 5%

T = 1 year

$\Rightarrow I = \dfrac{3,280 \times 5 \times 1}{100}\\[1em] = \dfrac{16,400}{100} \\[1em] = 164.$

A = P + I = ₹ 3,280 + ₹ 164 = ₹ 3,444.

Gain = Total balance - money invested

= ₹ 3,444 - (₹ 1,600 x 2)

= ₹ 3,444 - ₹ 3,200

= ₹ 244.

Hence, at the end of 2 years amount = ₹ 3,444 and the gain = ₹ 244.

Mr. Dubey borrows ₹100000 from State Bank of India at 11% per annum compound interest. He repays ₹41000 at the end of first year and ₹47700 at the end of second year. Find the amount outstanding at the beginning of the third year.

Answer

Principal for first year = ₹100000, rate = 11%.

Interest for first year = $\dfrac{100000 \times 11 \times 1}{100} = \dfrac{1100000}{100}$ = ₹11000.

Amount after first year = ₹100000 + ₹11000 = ₹111000.

Money refunded at the end of first year = ₹41000.

Principal for second year = ₹111000 - ₹41000 = ₹70000.

Interest for second year = $\dfrac{70000 \times 11 \times 1}{100} = \dfrac{770000}{100}$ = ₹7700.

Amount after second year = ₹70000 + ₹7700 = ₹77700.

Money refunded at the end of second year = ₹47700.

Principal for third year = ₹77700 - ₹47700 = ₹30000.

Hence, the amount outstanding at the beginning of third year = ₹30000.

Jaya borrowed ₹50000 for 2 years. The rates of interest for two successive years are 12% and 15% respectively. She repays ₹33000 at the end of first year. Find the amount she must pay at the end of second year to clear her debt.

Answer

Principal for first year = ₹50000, rate = 12%.

Interest for first year = $\dfrac{50000 \times 12 \times 1}{100} = \dfrac{600000}{100}$ = ₹6000.

Amount after first year = ₹50000 + ₹6000 = ₹56000.

Money refunded at the end of first year = ₹33000.

Principal for second year = ₹56000 - ₹33000 = ₹23000, rate = 15%.

Interest for second year = $\dfrac{23000 \times 15 \times 1}{100} = \dfrac{345000}{100}$ = ₹3450.

Amount after second year = ₹23000 + ₹3450 = ₹26450.

Hence, Jaya must pay ₹26450 at the end of second year to clear her debt.