Utilizing OP Malhotra Class 9 Solutions Chapter 2 Compound Interest Ex 2(A) as a study aid can enhance exam preparation.

## S Chand Class 9 ICSE Maths Solutions Chapter 2 Compound Interest Ex 2(A)

Find the compound interest in the following:

Question 1.

Principal = ₹ 10000, Rate % p.a = 12 %, Number of years = 2.

Solution:

Rate (R) = 12%

Period (T) = 2 years

Interest for the first year = \(\frac { PRT }{ 100 }\)

= \(\frac{10000 \times 12 \times 1}{100}\)

= ₹ 1200

Amount after first year = P + S.I.

= ₹ 10000 + ₹ 1200 = ₹ 11200

Principal for the second year = ₹ 11200

Interest for the second year

= \(\frac{11200 \times 12 \times 1}{100}\) = ₹ 1344

∴ Interest for 2 years = ₹ 1200 + ₹ 1344 = ₹ 2544

Hence compoind interest = ₹ 2544

Question 2.

Principal = ₹ 50000, Rate % p.a = 10 %, Number of years = 2.

Solution:

Principal (P) = ₹ 5000

Rate (R) = 10% p.a.

Period (T) = 2 years

∴ Interest after first year = \(\frac { PRT }{ 100 }\)

= \(\frac{5000 \times 10 \times 1}{100}\) = ₹ 500

∴ Amount after first year = P + S.I. = ₹ 5000 + ₹ 500 = ₹ 5500

∴ Principal for the second year = ₹ 5500

Interest for the second year

= \(\frac{5500 \times 10 \times 1}{100}\) = ₹ 550

∴ Interest for 2 years = ₹ 500 + ₹ 550 = ₹ 1050

Hence compound interest for 2 years = ₹ 1050

Question 3.

Principal = ₹ 2800, Rate % p.a = 10 %, Number of years = 1\(\frac { 1 }{ 2 }\).

Solution:

Principal (P) = ₹ 2800

Rate (R) = 10% p.a.

Period (T) = 1 \(\frac { 1 }{ 2 }\) years

Interest for the first year = \(\frac { PRT }{ 100 }\)

= \(\frac{2800 \times 10 \times 1}{100}\)

= ₹ 280

∴ Amount after first year = P + S.I.

= ₹ 2800 + ₹ 280 = ₹ 3080

Principal for the second year = ₹ 3080

Interest for the next 6 months (\(\frac { 1 }{ 2 }\) years)

= ₹ \(\frac{3080 \times 10 \times 1}{100 \times 2}\) = ₹ 154

Total interest for 1\(\frac { 1 }{ 2 }\) years = ₹ 280 + ₹ 154 = ₹ 434

Hence compound interest for 1 \(\frac { 1 }{ 2 }\) years = ₹ 434

Question 4.

Principal = ₹ 2000, Rate % p.a = 20 %, Number of years = 2.

Solution:

Principal (P) = ₹ 2000

Rate (R) = 20%

Period (T) = 2 years

∴ Interest for the first year = \(\frac { PRT }{ 100 }\)

= \(\frac{2000 \times 20 \times 1}{100}\)

= 400

Amount after one year = P + S.I.

= ₹ 2000 + ₹ 400 = ₹ 2400

or principal for the second year = ₹ 2400

Interest for the second year

= \(\frac{2400 \times 20 \times 1}{100}\)

= ₹ 480

∴ Total interest for 2 years = ₹ 400 + ₹ 480 = ₹ 880

or compound interest for 2 years = ₹ 880

Question 5.

Principal = ₹ 20480, Rate % p.a = 6\(\frac { 1 }{ 4 }\) %, Number of years = 2 years 73 days.

Solution:

Principal (P) = ₹ 20480

Rate (R) = 6 \(\frac { 1 }{ 4 }\) % = \(\frac { 25 }{ 4 }\) % p.a.

Period (T) = 2 years 73 days

Interest for the first year = \(\frac { PRT }{ 100 }\)

= \(\frac{20480 \times 25 \times 1}{100 \times 4}\)

= ₹ 1280

Amount after one year = P + S.I.

= ₹ 20480 + ₹ 1280 = ₹ 21760

or principal for the second year = ₹ 21760

Interest for the second year

= ₹ \(\frac{21760 \times 25 \times 1}{4 \times 100}\) = ₹ 1360

∴ Amount after 2 years = ₹ 21760 + ₹ 1360 = ₹ 23120

Principal for the next \(\frac { 1 }{ 5 }\) year

= ₹ 23120

Interest for \(\frac { 1 }{ 5 }\) year = ₹ \(\frac{23120 \times 25 \times 1}{100 \times 4 \times 5}\)

∴ Total interest for 2 \(\frac { 1 }{ 5 }\) years

= ₹ 1280 + ₹ 1360 + ₹ 289 = ₹ 2929

Question 6.

Find the amount and compound interest on a sum of ₹ 15625 at 4% per annum for 3 years compounded annually.

Solution:

Principal (P) = ₹ 15625

Rate (R) = 4% p.a.

Period (T) = 3 years

Interest for the first year = \(\frac { PRT }{ 100 }\)

= ₹ \(\frac{15625 \times 4 \times 1}{100}\) = ₹ 625

∴ Amount after one year P + S.I. = ₹ 15625 + ₹ 625 = ₹ 16250

Or principal for the second year = ₹ 16250

Interest for the second year

= ₹ \(\frac{16250 \times 4 \times 1}{100}\) = ₹ 650

Amount after second year = ₹ 16250 + 650 = ₹ 16900

or principal for the third year = ₹ 16900

Interest for the third year = \(\frac{16900 \times 4 \times 1}{100}\)

= ₹ 676

Amount after third year = ₹ 16900 + ₹ 676 = ₹ 17576

and compound interest = A – P = ₹ 17576 – ₹ 15625 = ₹ 1951.

Question 7.

To renovate his shop, Anurag obtained a loan of T8000 from a bank. If the rate of interest at 5% per annum is compounded annually, calculate the compound interest that Anurag will have to pay after 3 years.

Solution:

Amount of loan (Principal) (P) = ₹ 8000

Rate of interest (R) = 5% p.a.

Period (T) = 3 years

∴ Interest for the first year = \(\frac { PRT }{ 100 }\)

= ₹ \(\frac{8000 \times 5 \times 1}{100}\) = ₹ 400

Amount after one year = P + S.I.

= ₹ 8000 + ₹ 400 = ₹ 8400

or principal for the second year = Rs. 8400

Interest for the second year

= \(\frac{8400 \times 5 \times 1}{100}\)

= ₹ 420

Amount after two years = ₹ 8400 + ₹ 420 = ₹ 8820

or principal for the third year = ₹ 8820

Interest for the third year = ₹ \(\frac{8820 \times 5 \times 1}{100}\)

= ₹ 441

∴ Amount after 3 years = ₹ 8820 + ₹ 441 = ₹ 9261

∴ C.I. = A – P = 9261 – 8000 = ₹ 1261

Question 8.

Maria invests ₹ 93750 at 9.6% per annum for 3 years and the interest is compounded annually. Calculate.

(i) the amount standing to her credit at the end of the second year.

(ii) the interest for the 3rd year.

Solution:

Maria’s investment (P) = ₹ 93750

Rate of interest (R) = 9.6%

Period (T) = 3 years

Interest for the first year = \(\frac { PRT }{ 100 }\)

= ₹ \(\frac{93750 \times 9.6 \times 1}{100}\)

= ₹ \(\frac{93750 \times 96 \times 1}{100 \times 10}\)

= 9000

Amount after one year = P + S.I.

= ₹ 93750 + 9000 = ₹ 102750

or principal for the second year = ₹ 102750

= \(\frac{102750 \times 9.6 \times 1}{100}\)

= ₹ \(\frac{102750 \times 96 \times 1}{100 \times 10}\)

= ₹ 9864

(i) ∴ Amount after 2 years = ₹ 102750 + ₹ 9864 = ₹ 112614

or Principal for the third year = ₹ 112614

∴ Interest for the third year = \(\frac{112614 \times 9.6 \times 1}{100}\)

= \(\frac{112614 \times 96 \times 1}{100 \times 10}\) = ₹ 10810.94

(ii) Amount after third year = ₹ 112614 + 1081094 = ₹ 123424.94

Question 9.

A sum of ₹ 9,600 is invested for 3 years at 10% p.a. 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 the first 2 years.

(iv) Find the compound interest at the end of 3 years.

Solution:

Principal (P) = 9600

Rate of interest (R) = 10% p.a.

Period (T) = 3 years

Interest for the first year = \(\frac { PRT }{ 100 }\)

= \(\frac{9600 \times 10 \times 1}{100}\) = ₹ 960

(i) ∴ Amount after the first year = P + S.I. = ₹ 9600+ ₹ 960 = ₹ 10560

(ii) Principal for the second year = ₹ 10560

Interest for the second year = \(\frac{10560 \times 10 \times 1}{100}\)

= ₹ 1056

Amount after the second year = ₹ 10560 + ₹ 1056 = ₹ 11616

(iii) Compound interest for two years = ₹ 960 + ₹ 1056 = ₹ 2016

Principal for the third year = ₹ 11616

Interest for the third year = \(\frac{11616 \times 10 \times 1}{100}\)

= ₹ 1161.60

(iv) Compound interest after the end of third year = ₹ 2016 + ₹ 1161.60 = ₹ 3177.60

Question 10.

Shanker takes a loan of ₹ 10,000 at a compound interest rate of 10% per annum (p.a.)

(i) Find the compound interest after one year.

(ii) Find the compound interest for 2 years.

(iii) Find the sum of money required to clear the debt at the end of 2 years.

(iv) Find the difference between the compound interest and the simple interest at the same rate for 2 years.

Solution:

Amount of loan taken by Shanker (P)

= ₹ 10000

Rate of interest (R) = 10% p.a.

Period (T) = 2 years

(i) Interest for the first year = \(\frac { PRT }{ 100 }\)

\(\frac{10000 \times 10 \times 1}{100}\) = ₹ 1000

Amount after one year = P + S.I.

= ₹ 10000 + ₹ 1000 = ₹ 11000

Principal for the second year = ₹ 11000

Interest for the second year

\(\frac{11000 \times 10 \times 1}{100}\) = ₹ 1100

(ii) ∴ C.I. for 2 years = ₹ 1000 + ₹ 1100 = ₹ 2100

(iii) Amount after 2 years = ₹ 11000 + ₹ 1100 = ₹ 12100

(iv) Simple interest for 2 years

₹\(\frac{10000 \times 10 \times 2}{100}\) = ₹ 2000

Now difference between C.I. and S.I. for 2 years

= ₹ 2100 – ₹ 2000 = ₹ 100

Question 11.

Find compound interest on ₹ 5000 at 12% p.a. for 1 year, compounded half yearly.

Solution:

Principal (P) = ₹ 5000

Rate (R) = 12% p.a. or 6% half yearly

Period (T) = 1 year or 2 half years

Interest for the first half year = \(\frac { PRT }{ 100 }\)

= \(\frac{5000 \times 6 \times 1}{100}\) = ₹ 300

Amount after one half year = P + S.I.

= ₹ 5000 + ₹ 300 = ₹ 5300

Principal for the second half year = ₹ 5300

Interest for the second half year

= \(\frac{5300 \times 6 \times 1}{100}\) = ₹ 318

∴ Amount after 2 half years = ₹ 5300 + ₹ 318 = ₹ 5618

and C.I. for 2 half years = Rs. A – P = ₹ 5618 – ₹ 5000 = ₹ 618

Question 12.

Find the amount and the compound interest on ₹ 16000 for 1 \(\frac { 1 }{ 2 }\) years at 10% p.a. the interest being compounded half-yearly.

Solution:

Principal (P) = ₹ 16000

Rate (R) = 10% or 5% half yearly

Period (T) = 1 \(\frac { 1 }{ 2 }\) years or 3 half years

Interest for the first half year = \(\frac { PRT }{ 100 }\)

= \(\frac{16000 \times 5 \times 1}{100}\) = ₹ 800

Amount after first half year = P + S.I. = ₹ 16000 + ₹ 800 = ₹ 16800

or Principal for the second half year = ₹ 16800

Interest for the second half year

= \(\frac{16800 \times 5 \times 1}{100}\) = ₹ 840

Amount after 2 half years = ₹ 16800 + ₹ 840 = ₹ 17640

or Principal for the third half year = ₹ 17640

Interest for the third half year

= \(\frac{17640 \times 5 \times 1}{100}\) = ₹ 882

Amount after third half year = ₹ 17640 + ₹ 882 = ₹ 18522

∴ C.I. for 3 half years = A – P = ₹ 18522 – 16000 = ₹ 2522

Question 13.

Calculate the amount due and the compound interest on ₹ 40000 for 2 years when the rate of interest successive years is 7% and 8% respectively.

Solution:

Principal (P) = ₹ 40000

Period (T) = 2 years

Rate of interest for the first year (R_{1}) = 7%

and for the second year (R_{2}) = 8%

∴ Interest for the first year = \(\frac { PRT }{ 100 }\)

= ₹ \(\frac{40000 \times 7 \times 1}{100}\) = ₹ 2800

Amount after first year = P + S.I.

= ₹ 40000 + ₹ 2800 = ₹ 42800

Principal for the second year = ₹ 42800

Interest for the second year = \(\frac{42800 \times 8 \times 1}{100}\)

= ₹ 3424

∴ Amount after 2 years = ₹ 42800 + ₹ 3424 = ₹ 46224

and compound interest for 2 years = A – P = ₹ 46224 – 40000 = ₹ 6224

Question 14.

If the simple interest on a sum of money for 2 years at 5% per annum is ₹ 50, what will be the compound interest on the same sum at the same rate for the same time.

Solution:

S. Interest for 2 years

Period (T) = 2 years

Rate (R) = 5% p.a.

We know that S.I. and C.I. is same for the first year

∴ S. Interest for the first year = ₹ \(\frac { 50 }{ 2 }\) = ₹ 25

and S.I. for the second year = ₹ 25

∴ Principal (P) = \(\frac{\text { S.I. } \times 100}{R \times T}\)

= \(\frac{50 \times 100}{5 \times 2}\) = ₹ 500

Now amount after first year = P + S.I.

= ₹ 500 + ₹ 25 = ₹ 525

Principal for the second year = ₹ 525

∴ Interest for the second year

= \(\frac{525 \times 5 \times 1}{100}=\frac{2625}{100}\) = ₹ 26.25

C.I. for 2 years = ₹ 25 + ₹ 26.25 = ₹ 51.25

Question 15.

A man invests ₹ 46,875 at 4% per annum compound interest for 3 years. Calculate :

(i) the interest for the 1st year;

(ii) the amount standing to his credit at the end of the 2nd year;

(iii) the interest for the 3rd year.

Solution:

Investment (P) = ₹ 46875

Rate (R) = 4% p.a.

Period (T) = 3 years

(i) ∴ Interest for the first year = \(\frac { PRT }{ 100 }\)

= ₹ \(\frac{46875 \times 4 \times 1}{100}\) = ₹ 1875

(ii) Amout after one year = P + S.I.

= ₹ 46875 + ₹ 1875 = ₹ 48750

or Principal for the second year = ₹ 48750

Interest for the second year = \(\frac{48750 \times 4 \times 1}{100}\)

= ₹ 1950

∴ Amount after second year = ₹ 48750 + ₹ 1950 = ₹ 50700

(iii) or Principal for the third year = ₹ 50700

Interest for the third year = \(\frac{50700 \times 4 \times 1}{100}\)

= ₹ 2028