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 (R1) = 7%
and for the second year (R2) = 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