SChand CLASS 9 Chapter 2 Compound Interest Exercise 2(A)

 Exercise 2(A)

Question 1

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

Sol :

Principal=1000

Rate=12%

First year Interest$=\frac{PRT}{100}$

$\Rightarrow \frac{10000 \times 12 \times 1}{100}$

=100×12×1=1200

After first year Amount=P+S.I

=10000+1200=11200

∴Second year principal=11200

Second year Interest$=\frac{\text{Second PRT}}{100}$

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

=112×12=1334

∴Interest for 2 year=1200+1344=2544

Hence Compound Interest=2544

Question 2

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

Sol :

Principal (P)=5000

Rate=10%

Time=2 year

First Year Interest$=\frac{PRT}{100}$

$=\frac{5000 \times 10 \times 1}{100}$

=50×10×1=500

∴After First year amount=P+S.I

=5000+500=5500

∴Second year Principal=5500

Second year Interest$=\frac{\text{Second year PRT}}{100}$

$=\frac{5500 \times 10 \times 1}{100}$

=550

∴Interest for 2 year =500+550

=1050

Hence Compound Interest for 2 year

=1050

Question 3

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

Sol :

Principal=2800

Rate=10%

Time=$1 \frac{1}{2}$ years

First Year Interest$=\frac{PRT}{100}$

$=\frac{2800 \times 10 \times 1}{100}$

=28×10×1=280

∴After first year Amount=P+S.I.

=2800+280

=3080

Second Year Principal=3080

Next 6 month Interest $\left(\frac{1}{2}\right)$ year

$=\frac{\text{Second year }P \times R \times T}{100}$

$\frac{3080 \times 10 \times 1}{100 \times 2}$

=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.

Sol :

Principal=2000

Rate=20%

Time=2 years

∴First Year Interest$=\frac{PRT}{100}$

$=\frac{2000 \times 20 \times 1}{100}$

=20×20×1=400

After One year amount=P+S.I

=2000+400=2400

Second year Principal=2400

Second year Interest$=\frac{\text{Second year } P\times R\times T}{100}$

$=\frac{2400 \times 20 \times 1}{100}$

=24×20×1=480

∴Total Interest for 2 year

=400+480=880

Compound Interest for 2 year=880

Question 5

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

Sol :

Principal=20480

Rate$=6\frac{1}{4}%=\frac{25}{4}\%$ p.a.

Time=2 years 73 days

=2 years $\frac{73}{365}$

$=2\frac{1}{5}$ years

First Year Interest$\frac{PRT}{100}$

$=\frac{20480\times 25 \times 1}{100 \times 4}$

=1280

After One year amount=P+S.I.

=20480+1280

=21760

Second Year Principal=21760

Second Year Interest$=\frac{\text{Second Year}P\times R\times T}{100}$

$=\frac{21760 \times 25 \times 1}{4\times 100}$

=1280

After One Year amount=P+S.I

=20480+1280

=21760

Second year Principal=21760

Second Year Interest$=\frac{\text{Second year}P\times R\times T}{100}$

$=\frac{21760 \times 25 \times 1}{4 \times 100}$

=1360

After 2 year Amount=21760+1360

=23120

Next $\frac{1}{5}$ year principal=23120

Interest for $\frac{1}{5}$ year$=\frac{23120 \times 25 \times 1}{100 \times 4 \times 5}$

=289

Total Interest for $2\frac{1}{5}$ year

=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.

Sol :

Principal=15625

Rate=4%

Time=3 years

First Year Interest$=\frac{PRT}{100}$

$=\frac{15625 \times 4 \times 1}{100}$

=625

∴After One Year Amount=P+S.I

=15625+625=16250

Second Year Principal=16250

Second Year Interest$=\frac{\text{P of the second year} \times R \times T}{100}$

$=\frac{10250 \times 4 \times 1}{100}$

=650

After Second year amount=16250+650

=16900

Third Year Principal$=\frac{16900 \times 4 \times 1}{100}$

=169×4×1=676

∴After Third year amount=16900+676

=17576

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.

Sol :

Amount of loan (P)=8000

Rate=5%

Time (n)=3 years

∴First Year Interest$=\frac{PRT}{100}$

$=\frac{8000 \times 5 \times 1}{100}$

=80×5×1=400

∴After One Year amount=P+S.I

=8000+400=8400

Second Year Principal=8400

Second Year Interest$=\frac{\text{Second Year}P\times R \times T}{100}$

$=\frac{8400 \times 5 \times 1}{100}$

=420

After two year amount=8400+420=8820

Third year Principal=8820

Third Year Interest$=\frac{8820 \times 5 \times 1}{100}$

=441

∴After 3 year amount=8820+441=9261

Compound Interest=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.

Sol :

Maria's investment (P)=93750

Rate=9.6%

Time (n)= 3 year

First Year Interest$\frac{P R T}{150}$

$=\frac{93750 \times 9.6 \times 1}{1000}$

$=\frac{93750 \times 96 \times 1}{1000}$

=9000

∴After One Year Amount=P+S.I

=93750+9000=102750

Second Year Principal=102750

Second Year Interest$=\frac{\text{Second Year}P \times R \times T}{100}$

$=\frac{102750 \times 9.6 \times 1}{1000}$

$=\frac{102750 \times 96}{100}$

=9864

(i) After 2 year Amount=102750+9864

=112614

Third Year Principal=112614

Third Year Interest$=\frac{112614 \times 9.8 \times 1}{1000}$

$=\frac{112614 \times 98 \times 1}{1000}$

=10810.94

(ii) ∴After Third Year Amount

=112614+10810.94

=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.

Sol :

Principal=9600

Rate=10%

Time=3 year

First Year Interest$=\frac{ P \times R\times T}{100}$

$=\frac{9600 \times 10 \times 1}{100}$

=960

(i) ∴After First Year Amount=P+S.I

=9600+960=10560

(ii) Second Year Principal=10560

Second Year Interest$=\frac{10560 \times 10 \times 1}{100}$

=1056

After Second year Amount=10560+1056

=11616

(iii) Two Year Compound Interest

=960+1056

=2016

Principal of Third Year=11616

Third Year Interest$=\frac{11616 \times 10 \times 1}{100}$

=1161.60 

(iv) After the end of third year Compound Interest

=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.

Sol :

Amount of loan (P)=10000

Rate=10%

Time (n)=2 year

(i) First Year Interest$=\frac{P \times R \times T}{100}$

$=\frac{10000 \times 10 \times 1}{100}$

=1000

After One Year Amount=P+S.I.

=10000+1000=11000

Second Year Principal=11000

Second Year Interest$=\frac{\text{Second year} P \times R \times T}{100}$

$=\frac{11000 \times 10 \times 1}{100}$

=1100

(ii) ∴2 Year Compound Interest=1000+1100

=2100

(iii) After 2 Year Amount=11000+1100

=12100

(iv) Simple Interest For 2 Year 

$=\frac{1000 \times 10 \times 2}{100}$

=2000

2 year difference between C.I. and S.I

=2100-2000

=100

Question 11

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

Sol :

Principal=5000

Rate=12% or 6% half yearly

Time=1 year or 2 half year

First Half Year Amount=P+S.I.

$=\frac{5000 \times 6 \times 1}{100}$

=300

After One Year Principal=5300

=5000+300=5300

Second Half Year Principal=5300

Second Half Year Interest

$=\frac{5300\times 6 \times 1}{100}$

=318

∴After 2 Half Year Amount

=5300+318=5618

Compound Interest For 2 Half Year

=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.

Sol :

Principal=1600

Rate=10% or 5% half yearly

Time=$1 \frac{1}{2}$ year or 3 half years

First Half Year Interest$=\frac{P \times R \times T}{100}$

$=\frac{16000 \times 5 \times 1}{100}$

=800

After First Half Year Amount=P+S.I

=16000+800

=16800

Second Half Year Principal=16800

Second Half Year Interest$\frac{16000 \times 5 \times 1}{100}$

=840

After 2 Half Year Amount=16800+840=17640

Third Half Year Principal=17640

Third Half Year Interest

$=\frac{17640 \times 5 \times 1}{100}$

=882

After Third Year Compound Interest=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.

Sol :

Principal=40000

Time=2 year

First Year Rate of Interest (R1)=7%

Second Year Rate of Interest (R2)=8%

∴First Year Interest$=\frac{P R T}{100}$

$=\frac{4000 \times 7 \times 1}{100}$

=2800

After First Year Amount=P+S.I.

=40000+2800=42800

Second Year Principal =42800

Second Year Interest$=\frac{42800 \times 8 \times 1}{100}$

=3424

∴After Two Year Amount=42800+3424

=46224

2 Year Compound Interest=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.

Sol :

Simple Interest For 2 Year

T=2 Year 

R=5%

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

∴First year simple interest$=\frac{50}{2}$

=25

Second year simple interest=25

∴Principal$=\frac{S.I. \times 100}{R\times T}$

$=\frac{50 \times 100}{5 \times 2}$

=500 

After First Year Amount=P+S.I.

=500+25

=525

Second Year Principal=525

Second Year Interest$=\frac{525 \times 5 \times 1}{100}$

$=\frac{2625}{100}$

=26.25

2 Year Compound Interest

=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.

Sol :

Investment (P)=46875

Rate=4%

Time=3 year

(i) First Year Interest$=\frac{PRT}{100}$

$=\frac{46875 \times 4 \times 1}{100}$

=1875

(ii) After One Year Amount=P+S.I.

=46875+1875=48750

Second Year Principal=48750

Second Year Interest$=\frac{42750 \times 4 \times 1}{100}$

=1950

After Second Year Amount=48750+1950

=50700

(iii) Third Year Principal=50700

Third Year Interest$=\frac{50700 \times 4 \times 1}{100}$

=2028