# CLASS-9PROBLEM & SOLUTION ON COMPOUND INTEREST

PROBLEM & SOLUTION ON COMPOUND INTEREST -

Example.1) Richard invests \$ 10000, for four years at the rate of 10% per annum at compound interest, find the following queries –

a)  The sum due to Richard at the end of 1st year

b)  The interest he earns for the 2nd year

c)  The total amount due to him at the end of 4th year

Ans.) Principal for the 1st year \$ 10000

10

Interest for the 1st year = \$ ( 10000 X ------ X 1 ) = \$ 1000

100

Sum due to Richard at the end of 1st year = \$ ( 10000 + 1000 )

= \$ 11000………………….. (a)  (Ans.)

Principal for the 2nd year = \$ 11000

10

Interest for the 2nd year = \$ ( 11000 X ------ X 1 ) = \$ 1100

100

So, interest he earns at the end of 2nd year \$ 11000 ………………….(b)  (Ans.)

Amount at the end of 2nd year = \$ (11000 + 1100) =  \$ 12100

Principal for the 3rd year = \$ 12100

10

Interest for the 3rd year =  \$ ( 12100 X ------- X 1 ) = \$ 1210

100

Amount at the end of 3rd year = \$ ( 12100 + 1210 ) = \$ 13310

Principal for the 4th year = \$ 13310

10

Interest for the 4th year =  \$ ( 13310 X ------- X 1 ) =  \$ 1331

100

Amount at the end of 4th year = \$ ( 13310 + 1331 ) =  \$ 14641

So, total amount due to him at the end of the 4th year = \$ 1464 ..…….(c)   (Ans.)

Example.2) Calculate the amount and the compound interest on \$ 20000 for 2 years 6 months  at 12%  per annum, compounded annually.

Ans.)  Principal for the 1st year = \$ 20000

12

Interest for the 1st year = \$ ( 20000 X ------- X 1 ) = \$ 2400

100

Amount at the end of 1st year = \$ ( 20000 + 2400 ) = \$ 22400

Principal for the 2nd year = \$ 22400

12

Interest for the 2nd year = \$ ( 22400 X ------- X 1 ) = \$ 2688

100

At the end of 2nd year = \$ (22400 + 2688 ) = \$ 25088

Principal for the 3rd year \$ 25088, but money has been invested for 6 months.

12         6

So, Interest for 6 months = \$ ( 25088 X ------ X ------ ) = \$ 1505.28

100        12

So, amount at the end of 2 years 6 months = \$ (25088 + 1505.28)

= \$ 26593.28

Compound interest (C.I) after 2 years 6 months = \$ (26593.28 – 20000)

= \$ 6593.28

So, the amount \$ 26593.28 and the C.I is \$ 6593.28     (Ans.)

Example.3) Mr. Darby borrows \$ 100000 from bank at 10% per annum, compounded annually. He repays \$ 40000 at the end of 1st year and \$ 50000 at the end of the 2nd year. Find the amount outstanding against him at the beginning of the 3rd year.

Ans.) Principal for the 1st year = \$ 100000

10

Interest for the 1st year = \$ ( 100000 X ------- X 1 ) = \$ 10000

100

Amount after 1st year = \$ ( 100000 + 10000 )

= \$ 110000

Repayment at the end of 1st year = \$ ( 110000 – 40000) = \$ 70000

So, principal for 2nd year = \$ 70000

10

Interest for the 2nd year = \$ ( 70000 X -------- X 1 ) = \$ 7000

100

Amount after 2nd year = \$ (70000 + 7000 ) = \$ 77000

Repayment at the end of 2nd year = \$ ( 77000 – 50000 ) = \$ 27000

So, the required outstanding amount at the beginning of 3rd year is \$ 27000     (Ans.)

Example.4) Find the amount and the compound interest on \$ 8000 for

1

1----- years at 10%  per annum compounded half yearly.

2

Ans.) Here, rate = 10 % per annum = 5 % per half year.

1           3

And, time = 1 ----- = ( ----- X 2 ) half years = 3 half years

2           2

Principal for the 1st half year = \$ 8000                                                                                            5

Interest for the 1st half year = \$ ( 8000 X ------ X 1 ) = \$ 400

100

Amount at the end of 1st half year = \$ ( 8000 + 400 ) = \$ 8400

So, principal for the 2nd half year = \$ 8400

5

Interest for the 2nd half year = \$ ( 8400 X ------ X 1 ) = \$ 420

100

Amount at the end of 2nd half year = \$ ( 8400 + 420 ) = \$ 8820

5

Interest for the 3rd  half year = \$ ( 8820 X ------ X 1 ) = \$ 441

100

Amount at the end of 3rd half year = \$ ( 8820 + 441 ) = \$ 9261

So, the required amount = \$ 9261 and C.I = \$ ( 9261 – 8000 )

= \$ 1261       (Ans.)

Example.5) A man borrow \$ 40000 at 10% per annum, compounded half yearly, He pays back \$ 10000 at the end of every six months. Calculate the third payment, he has to make to clear the entire loan.

Ans.)   Sum borrowed = \$ 40000, Rate = 10%, Per annum = 5% half yearly.

Principal amount for the 1st half = \$ 40000

5

Interest for the 1st half year = \$ ( 40000 X ------- X 1 ) = \$ 2000

100

Amount at the end of 1st half = \$ ( 40000 + 2000 ) = \$ 42000

Repayment done on end of 1st half = \$ ( 42000 – 10000 ) = \$ 32000

Principal amount for the 2nd half \$ 32000

5

Interest for the 2nd half year = \$ ( 32000 X ------- X 1 ) = \$ 1600

100

Amount at the end of 2nd half  year = \$ ( 32000 + 1600 ) = \$ 33600

Repayment done on end of 2nd half = \$ ( 33600 – 10000 ) = \$ 23600

Principal amount for the 3rd half = \$ 23600

5

Interest for the 3rd half year = \$ ( 23600 X ------ X 1 ) = \$ 1180

100

Amount at the end of 3rd half  year = \$ ( 23600 + 1180 ) = \$ 24780

Hence the 3rd payment to clear the loan is \$ 24780        (Ans.)