CLASS-9FINDING COMPOUND INTEREST USING FORMULAE

COMPOUND INTEREST BY USING THE FORMULA

1) When interest is compounded annually

a) let Principal = \$ P, Rate = R % per annum, and Time = n years. Then,

R

Amount  = \$ { P ( 1 + ------- )ⁿ }

100

C.I  =  (Amount) – (Principal)

R

=  \$ [ P X { ( 1 + ------- )ⁿ - 1 } ]

100

Example.1) Calculate the amount and the compound interest on \$ 15000 for 2 years at 10% per annum, compounded annually.

Ans.)    Given P = \$ 15000, R = 10% p.a, n = 2 years

R

Amount =  P ( 1 + ------- )ⁿ

100

10

=  \$ { 15000 ( 1 + ------ )² }

100

110                            110         110

= \$  { 15000 X ( ------- )² } = \$ { 15000 X ------- X ------- }

100                            100         100

=  \$ ( 150 X 11 X 11 ) = \$ 18150

Compound Interest (C.I) = Amount (A) – Principal (P)

=  \$ ( 18150 – 15000 ) = \$ 3150

So, required amount (A) = \$ 18150, and Compound Interest (C.I) = \$ 3150         (Ans.)

Example.2) What sum of money will amount to \$ 9680 in 2 years at 10% per annum compounded annually.

Ans.)  As we know the formula –

R

Amount = \$ { P ( 1 + ------ )ⁿ }

100

Here, as per given condition Amount (A) = \$ 9680, Rate (R) = 10%, n = 2, and now we have to find principal (P) = ?

R

Amount =  \$ { P ( 1 + ------ )ⁿ }

100

10

9680 = P ( 1 + ------- )²

100

11

9680 =  P X ( ------- )²

10

( 11 X 11 ) P

9680  =  ---------------

10 X 10

9680 X 100

P  = ---------------  =  \$ 8000

121

So, here the required sum is \$ 8000.       (Ans.)

Example.3) The difference between the compound interest and the simple interest on a certain sum at 9% per annum for 2 years is \$ 324, find the sum.

Ans.) let the required sum be \$ a

Given, Rate (R) = 9% p.a, and Time = 2 hours

P X R X T                9              9a

As per the formula, S.I = ----------- = \$ ( a X ---- X 2 ) = -----

100                   100            50

9                         109 X 109

C.I =  \$ { a ( 1 + ------ )²- a } = \$ { a ( ------------ - a }

100                       100 X 100

(109 X 109)a  – (100 X 100) a

C.I =  \$ { ------------------------------ }

100 X 100

1881 a

=  \$ ------------

10000

1881 a        9 a           1881 a – 1800 a

C.I – S.I = \$ ( -------- - ------- ) = ( --------------- )

10000         50                 10000

81 a

= ---------

10000

But, this difference is given as \$ 324

81 a

So,     ---------- =  \$ 324

10000

324 X 10000

So,     a  =  ---------------  =  \$ 40000

81

Hence the required sum is \$ 40000           (Ans.)