Question

please help me solving this problem.

Answer 1

Let the sum invested be P.

Given that A certain sum of money invested at 10% p.a, the interest compounded annually.

We know that A = P(1 + r/100)^n.


Amount for 1st  year:

A = P(1 + 10/100)^n

   = P(1 + 1/10)^1

   = P(11/10)

   = 1.1P.

We know that CI = A - P.

Therefore the CI for 1st year = 1.1P - P

                                                = 0.1P.



Amount for 2nd year:

A = P(1 + r/100)^2

    = P(1 + 10/100)^2

    = P(11/10)^2

    = P(121/100)

    = 1.21P.


Compound Interest for 2 years = A - P

                                                     = 1.21P - P

                                                      = 0.21P


Therefore CI for 2nd year = 0.21P - 0.1P

                                             = 0.11P


Amount for the 3rd year:

A = P(1 + r/100)^3

   = P(1 + 10/100)^3

   = P(11/10)^3

   = P(1331/1000)

   = 1.331P.


Compound Interest of 3 years = A - P

                                                    = 1.331P - 0.21P

                                                    = 0.331P


Compound Interest for the 3rd year = 0.331P - 0.21P

                                                             = 0.121P


Now,

Given that difference between the Compound Interests of the 1st year & 3rd year = 1105.

0.121P - 0.1P = 1105

0.021P = 1105

P = 1105/0.021

P = 52619.04

Therefore the sum invested = 52619.04.


Hope this helps!