 Hello and welcome to the session. In this session we are going to discuss the following question and the question says that the compound interest calculated yearly on a certain sum of money for the second year is $600 and for the third year is $660. Calculate the rate of interest and the sum of money. As we know simple interest is equal to P into R into T upon 100 where P is equal to the principal, R is equal to the rate of interest and T is equal to time. With this key idea we shall proceed with the solution. First we shall calculate the difference of two interests which will give us the simple interest for the last period. So the difference of two interests is equal to $660 minus $600 which is equal to $60. Now as simple interest is equal to P into R into T upon 100 so for the last period simple interest will be equal to $60. The principal P is equal to $600 and the time T is equal to one year. Let R be equal to the rate of interest per annum which we have to calculate. So for the last period we have $60 is equal to $600 into R into one upon 100. This implies R is equal to $60 upon 6 which is equal to 10% per annum. Now suppose P was the original sum of money. Let P1 be the amount at the end of first year. Then in the second year 600 is equal to P1 into 10 into 1 upon 100 which implies P1 is equal to 600 into 100 upon 10. This implies that P1 is equal to $6000. Also as P1 is the amount at the end of first year. So P1 is equal to P plus interest that is $6000 is equal to P plus P into 10 into 1 upon 100. This implies $6000 is equal to 100 P plus 10 P upon 100. This implies $6000 is equal to 110 P upon 100 which is equal to 11 P upon 10. So $60,000 is equal to 11 P which implies P is equal to $60,000 upon 11 which is equal to $5,454.54. Hence the rate of interest R is equal to 10% per annum and the principal P is equal to $5,454.54. This is our answer. This completes our session. Hope you enjoyed the session.