 Hello and welcome to the session. In this session we are going to discuss the following question and the question says that if the population of a town decreases 7% annually and the present population is 500,000, find its population after 2 years. We know that the compound interest formula is amount a equal to p into 1 plus r upon 100 raise to power n. This formula can be applied to any quantity that increases or decreases. The value of both factor after n years is 1 plus r upon 100 raise to power n. And the value of decay factor after n years is 1 minus r upon 100 raise to power n. With this key idea let us proceed with the solution. According to the question the population of a town is 500,000 and it decreases 7% annually where we have to find the population after 2 years. Let the present population be p. It is given to us as 500,000. The decrease rate of population r is equal to 7% annually. The time n is equal to 2 years. So as the population of a town is the quantity that may increase or decrease so we may apply the compound interest formula. In this case the population decreases so we will use the decay factor to calculate the population after 2 years. Therefore according to the formula the amount a will be equal to p into 1 minus r upon 100 raise to power n. This implies a is equal to 500,000 into 1 minus 7 upon 100 raise to power 2 where a is the population after 2 years. So this will be equal to 500,000 into 100 minus 7 upon 100 raise to power 2 which is equal to 500,000 into 93 upon 100 raise to power 2. Which can be written as 500,000 into 93 upon 100 into 93 upon 100. Now after cancelling all the zeros we will obtain 50 into 93 into 93 which is equal to 432,450. Hence the population of the town after 2 years is equal to 432,450 which is our answer. This completes our session. Hope you enjoyed the session.