 Hello and welcome to the session. In this session, first we will discuss about the population problems. We have the compound interest law. In this, the formula is the amount A is equal to the principal P into 1 plus the rate of interest R upon 100, this whole to the power of n. And this formula applies to any quantity which increases or decreases so that the amount at the end of each period of constant length bears a constant ratio to the amount at the beginning of the period. And at this specific ratio, which is the ratio of the amount at the end of each period to the amount at the beginning of each period is greater than 1. In this case, the ratio is called the growth factor and in this case the law applied is A is equal to P into 1 plus D upon 100 whole to the power of n. This P is the present population, D is the growth rate and this n is the time in years. And also, if the same ratio is less than 1, then this ratio is called the decay factor. The law obeyed in this case is A equal to P into 1 minus D upon 100 whole to the power of n. Where again, this P is the present population, this D is the decay rate in this case and n is the time in years. If suppose we are given the present population P is equal to 40,260,000 and this population increases at 5% annually and we are supposed to find the population at the end of 2 years. So in this case, n would be equal to 2 as we have to find the population at the end of 2 years. Now as the population increases, so the ratio between the amount at the end of each period to the amount at the beginning of each period would be greater than 1. And so as the ratio would be greater than 1, then it would be a growth factor and in that case this law would be obeyed that is A is equal to P into 1 plus D upon 100 whole to the power of n. So to find out the population at the end of 2 years we apply the formula A equal to the present population P into 1 plus the growth rate which is D upon 100 whole to the power of n which is the time. So putting the respective values we get A is equal to the present population P which is 40,260,000 into 1 plus 5 upon 100 this whole to the power of 2 which is further equal to 40,260,000 into 1 plus 5 upon 100. Now we solve further to get A is equal to 44,386,650. So now the population at the end of 2 years is equal to 44,386,650. So this is how we solve the population problems. So if in case we are given the annual birth rate and the annual death rate then the net rate of increase is given by subtracting the annual death rate from the annual birth rate. Next let's discuss appreciation and depreciation if the value of an article increase with the passage of time the article take to appreciate and if the value of an article decreases with the passage of time the article is said to depreciate. Suppose that a man purchased a car for 16,000 dollars so the present value of the car is equal to 16,000 dollars and it's given that the cost of the car depreciating at percent we need to find the value of the car after the value of the car same how would be applied as for the population. So we would use the formula A is equal to P into 1 plus R upon 100 this whole to the power of n. Now as the cost of the car is depreciating so we take the value of R as negative that is R would be equal to minus 10 percent and n in this case would be equal to 2 as we have to find the value of the car after 2 years and P would be the present value of the car equal to 16,000 dollars so A is equal to 16,000 into 1 minus 10 upon 100 this whole to the power of 2. So further we have A is equal to 16,000 into 90 upon 100 into 90 upon 100. Now further solving we get amount A is equal to 12,960 dollars that is value of the car after 2 years is equal to 12,960 dollars. In the same way we can find out the value of an article after n number of years given its appreciation rate. So this completes the session hope you understood the population problems and appreciation and depreciation problems.