 Hello and welcome to the session. In this session we will discuss the following question which says, a machine costs $100,000 and its effective life is estimated to be 11 years. If the scrap value is $2,500 only, what amount should we retain out of profits at the end of each year to accumulate at 6% per annum compound interest? Let us first discuss the amount of annuity in case of the immediate annuity. In case of the immediate annuity, the payment is done at the end of the period. So in this case, the amount of annuity given by capital A is equal to small a upon i into 1 plus i to the power of n minus 1, behold. The small a is the annual payment of each installment. Then this n is the number of periods of annuity. This r percent is the rate of interest per period and i is calculated as r upon 100. So this is the key idea that we use in this question. Let us now move on to the solution. In the question we have that the cost of the machine is $100,000 and its effective life is estimated to be 11 years. Also, the scrap value of the machine is given as $2,500. We have to find the amount that should be retained out of profits at the end of each year to accumulate at 6% per annum compound interest. Now as the amount is accumulated at the end of each year, so this is the case of immediate annuity and so we would use this formula to find out the amount of annuity. Now from the question we have that the amount of annuity is to continue for 11 years. So the balance amount to be retained will be given by subtracting the scrap value from the cost of the machine. That is, we subtract $2,500 from $100,000. So this is equal to $97,500. This is the balance amount to be retained. So we would consider capital A that is the amount of annuity as $97,500. Then n would be equal to 11 as the amount of annuity is to continue for 11 years. Then next we have r% is given as 6% per annum. So this is equal to 6% per annum and so from here we have i equal to r upon 100 that is equal to 6 upon 100 which is equal to 0.06 is equal to i. Now from the formula we have a is equal to small a upon i into 1 plus i to the power of n minus 1 where this a which is the annual payment of each installment is to be found. So now putting the respective values we get 97500 is equal to small a upon 0.06 which is i into 1 plus 0.06 which is the value of i this whole to the power of 11 minus 1 the whole. So further we get 97500 into 0.06 is equal to a into 1.06 to the power of 11 minus 1 the whole. On removing this decimal we write here 100 these to 0 cancel with these to 0 and so we get 975 multiplied by 6 as 5850 this is equal to a into 1.06 to the power of 11 minus 1 the whole. Now we find the value of 1.06 to the power of 11 for this we suppose let x be equal to 1.06 to the power of 11. So now log x is equal to 11 into log of 1.06. So further we get log x is equal to 11 into value of log 1.06 is 0.0253 the 11 multiplied by 0.0253 is equal to 0.2783. So log x is equal to 0.2783 and from here we get x is equal to empty log of 0.2783 this gives x equal to 1.898. So we have 1.06 to the power of 11 is equal to 1.898 putting this value in place of 1.06 to the power of 11 we get 5850 is equal to a into 1.898 minus 1 the whole. So further we have 5850 is equal to a into 0.898 that is a is equal to 5850 upon 0.898. So from here we have a is equal to 6514.47 approximately so value of a is 6514.47 dollars approximately. So now the required amount we retained out of profits at the end of each year is equal to 6514.47 dollars approximately. This is our final answer this completes the session hope you have understood the solution of this question.