Finding the accumulated profit.

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The above project is financed by a loan. The company pays 6.25% on the money borrowed and earns 4% on money invested in its deposit account.(Spare funds can't be used to repay the loan at any time)

Cashflow can be understood as :

enter image description here (1) At the start i.e at t=0 , the loan amount is 80,000 , at t=1 , income is 10,000 and another outlay costs 20,000 . So , the total amount to be paid is : (Interest on 80,000)+(20,000 outlay) = (5,000 + 20,000) = 25,000. Income is 10,000 , so additional 15,000 are required. Thus , 15,000 are borrowed additionally. So the new loan amount becomes : Rs. 95,000.

(2) Coming at t=2 , payments to be made are : (Interest on 95,000) + (5,000 Outlay) =(0.0625x95,000)+(5000)= (5937.5 + 5000) = 10,937.5 . Income is 30,000 so payments are paid and the remaining amount left for investment is : ( 30,000 - 10,937.5) = 19,062.5

(3) Coming at t = 3 , payments to be made : (Interest on 95,000)=(0.0625x95,000) = 5,937.5 . Income is 87,000 , so , payments are paid and the remaining amount for investment is : 81,062.5

So , we have spare funds of 19,062.5 at t=2 and 81,062.5 at t=3. Now we need to calculate the accumulated profit at t=5.

Now according to me it should be : $ [\underline{81,062.5} + 19,062.5(1+0.04)^{1} -95,000(1+0.0625)^{1}] (1 + 0.04)^{2}$

But in the solution instead of $\underline{81,062.5}$ , there is $\underline{87,000}$. How is that ? The interest to be paid has to be deducted from the income and the remaining goes for the investment. So , (87,000) how ? Can anyone explain ?

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Answer:

The correct alternatives are (1) and (2):

$ [\underline{81,062.5} + 19,062.5(1+0.04)^{1} -95,000] (1 + 0.04)^{2}\tag{1}$

$ [\boxed{87000} + 19,062.5(1+0.04)^{1} -95,000(1+{\boxed{0.0625}})^{1}] (1 + 0.04)^{2}\tag{2}$

This is because, you are adding the interest for the third year in the second solution where as you are subtracting the interest for the third year in the first term in the first solution. You cannot substract interest for the third year twice as below:

$ [\underline{81,062.5} + 19,062.5(1+0.04)^{1} -95,000(1+\underline{0.0625})^{1}] (1 + 0.04)^{2}\tag{3}$

Hence the first two forms (1) and (2) of the solutions are correct whereas the third (3) is wrong.