# Compound Interest : Clarification Of The Given Solution To This Problem.

I was solving some problems on CI from my textbook, there's a problem whose solution is given in my book but I can't understand it. Here is the problem -

A man borrowed a sum of money and agrees to pay it off by paying 43,200 at the end of the first year and 34,992 at the end of second year. If the reate of compound interest is 8% P.A., find the sum borrowed.

Given Solution -

For the payment of 43,200 at the end of the first year :

A = 43,200; n = 1 yr; r = 8%

$A = P(1 + R/100)^n$ => $43200 = P(1+8/100)^1$

=> $43200 * 100/108 = 40,000$

For the payment of 34,992 at the end of the second year :

$34,992 = P(1 + 8/100)^2$ => $P = 34,992 * (100/108)^2$ = 30,000

Sum borrowed = 40,000 + 30,000 = 70,000

I can't understand the first step. Why 43,200 is taken as amount? As far as I know the actual amount of the first year should be more than what he pays but according to the solution they are considering the amount at the end of 1 year to be exactly equal to what he repays. If he repays total amount by the first time then the principal for the second year must be zero, but according to the question that is not the case. Finally, when they add both the principals to calculate final one also seems confusing to me, probably there's something in the question I am missing. Can someone clarify the solution to me?

• The problem is to find the sum borrowed originally. You are given two future values of 43, 2000 (y1) and 34992 (y2). You find their present value (year 0) and add to give the total sum borrowed. – Paul Dec 14 '15 at 13:47

In general the sum of the present values of the payments at year i ($P_i$) has to be equal to the borrowed money ($B_0$).

The first payment is made at the end of the first year. Therefore the first payment has to be discounted once. The second payment is made at the end of the second year. Thus the second payment has to be discounted twice.

$B_0=\frac{P_1}{1+i}+\frac{P_2}{(1+i)^2}$

$B_0=\frac{43,200}{1+0.08}+\frac{34,992}{(1+0.08)^2}$

$B_0=40,000+30,000$

The payment 43,200 has an value of 40,000 at the beginning of the first year. Or the other way round. 40,000 can be saved at the beginning of the first year. After one year (end of the first year) you get $0.08\cdot 40,000$ interest. In total after one year you have $3200+40,000=43,200$.

Therefore 40,000 at the beginning of the year is equivalent to 43,200 at the end of the year. The assumption is that the borrowing rate is equal to investment rate.