Question

A man borrows Rs.30000 at 12% per annum compound interest from a bank and promises to pay off the loans in 20 annual installments beginning at the end of the first year. What is the annual payment necessary?
(A). 4016.76
(B). 3013.54
(C). 4065.24
(D). 1034.54

Answer 1

Hint: The given value is the percent value, thus find the amount which has to be paid at the end of the year. Thus with rate of interest and period substitute the values in the formula.

Complete step-by-step solution -
It is said that a man borrows Rs.30000 at 12% per annum.
It is said that he will pay back in 20 annual installments i.e. n = 20.
We need to find the annual payment, at the end of each year. We can need the formula for ordinary annuity. Ordinary annuity is a series of equal payments made at the end of a consecutive period over a fixed length of time. Now the payment of ordinary annuity can be made as frequently as every week, or monthly, quarterly or annually. Now in this question the payment is paid off by 20 annual installments. Ordinary Annuity is given as ‘A’.
Now the rate of interest is 12%, which is yearly in our case.
Thus we can take, V = Rs.30000, i.e. the present value of money.
 r = 12% = \[\dfrac{12}{100}=0.12\].
The formula for ordinary annuity is,
Ordinary Annuity = r \[\times \] Present value / \[\left[ 1-{{\left( 1+r \right)}^{-n}} \right]\].
\[\Rightarrow A=\dfrac{Vr}{\left[ 1-{{\left( 1+r \right)}^{-n}} \right]}=\dfrac{30000\times 0.12}{\left[ 1-{{\left( 1+0.12 \right)}^{-20}} \right]}\]
Substitute the value of V and r and simplify it.
\[\begin{align}
  & =\dfrac{3600}{\left( 1-0.1036 \right)}=\dfrac{3600}{1-0.1036} \\
 & =\dfrac{3600}{0.8964}=4016.06 \\
\end{align}\]
Thus we got the annual payment as Rs.4016.06.
\[\therefore \] Option (a) is the correct answer.

Note: The pay in the beginning of the period that is you pay year installments in advance. Zero means you pay at the end of period, you pay at the end of the year. Thus after 20 annual installments the loan is settled.