Question

On January 1, 1985, Marc has the following two options for repaying a loan:
i. Sixty monthly payments of 100 commencing February 1, 1985.
ii. A single payment of 6000 at the end of K months. Interest is at a nominal annual rate of 12 percent compounded monthly.
The two options have the same present value. Determine K.

Answer 1

Answer:

The correct answer is

K = 29.0123 months

Step-by-step explanation:

The given variables are interest at 12 % monthly compounded

to find the number of months

Compound interest = $A=P(1+\frac{r}{n})^{nt}$and

A = Future value.

P= Principal, initial value amount,  

r= Interest rate

n= number of time units,

t = time

and  $F = P(1+i)^{n}$where

P = principal,

i = interest,

F = Future sum payment

n = number of payment time units

The two options have the same eventual value

Therefore we have

$100(1+\frac{r^{-60} }{n}) = 6000*((1+.01)^{-K})$

$100(\frac{1-1.01^{-60} }{.01}) = 6000*((1+.01)^{-K})$

Taking natural logarithm of both sides we have

K= -(㏑(10/6)+$ln(1-1.01^{-60})$)/㏑(1.01)  = 29.0123 months

Alternatively we have