Question

1) Suppose you invest $ 1000in an account paying 6 %interest per year.  

a. What is the balance in the account after 2 years? How much of this balance corresponds to​ "interest on​ interest"?

b. What is the balance in the account after 34 years? How much of this balance corresponds to​ "interest on​ interest"?

1. What is the balance in the account after 2 years? The balance in the account​ (with compounded​ interest) after 2 years is $___ (Round to the nearest​ cent.)

2) You have a loan outstanding. It requires making 6 annual payments at the end of the next 6 years of $6,000 each. Your bank has offered to allow you to skip making the next 5 payments in lieu of making one large payment at the end of the​ loan's term in 6 years. If the interest rate on the loan is 8.16%, what final payment will the bank require you to make so that it is indifferent between the two forms of​ payment?

3) Your friend in mechanical engineering has invented a money machine. The main drawback of the machine is that it is slow. It takes one year to manufacture $500 . ​However, once​ built, the machine will last forever and will require no maintenance. The machine can be built​ immediately, but it will cost $5,000 to build. Your friend wants to know if he should invest the money to construct it. If the interest rate is 14.5% per​ year, what should your friend​ do?

Answer 1

Q1
Future Value=Present Value*(1+r)^n

Interest on Interest=Future Value-Present Value*(1+r*n)

a).
Balance after 2 years=1000*1.06^2=1123.6000

Interest on interest=1000*1.06^2-1000*(1+6%*2)=3.6000

b).
Balance after 34 years=1000*1.06^34=7251.0253

Interest on interest=1000*1.06^34-1000*(1+6%*34)=4211.0253

Q2
=Future Value of six payments
=Future Value of annuity
=Payment/rate*((1+r)^n-1)
=6000/8.16%*(1.0816^6-1)
=44193.5455

Q3
NPV=-5000+500/14.5%=-1551.7241

As NPV is negative, do not invest