M203J Practice Quiz 4 Solutions	Covering Chapter 10

Instructions: Answer each question, and when required explain your answer. 
Your explanation must be clear and complete. You may refer to your book, 
your notes and your homework papers.

[1] Which is the best deal over a 3-year period and which is the worst: investing at 7.2% compounded annually, investing at 7% compounded monthly or investing at 6.8% compounded daily? Show how you determine the answer.

Compute the interest on a dollar over a year (the 3-year period is irrelevant). Or even simpler just find out how much $1 grows to over the course of a year (if you subtract that $1 out at the end you have the effective rate or APY). For an annual rate of 7.2% compounded annually, $1 grows to (1+(.072/1))^1=1.072. For an annual rate of 7% compounded monthly, $1 grows to (1+(.07/12))^12=1.0722. For an annual rate of 6.8% compounded daily, $1 grows to (1+(.068/365))^365=1.07036. Thus 7% compounded monthly is best and 6.8% compounded daily is worst.

[2] How much money would have to be invested in an account at 4.25% annual interest to achieve a balance of $50,000 in 20 years in each of the following cases?

(a) The account pays simple interest. Solve for P in F = P(1+rt) where F = $50,000, r = .0425 and t = 20. This gives $50,000 = P(1+.0425*20) or 50,000 = P(1.85) so P = 50,000/1.85 = $27,027.03. 

(b) The account compounds interest semi-annually. Solve for P in F = P(1+(r/m))^(mt) where F = $50,000, r = .0425, m = 2 and t = 20. This gives $50,000 = P(1+.0425/2)^(40) or 50,000 = P(1.02125)^(40) so P = 50,000/(1.02125)^(40) = 50,000/2.3189 = $21,561.91. 

[3] Use the appropriate formula to determine the monthly payment for a 60-month amortized loan of $25,495 at 4.5% interest, compounded monthly.

The formula to use is F = C((1 + (r/12))^(12t) - 1)/(r/12), where F = P(1 + (r/12))^(12t) is the future value of the loan payments, C is the payment, P = $25,495 is the present value of the loan (i.e., it's the amount of the loan), r = 0.045 and t = 5 years. Thus 
C = P(r/12)(1 + (r/12))^(12t)/((1 + (r/12))^(12t) - 1) = $475.30. 

[4] How much money should you put in the bank today, at an APR of 3% compounded monthly, if you wish to withdraw 60 payments of $100 each, at monthly intervals with the first withdrawal one month from today, the second withdrawal two months from today, etc.? (I.e., what is the present value of 60 monthly payments of $100 each?)

Again the formulas to use are F = C((1 + (r/12))^(12t) - 1)/(r/12) and F = P(1 + (r/12))^(12t), where C = 100, t = 5 years (since 60 monthly payments will take 5 years), r = 0.03 and P is what we want to know. Thus P = C((1 + (r/12))^(12t) - 1)/((r/12)(1 + (r/12))^(12t)) = $5565.24. Note that 60 payments of $100 each adds up to $6000, which is more than what we need to put into the account. The difference is made up by the interest earned by the money in the account.

[5] How much money will you have in the bank in 5 years if starting today you make 60 monthly payments of $100 each, assuming the account has an APR of 3% compounded monthly? (I.e., what is the future value of 60 monthly payments of $100 each?) Note that the last payment will be in the account for one month, while the first payment is in the account for 60 months.

This time the formula to use is F = C(1+(r/12))((1 + (r/12))^(12t) - 1)/(r/12) = $6480.83. Note that this is more than the $6000 that the payments add up to, since the payments accrue interest.

[6] Problem 84 on p. 397.

The redemption value of the bond, $10000, is the future value (after 7 years at an APR of 3.5%, compounded annually). The purchase price of the bond is the present value of the $10000. Thus F = 10000, where F = P(1 + r)^7 and r = 0.035. Thus P = 10000/1.035^7 = $7859.91.