• [1] (6 points) Problem 20 on p. 813: Which is the best deal over a 5-year period: investing at 8% compounded annually, investing at 7% compounded monthly or investing at 6.8% compounded continuously?
• Solution using EAR approach: Compute the interest on a dollar over a year. 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 EAR). For an annual rate of 8% compounded annually, \$1 grows to (1+(.08/1))1=1.08. 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 continuously, \$1 grows to e.068=1.0704 (using the formula F = Pert). Thus 8% compounded annually is best.
• Solution using doubling time: the deal with the shortest doubling time is best. For an annual rate of 8% compounded annually, the doubling time is t = log(2)/(1*log(1.08)) = 9.006 years. For an annual rate of 7% compounded monthly, the doubling time is t = log(2)/(12*log(1+(.07/12))) = 9.931 years. For an annual rate of 6.8% compounded continuously, the formula for doubling time is t = ln(2)/r, which gives t = 10.193 years. Thus 8% compounded annually is best since it gives the shortest doubling time.
• [2] (6 points) Problem 21 on p. 814: 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.
• (c) The account compounds continuously?
• Solve for P in F = Pert where F = \$50,000, r = .0425 and t = 20. This gives \$50,000 = Pe0.85 or 50,000 = P2.3396 so P = 50,000/2.3396 = \$21,370.74. Thus we see that the account which compounds most frequently needs the smallest initial balance to grow to \$50,000.
• Here is an alternative possibility for [2]: (6 points) Problem 22 on p. 814: How much money would have to be invested in an account at 3.98% annual interest to achieve a balance of \$25,000 in 17 years in each of the following cases?
• (a) The account pays simple interest?
• Solve for P in F = P(1+rt) where F = \$25,000, r = .0398 and t = 17. This gives \$25,000 = P(1+.0398*17) or 25,000 = P(1.6766) so P = 25,000/1.6766 = \$14911.13.
• (b) The account compounds interest quarterly
• Solve for P in F = P(1+(r/m))mt where F = \$25, r = .0398, m = 4 and t = 17. This gives \$25,000 = P(1+.0398/4)68 or 25,000 = P(1.00995)68 so P = 25,000/(1.00995)68 = 25,000/1.9606 = \$12,751.13.
• (c) The account compounds continuously?
• Solve for P in F = Pert where F = \$25,000, r = .0398 and t = 17. This gives \$25,000 = Pe0.6766 or 25,000 = P1.9672 so P = 25,000/1.9672 = \$12,708.56. Thus we see that the account which compounds most frequently needs the smallest initial balance to grow to \$25,000.
• [3] (4 points) Problem 48cd on p. 817.
• (c) Use the rule of 72 to find the approximate doubling time if an investment earns 4.5% compounded daily. Check your result by using the compound interest formula.
• By the rule of 72 it is 72/4.5 = 16 years. The exact answer is t = log(2)/(365*log(1+(.045/365))) = 15.4 years
• (d) Use the rule of 72 to find the approximate annual interest rate, compounded monthly, required for an investment to double in 1.5 years. Check your result by using the compound interest formula.
• By the rule of 72 it is 72/r = 1.5 so r% = 72/1.5 = 48%. To get the exact answer we can use the formula 2 = (1 + (r/12))12*1.5 with guess and check. We'll use r = 48% as our first guess: (1 + (0.48/12))18 = 2.026 is bigger than 2 so r needs to be a little smaller. For r = 0.47 we get (1 + (0.47/12))18 = 1.9968. This is too small so we make r a bit bigger: r = 0.475 gives (1 + (0.475/12))18 = 2.0113 so we try r = 0.4725 and get (1 + (0.4725/12))18 = 2.004. This is pretty close, so the answer is just a bit less than 47.25%. (The exact answer requires us to solve 2 = (1+r/12)12*1.5 for r. Taking logs we get log(2) = 12*1.5*log(1+r/12) or log(2) = 18*log(1+r/12) or log(2)/18 = log(1 + r/12). Exponentiating gives 1.03926 = 10log(2)/18 = 1 + r/12; solving for r gives r = (1.03926 - 1)*12 = 0.471111 or r = 47.1111%.)
• [4] (8 points) Problem 24 p. 831.
• (a) Use the monthly payment formula to determine the monthly payment for a 60-month amortized loan of \$25,495 at 4.5% interest.
• The formula is that the payment is PMT = P(r/12)(1 + (r/12))12t/((1 + (r/12))12t - 1), where P = \$25,495, r = 0.045 and t = 5 years, which gives PMT = \$475.30.
• (b) Use an amortization table to find the monthly payment for the loan from part (a), and compare the result with the monthly payment found in part (a).
• The entry in the table (on p. 824) for r = 4.5% and 5 years is 18.643019 is the payment for \$1000. Multiply by 25.495 to get our payment of \$475.30, which is what we got using the formula.

• [5] (6 points) Problem 36 p. 831: Tim needs to buy a car. After reviewing his budget, he decides he can afford \$250 a month for a car payment. If he pays no money down and gets financing for 5 years at 7.25% interest, how much can he afford to pay for a car? Round to the nearest dollar.
• Use PMT = P(r/12)(1 + (r/12))12t/((1 + (r/12))12t - 1) with PMT = 250, r = 0.0725 and t = 5 and solve for P. We get 250 = P0.019919 so P = 250/0.019919 = \$12,551.