AS/MATH2580 A Assignment AS/MATH2580 A Assignment #1 Answers
Worked solutions using the value K = 25. Numerical solutions for all values of K from 0 to 99 are also posted.

1. Terms 5/5, n/45 mean that: 14542(0.95)(1+((45-5)/365)r) = 14542.
Thus r = ((1/0.95)-1)(365/40) = 0.480263. The maximum simple interest rate at which the borrower can take advantage of the 5% discount is 48.0263%.

2. Interest at 8.5% on \$10000 for 35 days is 10000(35/365)(0.085) = \$81.51. Interest at 8.5% on \$7000 for (218-131) = 87 days is 7000(87/365)(0.085) = \$141.82. Interest at 9.5% on \$7000 for (282-218) = 64 days is 7000(64/365)(0.095) = \$116.60. Interest at 9.5% on \$6000 fir (354-282) = 72 days is 6000(72/365)(0.095) = \$112.44 The total interest paid is 81.51+141.82+116.60+112.44 = \$452.37. The final payment is \$7000 + (interest Dec 1 to 20) = 6000(1+(19/365)(0.095)) = \$6029.67.

3. The mature value of the note in Nov 18, 1999 is 4000(1+((322-93)/365)(0.15)) = \$4376.44. The discounted value (322-93)-35 = 194 days earlier is 4376.44(1+(194/365)(0.18))-1 = \$3994.30. The company pays \$3994.30 for the note.

4. 9500(1+(84/365)(0.10))-2025[1+(42/365)(0.10)+1+(28/365)(0.10)+1+(14/365)(0.10)] = 3957.03. The balance at 12 weeks is \$3957.03.

5. 6 weeks: 9500(1+(42/365)(0.10))-2025 = \$7584.32; 8 weeks: 7584.32(1+(14/365)(0.10))-2025 = \$5588.41; 10 weeks: 5588.41(1+(14/365)(0.10))-2025 = \$3584.84; 12 weeks: 3584.84(1+(14/365)(0.10)) = \$3598.60. The balance at 12 weeks is \$3598.60.

6. 4376.44(1-(194/365)(0.19)) = \$3934.48. The company pays \$3934.48.

7. (a) j4 = 6.5% so 1+j1 = (1+(0.065/4))4 = 1.0666012. Thus the effective annual rate is 6.6602%.

(b) (1+(j365/365))365 = (1+(0.065/4))4 so j365 = 365((1.01625)4/365-1) = 6.4483%. The nominal rate compounded daily is 6.4483%.

8. (a) I = 10000(73/12)(0.08) = \$4866.67. The simple interest is \$5475.

(b) I = S-P = 10000(1+(0.08/12))73-10000 = \$6242.59. The compound interest is \$7253.97.

9. (1+j2/2)2 = 1+j1 so (1+x/2)2 = 1+x+0.0064. Thus 1+x+x2/4 = 1+x+0.0064 so x2/4 = 0.0064 and x = 0.16. Hence x is 0.16 or 16%.

10. On Sept. 1, 2000 her balance is 5000(1+((244-227)/365)(0.07)) = \$5016.30. Let it amount to \$10000 in n months at 0.07/12 = 0.00583333333 per month. Then 5016.30(1.0083333333)n = 10000. Thus n = (log(10000/5016.30))/(log(1.005833333) = 118.61. Take 118 months (= 9 years plus 10 months = July 1, 2010). At that time the balance is 5016.30(1.00583333)118 = 9964.47. After D more days the balance is 9964.47(1+(D/365)(0.07)) = 10000. Thus D = 18.59 days and thus we need 19 more days. Thus on July 20, 2010 she can for the first time withdraw \$10000. [Note: We could also use the simpler calculation 0.61(365/12) = 18.55 days, so use 19 days.]

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On 13 Oct 2000, 14:42.