Worked solutions using the value K = 25. Numerical solutions for all values of K from 0 to 99 are also posted.

- 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%. - 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.
- 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. - 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.
- 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.
- 4376.44(1-(194/365)(0.19)) = $3934.48. The
company pays $3934.48.
- (a)
*j*_{4}= 6.5% so 1+*j*_{1}= (1+(0.065/4))^{4}= 1.0666012. Thus the effective annual rate is 6.6602%.(b) (1+(

*j*_{365}/365))^{365}= (1+(0.065/4))^{4}so*j*_{365}= 365((1.01625)^{4/365}-1) = 6.4483%. The nominal rate compounded daily is 6.4483%. - (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. - (1+
*j*_{2}/2)^{2}= 1+*j*_{1}so (1+*x*/2)^{2}= 1+*x*+0.0064. Thus 1+*x*+*x*^{2}/4 = 1+*x*+0.0064 so*x*^{2}/4 = 0.0064 and*x*= 0.16. Hence*x*is 0.16 or 16%. - 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.]

File translated from T

On 13 Oct 2000, 14:42.