18.9: Finance
1. \(A=200+.05(200)=\$ 210\)
3. \(\mathrm{I}=200 . \mathrm{t}=13 / 52(13 \text { weeks out of } 52 \text { in a year }) . \mathrm{P}_{0}=9800\)
\(200=9800(\mathrm{r})(13 / 52) \mathrm{r}=0.0816=8.16 \%\) annual rate
5. \(P_{10}=300(1+.05 / 1)^{10(1)}=\$ 488.67\)
7.
- \(P_{20}=2000(1+.03 / 12)^{20(12)}=\$ 3641.51\) in 20 years
- \(3641.51-2000=\$ 1641.51\) in interest
9. \(P_{8}=P_{0}(1+.06 / 12)^{8(12)}=6000 \cdot P_{0}=\$ 3717.14\) would be needed
11.
- \(P_{30}=\frac{200\left((1+0.03 / 12)^{30(12)}-1\right)}{0.03 / 12}=\$ 116,547.38\)
- \(200(12)(30)=\$ 72,000\)
- \(\$ 116,547.40-\$ 72,000=\$ 44,547.38\) of interest
13.
- \(P_{30}=800,000=\frac{d\left((1+0.06 / 12)^{30(12)}-1\right)}{0.06 / 12} \mathrm{d}=\$ 796.40\) each month
- \(\$ 796.40(12)(30)=\$ 286,704\)
- \(\$ 800,000-\$ 286,704=\$ 513,296\) in interest
15.
- \(P_{0}=\frac{30000\left(1-(1+0.08 / 1)^{-25(1)}\right)}{0.08 / 1}=\$ 320,243.29\)
- \(30000(25)=\$ 750,000\)
- \(\$ 750,000-\$ 320,243.29=\$ 429,756.71\)
17. \(P_{0}=500,000=\frac{d\left(1-(1+0.06 / 12)^{-20(12)}\right)}{0.06 / 12} \mathrm{d}=\$ 3582.16\) each month
19.
\(P_{0}=\frac{700\left(1-(1+0.05 / 12)^{-30(12)}\right)}{0.05 / 12}=\) a \(\$ 130,397.13\) loan
\(700(12)(30)=\$ 252,000\)
\(\$ 252,200-\$ 130,397.13=\$ 121,602.87\) in interest
21. \(P_{0}=25,000=\frac{d\left(1-(1+0.02 / 12)^{-48}\right)}{0.02 / 12}=\$ 542.38\) a month
23.
Down payment of \(10 \%\) is \(\$ 20,000\), leaving \(\$ 180,000\) as the loan amount
\(P_{0}=180,000=\frac{d\left(1-(1+0.05 / 12)^{-30(12)}\right)}{0.05 / 12} \mathrm{d}=\$ 966.28 \mathrm{amonth}\)
\(P_{0}=180,000=\frac{d\left(1-(1+0.06 / 12)^{-30(12)}\right)}{0.06 / 12} \mathrm{d}=\$ 1079.19\) a month
25. First we find the monthly payments:
\(P_{0}=24,000=\frac{d\left(1-(1+0.03 / 12)^{-5(12)}\right)}{0.03 / 12} \cdot d=\$ 431.25\)
Remaining balance: \(P_{0}=\frac{431.25\left(1-(1+0.03 / 12)^{-2(12)}\right)}{0.03 / 12}=\$ 10,033.45\)
27. \(6000(1+0.04 / 12)^{12 N}=10000\)
\((1.00333)^{12 N}=1.667\)
\(\log \left((1.00333)^{12 N}\right)=\log (1.667)\)
\(12 N \log (1.00333)=\log (1.667)\)
\(N=\frac{\log (1.667)}{12 \log (1.00333)}=\) about 12.8 years
29. \(3000=\frac{60\left(1-(1+0.14 / 12)^{-12 N}\right)}{0.14 / 12}\)
\(3000(0.14 / 12)=60\left(1-(1.0117)^{-12 N}\right)\)
\(\frac{3000(0.14 / 12)}{60}=0.5833=1-(1.0117)^{-12 N}\)
\(0.5833-1=-(1.0117)^{-12 N}\)
\(-(0.5833-1)=(1.0117)^{-12 N}\)
\(\log (0.4167)=\log \left((1.0117)^{-12 N}\right)\)
\(\log (0.4167)=-12 N \log (1.0117)\)
\(N=\frac{\log (0.4167)}{-12 \log (1.0117)}=\) about 6.3 years
31. First 5 years: \(P_{5}=\frac{50\left((1+0.08 / 12)^{5(12)}-1\right)}{0.08 / 12}=\$ 3673.84\)
Next 25 years: \(3673.84(1+.08 / 12)^{25(12)}=\$ 26,966.65\)
33. Working backwards, \(P_{0}=\frac{10000\left(1-(1+0.08 / 4)^{-10(4)}\right)}{0.08 / 4}=\$ 273,554.79\) needed at retirement. To end up with that amount of money, \(273,554.70=\frac{d\left((1+0.08 / 4)^{15(4)}-1\right)}{0.08 / 4}\). He’ll need to contribute \(d=\$ 2398.52\) a quarter.