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18.9: Finance

Last updated

Jan 2, 2021
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- 18.8: Growth Models
- Back Matter

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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. $P_{10}=300(1+.05 / 1)^{10(1)}=\$ 488.67$

$P_{20}=2000(1+.03 / 12)^{20(12)}=\$ 3641.51$ 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.

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