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# 18.9: Finance

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

7.

1. $$P_{20}=2000(1+.03 / 12)^{20(12)}=\ 3641.51$$ in 20 years
2. $$3641.51-2000=\ 1641.51$$ in interest

9. $$P_{8}=P_{0}(1+.06 / 12)^{8(12)}=6000 \cdot P_{0}=\ 3717.14$$ would be needed

11.

1. $$P_{30}=\frac{200\left((1+0.03 / 12)^{30(12)}-1\right)}{0.03 / 12}=\ 116,547.38$$
2. $$200(12)(30)=\ 72,000$$
3. $$\ 116,547.40-\ 72,000=\ 44,547.38$$ of interest

13.

1. $$P_{30}=800,000=\frac{d\left((1+0.06 / 12)^{30(12)}-1\right)}{0.06 / 12} \mathrm{d}=\ 796.40$$ each month
2. $$\ 796.40(12)(30)=\ 286,704$$
3. $$\ 800,000-\ 286,704=\ 513,296$$ in interest

15.

1. $$P_{0}=\frac{30000\left(1-(1+0.08 / 1)^{-25(1)}\right)}{0.08 / 1}=\ 320,243.29$$
2. $$30000(25)=\ 750,000$$
3. $$\ 750,000-\ 320,243.29=\ 429,756.71$$

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.

18.9: Finance is shared under a CC BY-SA 3.0 license and was authored, remixed, and/or curated by David Lippman via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.