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Mathematics LibreTexts

18.9: Finance

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    41799
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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.

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