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12.7: Capítulo 7 Soluciones para ejercicios

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    112983
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    Capítulo 7: Finanzas

    1. \(A = 200 + .05(200) = $210\)

    3. \(I=200\). \(t = \dfrac{13}{52}\)(\(13\)semanas fuera\(52\) de un año). \(P_0 = 9800\)

    \(200 = 9800(r)(\dfrac{13}{52})\)

    \(r = 0.0816 = 8.16 \%\)tasa anual

    5. \(P_{10} = 300 \left(1+ \dfrac{.05}{1} \right)^{10(1)} = $488.67\)

    7. a.\(P_{20} = 2000(1+ \dfrac{.03}{12})^{20(12)} = $3641.51\) en\(20\) años

    b.\(3641.51 – 2000 = $1641.51\) en interés

    9. \(P_8 = P_0\left (1 + \dfrac{.06}{12} \right)^{8(12)} = 6000\). \(P_0 = $3717.14\)sería necesario

    11. a.\(P_{30} = \dfrac{200((1 + \dfrac{0.03}{12})^{30(12)}-1)}{\dfrac{0.03}{12}} = $116,547.38\)

    b.\(200(12)(30) = $72,000\)

    c.\($116,547.40 - $72,000 = $44,547.38\) de interés

    13. a.\(P_{30} = 800000 = \dfrac{d((1 + \dfrac{0.06}{12})^{30(12)}-1)}{\dfrac{0.06}{12}}\;\; d = $796.40\) cada mes

    b.\($796.40(12)(30) = $286,704\)

    c.\($800,000 - $286,704 = $513,296\) en interés

    15. a.\(P_{0} = \dfrac{30000(1 - (1 + \dfrac{0.08}{1})^{-25(1)})}{\dfrac{0.08}{1}} = $320,253.29\)

    b.\(30000(25) = $750,000\)

    c.\($750,000 - $320,253.29 = $429,756.71\)

    17. \(P_{0} = 500000 = \dfrac{d(1 - (1 + \dfrac{0.06}{12})^{-20(12)})}{\dfrac{0.06}{12}}\)\(d = $3582.16\)cada mes

    19. a.\(P_{0} = 500000 = \dfrac{700(1 - (1 + \dfrac{0.05}{12})^{-30(12)})}{\dfrac{0.05}{12}} =\; \text{a}\; $130,397.13 \text{ loan}\)

    b.\(700(12)(30) = $252,000\)

    c.\($252,200 - $130,397.13 = $121,602.87\) en interés

    21. \(P_{0} = 25000 = \dfrac{d(1 - (1 + \dfrac{0.02}{12})^{-48})}{\dfrac{0.02}{12}} = $542.38\)al mes.

    23. a. Pago inicial de\(10 \%\) es\($20,000\), dejando\($180,000\) como monto del préstamo

    b.\(P_{0} = 180000 = \dfrac{d(1 - (1 + \dfrac{0.05}{12})^{-30(12)})}{\dfrac{0.05}{12}}\; d = $966.28\) un mes.

    c.\(P_{0} = 180000 = \dfrac{d(1 - (1 + \dfrac{0.06}{12})^{-30(12)})}{\dfrac{0.06}{12}}\; d = $1079.19\) un mes.

    25. Primero encontramos los pagos mensuales:

    \(P_{0} = 24000 = \dfrac{d(1 - (1 + \dfrac{0.03}{12})^{-5(12)})}{\dfrac{0.03}{12}}.\; d = $431.25\)

    Saldo restante:\(P_{0} = 24000 = \dfrac{431.25(1 - (1 + \dfrac{0.03}{12})^{-2(12)})}{\dfrac{0.03}{12}}.\; d = $10033.45\)

    27. \(6000(1 + \dfrac{0.04}{12})^{12N} = 10000\)

    \((1.00333)^{12N} = 1.667\)

    \(\log\left((1.00333)^{12N}\right) = \log(1.667)\)

    \(12N \log(1.00333) = \log(1.667)\)

    \(N = \dfrac{\log(1.667)}{12 \log(1.00333)} =\)acerca de\(12.8\) años

    29. \(3000 = \dfrac{60(1 - (1 + \dfrac{0.14}{12})^{-12N})}{\dfrac{0.14}{12}} \)

    \(3000(\dfrac{0.14}{12}) = 60(1 - (1.0117)^{-12N}) \)

    \(\dfrac{3000(\dfrac{0.14}{12})}{60} = 0.5833 = 1 - (1.0117)^{-12N} \)

    \(0.5833 - 1 = -(1.0117)^{-12N} \)

    \(-(0.5833 - 1) = (1.0117)^{-12N} \)

    \(\log(0.4167) = \log((1.0117)^{-12N})\)

    \(\log(0.4167) = -12N \log(1.0117)\)

    \(N = \dfrac{\log(0.4167)}{-12N \log(1.0117)} = \)acerca de\(6.3\) años.

    31. Primeros\(5\) años:\(P_{5} = \dfrac{50((1 + \dfrac{0.08}{12})^{5(12)}-1)}{\dfrac{0.08}{12}} = $3673.84\)

    Los próximos\(25\) años:\(P_{25} = 3673.84(1 + \dfrac{0.08}{12})^{25(12)} = $26,966.65\)

    32. Trabajando hacia atrás,

    \(P_{0} = \dfrac{10000(1 - (1 + \dfrac{0.08}{4})^{-10(4)})}{\dfrac{0.08}{4}} = $273,554.79\)necesarios al jubilarse.

    Para terminar con esa cantidad de dinero,

    \(273,554.70 = \dfrac{d((1 + \dfrac{0.08}{4})^{15(4)} -1)}{\dfrac{0.08}{4}}\).

    Tendrá que aportar\(d = $2398.52\) una cuarta parte.


    This page titled 12.7: Capítulo 7 Soluciones para ejercicios is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Darlene Diaz (ASCCC Open Educational Resources Initiative) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.