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# 12.7: Capítulo 7 Soluciones para ejercicios

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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.