12.7: Capítulo 7 Soluciones para ejercicios
- Page ID
- 112983
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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.