12.7: Chapter 7 Exercise Solutions
- Page ID
- 60001
Chapter 7: Finance
1. \(A = 200 + .05(200) = $210\)
3. \(I=200\). \(t = \dfrac{13}{52}\) (\(13\) weeks out of \(52\) in a year). \(P_0 = 9800\)
\(200 = 9800(r)(\dfrac{13}{52})\)
\(r = 0.0816 = 8.16 \%\) annual rate
5. \(P_{10} = 300 \left(1+ \dfrac{.05}{1} \right)^{10(1)} = $488.67\)
7. a. \(P_{20} = 2000(1+ \dfrac{.03}{12})^{20(12)} = $3641.51\) in \(20\) years
b. \(3641.51 – 2000 = $1641.51\) in interest
9. \(P_8 = P_0\left (1 + \dfrac{.06}{12} \right)^{8(12)} = 6000\). \(P_0 = $3717.14\) would be needed
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. \($116,547.40 - $72,000 = $44,547.38\) of interest
13. a. \(P_{30} = 800000 = \dfrac{d((1 + \dfrac{0.06}{12})^{30(12)}-1)}{\dfrac{0.06}{12}}\;\; d = $796.40\) each month
b. \($796.40(12)(30) = $286,704\)
c. \($800,000 - $286,704 = $513,296\) in interest
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. \(P_{0} = 500000 = \dfrac{d(1 - (1 + \dfrac{0.06}{12})^{-20(12)})}{\dfrac{0.06}{12}}\) \(d = $3582.16\) each month
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. \($252,200 - $130,397.13 = $121,602.87\) in interest
21. \(P_{0} = 25000 = \dfrac{d(1 - (1 + \dfrac{0.02}{12})^{-48})}{\dfrac{0.02}{12}} = $542.38\) a month.
23. a. Down payment of \(10 \%\) is \($20,000\), leaving \($180,000\) as the loan amount
b. \(P_{0} = 180000 = \dfrac{d(1 - (1 + \dfrac{0.05}{12})^{-30(12)})}{\dfrac{0.05}{12}}\; d = $966.28\) a month.
c. \(P_{0} = 180000 = \dfrac{d(1 - (1 + \dfrac{0.06}{12})^{-30(12)})}{\dfrac{0.06}{12}}\; d = $1079.19\) a month.
25. First we find the monthly payments:
\(P_{0} = 24000 = \dfrac{d(1 - (1 + \dfrac{0.03}{12})^{-5(12)})}{\dfrac{0.03}{12}}.\; d = $431.25\)
Remaining balance: \(P_{0} = 24000 = \dfrac{431.25(1 - (1 + \dfrac{0.03}{12})^{-2(12)})}{\dfrac{0.03}{12}}.\; d = $10033.45\)
27. \(6000(1 + \dfrac{0.04}{12})^{12N} = 10000\)
\((1.00333)^{12N} = 1.667\)
\(\log\left((1.00333)^{12N}\right) = \log(1.667)\)
\(12N \log(1.00333) = \log(1.667)\)
\(N = \dfrac{\log(1.667)}{12 \log(1.00333)} =\) about \(12.8\) years
29. \(3000 = \dfrac{60(1 - (1 + \dfrac{0.14}{12})^{-12N})}{\dfrac{0.14}{12}} \)
\(3000(\dfrac{0.14}{12}) = 60(1 - (1.0117)^{-12N}) \)
\(\dfrac{3000(\dfrac{0.14}{12})}{60} = 0.5833 = 1 - (1.0117)^{-12N} \)
\(0.5833 - 1 = -(1.0117)^{-12N} \)
\(-(0.5833 - 1) = (1.0117)^{-12N} \)
\(\log(0.4167) = \log((1.0117)^{-12N})\)
\(\log(0.4167) = -12N \log(1.0117)\)
\(N = \dfrac{\log(0.4167)}{-12N \log(1.0117)} = \) about \(6.3\) years.
31. First \(5\) years: \(P_{5} = \dfrac{50((1 + \dfrac{0.08}{12})^{5(12)}-1)}{\dfrac{0.08}{12}} = $3673.84\)
Next \(25\) years: \(P_{25} = 3673.84(1 + \dfrac{0.08}{12})^{25(12)} = $26,966.65\)
32. Working backwards,
\(P_{0} = \dfrac{10000(1 - (1 + \dfrac{0.08}{4})^{-10(4)})}{\dfrac{0.08}{4}} = $273,554.79\) needed at retirement.
To end up with that amount of money,
\(273,554.70 = \dfrac{d((1 + \dfrac{0.08}{4})^{15(4)} -1)}{\dfrac{0.08}{4}}\).
He’ll need to contribute \(d = $2398.52\) a quarter.