12.7: Chapter 7 Exercise Solutions

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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.

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