Personal Finance: Get Stronger Answer Key

  1. 1. [latex]A = 200+.05(200)= $210[/latex]
  2. 3. [latex]I = 200, t=\frac{13}{52}[/latex] ([latex]13[/latex] weeks out of [latex]52[/latex] in a year), [latex]P_{0} = 9800[/latex]
    [latex]200 = 9800(r)(\frac{13}{52})[/latex]
    [latex]r = 0.0816 = 8.16\%[/latex] annual rate
  3. 5. [latex]P_{10} = 300(1+\frac{.05}{1})^{10(1)} = $488.67[/latex]
  4. 7.
    1. [latex]P_{20} = 2000(1+\frac{.03}{12})^{20(12)} = $3641.51[/latex] in [latex]20[/latex] years
    2. [latex]3641.51−2000=$1641.51[/latex] in interest
  5. 9. [latex]P_{8} = P_{0}(1+\frac{.06}{12})^{8(12)} = 6000 ⋅ P_{0} = $3717.14[/latex] would be needed
  6. 11.
    1. [latex]P_{30} = \frac{200((1+\frac{0.03}{12})^{30(12)}−1)}{\frac{0.03}{12}} = $116,547.38[/latex]
    2. [latex]200(12)(30)=$72,000[/latex]
    3. [latex]$116,547.40−$72,000=$44,547.38[/latex] of interest
  7. 13.
    1. [latex]P_{30} = 800,000 = \frac{d((1+\frac{0.06}{12})^{30(12)}−1)}{\frac{0.06}{12}}d = $796.40[/latex] each month
    2. [latex]$796.40(12)(30)= $286,704[/latex]
    3. [latex]$800,000−$286,704=$513,296[/latex] in interest
  8. 15.
    1. [latex]P_{0} = \frac{30000(1−(1+\frac{0.08}{1})^{−25(1)})}{\frac{0.08}{1}} = $320,243.29[/latex]
    2. [latex]300000(25)=$750,000[/latex]
    3. [latex]$750,000−$320,243.29=$429,756.71[/latex]
  9. 17. [latex]P_{0} = 500,000 = \frac{d(1−(1+\frac{0.06}{12})^{−20(12)})}{\frac{0.06}{12}}d = $3582.16[/latex] each month
  10. 19. [latex]P_{0} = \frac{700(1−(1+\frac{0.05}{12})^{−30(12)})}{\frac{0.05}{12}} = $130,397.13[/latex] loan
    [latex]700(12)(30) = $252,000[/latex]
    [latex]$252,200 − $130,397.13 = $121,602.87[/latex] in interest
  11. 21. [latex]P_{0} = 25,000 = \frac{d(1−(1+\frac{0.02}{12})^{−48})}{\frac{0.02}{12}} = $542.38[/latex] a month
  12. 23. A down payment of [latex]10\%[/latex] is [latex]$20,000[/latex], leaving [latex]$180,000[/latex] as the loan amount.
    [latex]P_{0} = 180,000 = \frac{d(1−(1+\frac{0.05}{12})^{−30(12)})}{\frac{0.05}{12}}d = $966.28[/latex] a month
    [latex]P_{0} = 180,000 = \frac{d(1−(1+\frac{0.06}{12})^{−30(12)})}{\frac{0.06}{12}}d = $1079.19[/latex] a month
  13. 25. First we find the monthly payments: [latex]P_{0} = 24,000 = \frac{d(1−(1+\frac{0.03}{12})^{−5(12)})}{\frac{0.03}{12}}⋅d = $431.25[/latex]
    Remaining balance: [latex]P_{0} = \frac{431.25(1−(1+\frac{0.03}{12})^{−2(12)})}{\frac{0.03}{12}} = $10,033.45[/latex]
  14. 27. [latex]6000(1+\frac{0.04}{12})^{12N} = 10000[/latex]
    [latex](1.00333)^{12N} = 1.667[/latex]
    [latex]\log( (1.00333)^{12N}) = \log(1.667)[/latex]
    [latex]12N \log(1.00333) = \log(1.667)[/latex]
    [latex]N = \frac{\log(1.667)}{12\log(1.00333)} =[/latex] about [latex]12.8[/latex] years
  15. 29. [latex]3000 = \frac{60(1−(1+\frac{0.14}{12})^{−12N})}{\frac{0.14}{12}}[/latex]
    [latex]3000(\frac{0.14}{12}) = 60(1−(1.0117)^{−12N})[/latex]
    [latex]\frac{3000(\frac{0.14}{12})}{60} = 0.5833 = 1−(1.0117)^{−12N}[/latex]
    [latex]0.5833−1 = −(1.0117)^{−12N}[/latex]
    [latex]−(0.5833−1) = (1.0117)^{−12N}[/latex]
    [latex]\log(0.4167) = \log((1.0117)^{−12N})[/latex]
    [latex]\log(0.4167) = −12N \log(1.0117)[/latex]
    [latex]N = \frac{\log(0.4167)}{−12\log(1.0117)} =[/latex] about [latex]6.3[/latex] years
  16. 31. First [latex]5[/latex] years: [latex]P_{5} = \frac{50((1+\frac{0.08}{12})^{5(12)}−1)}{\frac{0.08}{12}} = $3673.84[/latex]
    Next [latex]25[/latex] years: [latex]3673.84(1+\frac{.08}{12})^{25(12)} = $26,966.65[/latex]
  17. 33. Working backwards, [latex]P_{0} = \frac{10000(1−(1+\frac{0.08}{4})^{−10(4)})}{\frac{0.08}{4}} = $273,554.79[/latex] needed at retirement.
    To end up with that amount of money, [latex]273,554.70 = \frac{d((1+\frac{0.08}{4})^{15(4)}−1)}{\frac{0.08}{4}}[/latex].
    He’ll need to contribute [latex]d = $2398.52[/latex] a quarter.