{"id":48,"date":"2025-01-02T23:03:49","date_gmt":"2025-01-02T23:03:49","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/chapter\/personal-finance-get-stronger-answer-key\/"},"modified":"2025-01-02T23:03:49","modified_gmt":"2025-01-02T23:03:49","slug":"personal-finance-get-stronger-answer-key","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/chapter\/personal-finance-get-stronger-answer-key\/","title":{"raw":"Personal Finance: Get Stronger Answer Key","rendered":"Personal Finance: Get Stronger Answer Key"},"content":{"raw":"\n
    \n\t
  1. 1. [latex]A = 200+.05(200)= $210 [\/latex]<\/li>\n\t
  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]
    \n[latex]200 = 9800(r)(\\frac{13}{52})[\/latex]
    \n[latex]r = 0.0816 = 8.16\\%[\/latex] annual rate<\/li>\n\t
  3. 5. [latex]P_{10} = 300(1+\\frac{.05}{1})^{10(1)} = $488.67[\/latex]<\/li>\n\t
  4. 7. \n
      \n
    1. [latex]P_{20} = 2000(1+\\frac{.03}{12})^{20(12)} = $3641.51[\/latex] in [latex]20[\/latex] years<\/li>\n
    2. [latex]3641.51\u22122000=$1641.51[\/latex] in interest<\/li>\n<\/ol>\n<\/li>\n\t
    3. 9. [latex]P_{8} = P_{0}(1+\\frac{.06}{12})^{8(12)} = 6000 \u22c5 P_{0} = $3717.14[\/latex] would be needed<\/li>\n\t
    4. 11. \n
        \n
      1. [latex]P_{30} = \\frac{200((1+\\frac{0.03}{12})^{30(12)}\u22121)}{\\frac{0.03}{12}} = $116,547.38[\/latex]<\/li>\n
      2. [latex]200(12)(30)=$72,000[\/latex]<\/li>\n
      3. [latex]$116,547.40\u2212$72,000=$44,547.38[\/latex] of interest<\/li>\n<\/ol>\n<\/li>\n\t
      4. 13. \n
          \n
        1. [latex]P_{30} = 800,000 = \\frac{d((1+\\frac{0.06}{12})^{30(12)}\u22121)}{\\frac{0.06}{12}}d = $796.40[\/latex] each month<\/li>\n
        2. [latex]$796.40(12)(30)= $286,704[\/latex]<\/li>\n
        3. [latex]$800,000\u2212$286,704=$513,296[\/latex] in interest<\/li>\n<\/ol>\n<\/li>\n\t
        4. 15. \n
            \n
          1. [latex]P_{0} = \\frac{30000(1\u2212(1+\\frac{0.08}{1})^{\u221225(1)})}{\\frac{0.08}{1}} = $320,243.29[\/latex]<\/li>\n
          2. [latex]300000(25)=$750,000[\/latex]<\/li>\n
          3. [latex]$750,000\u2212$320,243.29=$429,756.71[\/latex]<\/li>\n<\/ol>\n<\/li>\n\t
          4. 17. [latex]P_{0} = 500,000 = \\frac{d(1\u2212(1+\\frac{0.06}{12})^{\u221220(12)})}{\\frac{0.06}{12}}d = $3582.16[\/latex] each month<\/li>\n\t
          5. 19. [latex]P_{0} = \\frac{700(1\u2212(1+\\frac{0.05}{12})^{\u221230(12)})}{\\frac{0.05}{12}} = $130,397.13[\/latex] loan
            \n[latex]700(12)(30) = $252,000[\/latex]
            \n[latex]$252,200 \u2212 $130,397.13 = $121,602.87[\/latex] in interest\n<\/li>\n\t
          6. 21. [latex]P_{0} = 25,000 = \\frac{d(1\u2212(1+\\frac{0.02}{12})^{\u221248})}{\\frac{0.02}{12}} = $542.38[\/latex] a month<\/li>\n\t
          7. 23. A down payment of [latex]10\\%[\/latex] is [latex]$20,000[\/latex], leaving [latex]$180,000[\/latex] as the loan amount.
            \n[latex]P_{0} = 180,000 = \\frac{d(1\u2212(1+\\frac{0.05}{12})^{\u221230(12)})}{\\frac{0.05}{12}}d = $966.28[\/latex] a month
            \n[latex]P_{0} = 180,000 = \\frac{d(1\u2212(1+\\frac{0.06}{12})^{\u221230(12)})}{\\frac{0.06}{12}}d = $1079.19[\/latex] a month\n<\/li>\n\t
          8. 25. First we find the monthly payments: [latex]P_{0} = 24,000 = \\frac{d(1\u2212(1+\\frac{0.03}{12})^{\u22125(12)})}{\\frac{0.03}{12}}\u22c5d = $431.25[\/latex]
            \nRemaining balance: [latex]P_{0} = \\frac{431.25(1\u2212(1+\\frac{0.03}{12})^{\u22122(12)})}{\\frac{0.03}{12}} = $10,033.45[\/latex]<\/li>\n\t
          9. 27. [latex]6000(1+\\frac{0.04}{12})^{12N} = 10000[\/latex]
            \n[latex] (1.00333)^{12N} = 1.667[\/latex]
            \n[latex] \\log( (1.00333)^{12N}) = \\log(1.667) [\/latex]
            \n[latex] 12N \\log(1.00333) = \\log(1.667) [\/latex]
            \n[latex] N = \\frac{\\log(1.667)}{12\\log(1.00333)} = [\/latex] about [latex]12.8[\/latex] years<\/li>\n\t
          10. 29. [latex]3000 = \\frac{60(1\u2212(1+\\frac{0.14}{12})^{\u221212N})}{\\frac{0.14}{12}} [\/latex]
            \n[latex] 3000(\\frac{0.14}{12}) = 60(1\u2212(1.0117)^{\u221212N}) [\/latex]
            \n[latex] \\frac{3000(\\frac{0.14}{12})}{60} = 0.5833 = 1\u2212(1.0117)^{\u221212N} [\/latex]
            \n[latex] 0.5833\u22121 = \u2212(1.0117)^{\u221212N} [\/latex]
            \n[latex] \u2212(0.5833\u22121) = (1.0117)^{\u221212N} [\/latex]
            \n[latex] \\log(0.4167) = \\log((1.0117)^{\u221212N}) [\/latex]
            \n[latex] \\log(0.4167) = \u221212N \\log(1.0117) [\/latex]
            \n[latex] N = \\frac{\\log(0.4167)}{\u221212\\log(1.0117)} = [\/latex] about [latex]6.3[\/latex] years<\/li>\n\t
          11. 31. First [latex]5[\/latex] years: [latex]P_{5} = \\frac{50((1+\\frac{0.08}{12})^{5(12)}\u22121)}{\\frac{0.08}{12}} = $3673.84[\/latex]
            \nNext [latex]25[\/latex] years: [latex]3673.84(1+\\frac{.08}{12})^{25(12)} = $26,966.65[\/latex]<\/li>\n\t
          12. 33. Working backwards, [latex]P_{0} = \\frac{10000(1\u2212(1+\\frac{0.08}{4})^{\u221210(4)})}{\\frac{0.08}{4}} = $273,554.79[\/latex] needed at retirement.
            \nTo end up with that amount of money, [latex] 273,554.70 = \\frac{d((1+\\frac{0.08}{4})^{15(4)}\u22121)}{\\frac{0.08}{4}} [\/latex].
            \nHe\u2019ll need to contribute [latex] d = $2398.52 [\/latex] a quarter.<\/li>\n<\/ol>\n\n","rendered":"
              \n
            1. 1. [latex]A = 200+.05(200)= $210[\/latex]<\/li>\n
            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]
              \n[latex]200 = 9800(r)(\\frac{13}{52})[\/latex]
              \n[latex]r = 0.0816 = 8.16\\%[\/latex] annual rate<\/li>\n
            3. 5. [latex]P_{10} = 300(1+\\frac{.05}{1})^{10(1)} = $488.67[\/latex]<\/li>\n
            4. 7.\n
                \n
              1. [latex]P_{20} = 2000(1+\\frac{.03}{12})^{20(12)} = $3641.51[\/latex] in [latex]20[\/latex] years<\/li>\n
              2. [latex]3641.51\u22122000=$1641.51[\/latex] in interest<\/li>\n<\/ol>\n<\/li>\n
              3. 9. [latex]P_{8} = P_{0}(1+\\frac{.06}{12})^{8(12)} = 6000 \u22c5 P_{0} = $3717.14[\/latex] would be needed<\/li>\n
              4. 11.\n
                  \n
                1. [latex]P_{30} = \\frac{200((1+\\frac{0.03}{12})^{30(12)}\u22121)}{\\frac{0.03}{12}} = $116,547.38[\/latex]<\/li>\n
                2. [latex]200(12)(30)=$72,000[\/latex]<\/li>\n
                3. [latex]$116,547.40\u2212$72,000=$44,547.38[\/latex] of interest<\/li>\n<\/ol>\n<\/li>\n
                4. 13.\n
                    \n
                  1. [latex]P_{30} = 800,000 = \\frac{d((1+\\frac{0.06}{12})^{30(12)}\u22121)}{\\frac{0.06}{12}}d = $796.40[\/latex] each month<\/li>\n
                  2. [latex]$796.40(12)(30)= $286,704[\/latex]<\/li>\n
                  3. [latex]$800,000\u2212$286,704=$513,296[\/latex] in interest<\/li>\n<\/ol>\n<\/li>\n
                  4. 15.\n
                      \n
                    1. [latex]P_{0} = \\frac{30000(1\u2212(1+\\frac{0.08}{1})^{\u221225(1)})}{\\frac{0.08}{1}} = $320,243.29[\/latex]<\/li>\n
                    2. [latex]300000(25)=$750,000[\/latex]<\/li>\n
                    3. [latex]$750,000\u2212$320,243.29=$429,756.71[\/latex]<\/li>\n<\/ol>\n<\/li>\n
                    4. 17. [latex]P_{0} = 500,000 = \\frac{d(1\u2212(1+\\frac{0.06}{12})^{\u221220(12)})}{\\frac{0.06}{12}}d = $3582.16[\/latex] each month<\/li>\n
                    5. 19. [latex]P_{0} = \\frac{700(1\u2212(1+\\frac{0.05}{12})^{\u221230(12)})}{\\frac{0.05}{12}} = $130,397.13[\/latex] loan
                      \n[latex]700(12)(30) = $252,000[\/latex]
                      \n[latex]$252,200 \u2212 $130,397.13 = $121,602.87[\/latex] in interest\n<\/li>\n
                    6. 21. [latex]P_{0} = 25,000 = \\frac{d(1\u2212(1+\\frac{0.02}{12})^{\u221248})}{\\frac{0.02}{12}} = $542.38[\/latex] a month<\/li>\n
                    7. 23. A down payment of [latex]10\\%[\/latex] is [latex]$20,000[\/latex], leaving [latex]$180,000[\/latex] as the loan amount.
                      \n[latex]P_{0} = 180,000 = \\frac{d(1\u2212(1+\\frac{0.05}{12})^{\u221230(12)})}{\\frac{0.05}{12}}d = $966.28[\/latex] a month
                      \n[latex]P_{0} = 180,000 = \\frac{d(1\u2212(1+\\frac{0.06}{12})^{\u221230(12)})}{\\frac{0.06}{12}}d = $1079.19[\/latex] a month\n<\/li>\n
                    8. 25. First we find the monthly payments: [latex]P_{0} = 24,000 = \\frac{d(1\u2212(1+\\frac{0.03}{12})^{\u22125(12)})}{\\frac{0.03}{12}}\u22c5d = $431.25[\/latex]
                      \nRemaining balance: [latex]P_{0} = \\frac{431.25(1\u2212(1+\\frac{0.03}{12})^{\u22122(12)})}{\\frac{0.03}{12}} = $10,033.45[\/latex]<\/li>\n
                    9. 27. [latex]6000(1+\\frac{0.04}{12})^{12N} = 10000[\/latex]
                      \n[latex](1.00333)^{12N} = 1.667[\/latex]
                      \n[latex]\\log( (1.00333)^{12N}) = \\log(1.667)[\/latex]
                      \n[latex]12N \\log(1.00333) = \\log(1.667)[\/latex]
                      \n[latex]N = \\frac{\\log(1.667)}{12\\log(1.00333)} =[\/latex] about [latex]12.8[\/latex] years<\/li>\n
                    10. 29. [latex]3000 = \\frac{60(1\u2212(1+\\frac{0.14}{12})^{\u221212N})}{\\frac{0.14}{12}}[\/latex]
                      \n[latex]3000(\\frac{0.14}{12}) = 60(1\u2212(1.0117)^{\u221212N})[\/latex]
                      \n[latex]\\frac{3000(\\frac{0.14}{12})}{60} = 0.5833 = 1\u2212(1.0117)^{\u221212N}[\/latex]
                      \n[latex]0.5833\u22121 = \u2212(1.0117)^{\u221212N}[\/latex]
                      \n[latex]\u2212(0.5833\u22121) = (1.0117)^{\u221212N}[\/latex]
                      \n[latex]\\log(0.4167) = \\log((1.0117)^{\u221212N})[\/latex]
                      \n[latex]\\log(0.4167) = \u221212N \\log(1.0117)[\/latex]
                      \n[latex]N = \\frac{\\log(0.4167)}{\u221212\\log(1.0117)} =[\/latex] about [latex]6.3[\/latex] years<\/li>\n
                    11. 31. First [latex]5[\/latex] years: [latex]P_{5} = \\frac{50((1+\\frac{0.08}{12})^{5(12)}\u22121)}{\\frac{0.08}{12}} = $3673.84[\/latex]
                      \nNext [latex]25[\/latex] years: [latex]3673.84(1+\\frac{.08}{12})^{25(12)} = $26,966.65[\/latex]<\/li>\n
                    12. 33. Working backwards, [latex]P_{0} = \\frac{10000(1\u2212(1+\\frac{0.08}{4})^{\u221210(4)})}{\\frac{0.08}{4}} = $273,554.79[\/latex] needed at retirement.
                      \nTo end up with that amount of money, [latex]273,554.70 = \\frac{d((1+\\frac{0.08}{4})^{15(4)}\u22121)}{\\frac{0.08}{4}}[\/latex].
                      \nHe\u2019ll need to contribute [latex]d = $2398.52[\/latex] a quarter.<\/li>\n<\/ol>\n","protected":false},"author":6,"menu_order":13,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":3,"module-header":"","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/48"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":0,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/48\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/parts\/3"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/48\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/media?parent=48"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapter-type?post=48"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/contributor?post=48"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/license?post=48"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}