{"id":461,"date":"2020-04-27T20:08:54","date_gmt":"2020-04-28T00:08:54","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/?post_type=chapter&p=461"},"modified":"2021-07-12T14:15:15","modified_gmt":"2021-07-12T18:15:15","slug":"solutions-to-chapter-3-learning-activities","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/chapter\/solutions-to-chapter-3-learning-activities\/","title":{"raw":"Solutions to Chapter 3 Knowledge Checks","rendered":"Solutions to Chapter 3 Knowledge Checks"},"content":{"raw":"

Knowledge Check 3.1<\/h2>\r\n
    \r\n \t
  1. P\u00a0 = $3,000;\u00a0 r<\/em>\u00a0 =\u00a0 6%\u00a0 = 0.06;\u00a0 t<\/em>\u00a0 =180\/365 years [latex]I=Prt=$3,000\\times0.06\\times\\frac{180}{365}=$88.77[\/latex]<\/li>\r\n \t
  2. I=$55; r<\/em> =5.5%=0.055; t =125\/365 years [latex]P=\\frac{I}{rt}=\\frac{$55}{0.055\\left(\\frac{125}{365}\\right)} =$2920.00[\/latex]<\/li>\r\n \t
  3. P = $900,\u00a0\u00a0 I = $65,\u00a0\u00a0 r<\/em> = 7.5% = 0.075<\/li>\r\n<\/ol>\r\n[latex]t=\\frac{I}{Pr}=\\frac{$65}{$900\\times 0.075}=0.962963\\text{ years}[\/latex]\r\n\r\nTo convert this answer to days, you must multiply by 365. To eliminate a rounding error, be sure to use the exact value from your calculator, i.e., just multiply the above value by 365 without re-entering the displayed value .\r\n

    [latex]\u2234 t =\\frac{$65}{$900\\times0.075}\\text{years}\\times \\frac{365\\text{ days}}{1\\text{ years}} = 351.48\\text{ days}[\/latex]<\/p>\r\nThis should now be rounded up to 352 days.\r\n

      \r\n \t
    1. \u00a0P\u00a0 =\u00a0 $975, I\u00a0 = $36.73, t<\/em>\u00a0 =220\/365\u00a0 years<\/li>\r\n<\/ol>\r\n

      [latex]r=\\frac{I}{Pt}=\\frac{$36.73}{$975\\left(\\frac{220}{365}\\right)} = 0.062500932 = 6.25%[\/latex]<\/p>\r\n \r\n

      Knowledge Check 3.2<\/h2>\r\n
        \r\n \t
      1. P = $4,000, r\u00a0 = 8%\u00a0 = 0.08,\u00a0 t = 210\/365\u00a0 years<\/li>\r\n<\/ol>\r\n\"Timeline\r\n\r\n[latex]FV=$4,000 \\left[1 +0.08\\left(\\frac{210}{365}\\right)\\right][\/latex]\r\n
          \r\n \t
        1. P = $1,250,\u00a0 r<\/em> = 6.75% = 0.0675; [latex]t = (251 - 69)+365 =\\frac{182}{365}[\/latex] years (from Table 3- 1)<\/li>\r\n<\/ol>\r\n

          \"Timeline<\/p>\r\n[latex]FV=$1,250\\left[1 +0.0675\\left(\\frac{182}{365}\\right)\\right]=$1,292.07[\/latex]\r\n

            \r\n \t
          1. P = $2,500, r = 3.75% = 0.0375, t = 2 years<\/li>\r\n<\/ol>\r\n\"Timeline\r\n\r\n[latex]FV =$2,500 (1 +0.0375\\times 2)= $2,687.50[\/latex]\r\n\r\n \r\n

            Knowledge Check 3.3<\/h2>\r\n
              \r\n \t
            1. P = $2,000, FV = $2,210, t<\/em> = 1.5 years<\/li>\r\n<\/ol>\r\n\"Timeline\r\n

              Either of the two following approaches is acceptable:<\/p>\r\nApproach A:\r\n

              [latex]I = FV - P =$2,210 -$2,000 =$210[\/latex]<\/p>\r\nSo\r\n

              [latex]r=\\frac{I}{Pt}=\\frac{$210}{$2,000\\times1.5}=0.07=7%[\/latex]<\/p>\r\nApproach B:\r\n

              [latex] r = \\frac{\\frac{FV}{P}-1}{t} = \\frac{\\frac{$2,210}{$2,000}-1}{1.5}= 0.07[\/latex]<\/p>\r\n\r\n

                \r\n \t
              1. FV = $1,871.25, r<\/em> = 9% = 0.09, t<\/em> = 33 months =33\/12\u00a0 years.<\/li>\r\n<\/ol>\r\n\"Timeline\r\n

                [latex]P=\\frac{FV}{1+rt}=\\frac{$1,871.25}{1+0.09\\times \\left(\\frac{33}{12}\\right)}=$1,500.00[\/latex]<\/p>\r\n \r\n

                Knowledge Check 3.4<\/h2>\r\na. Find FV at 7%:\r\n\r\n\"Timeline\r\n

                [latex]FV = P(1+rt) = $10,000 \\left(1+0.07\u00d7\\frac{6}{12}\\right) = $10,350[\/latex]<\/p>\r\nConclusion: The value of $10,000 now, in six months\u2019 time is $10,350. Since this is $25 greater than $10,325, you would prefer $10,000 now from a purely financial point of view.\r\n\r\nb. Find FV at 6%:\r\n\r\n\"Timeline\r\n\r\n \r\n

                [latex]FV = P(1+rt) = $10,000 \\left(1+0.06\u00d7\\frac{6}{12}\\right) = $10,300[\/latex]<\/p>\r\nConclusion: The value of $10,000 now, in six months time is $10,300. Since this is less than $10,325, you would prefer $10,325 in six months time.\r\n

                Knowledge Check 3.5<\/h2>\r\n\"Timeline\r\n\r\n[latex]\\begin{align*}\r\n\r\n&FV_1=$20,000\\left(1+0.08\\times\\frac{15}{12}\\right)=$22,000.00\\\\\r\n\r\n&FV_2=$5,000\\left(1+0.08\\times\\frac{10}{12}\\right)=$5,333.33\\\\\r\n\r\n&FV_3=$10,000\\left(1+0.08\\times \\frac{6}{12}\\right)=$10,400.00\\\\\r\n\r\n\\end{align*}[\/latex]\r\n\r\nTotal Debt Outstanding = $37,733.33(6 months from now)\r\n\r\n \r\n

                Knowledge Check 3.6<\/h2>\r\n1.\r\n\r\n\"Timeline\r\n\r\n \r\n

                [latex]PV=\\frac{$5,000}{1+0.09\\times \\frac{19}{12}}=$4,376.37[\/latex]<\/p>\r\n\r\n

                  \r\n \t
                1. <\/li>\r\n<\/ol>\r\n\"Timeline\r\n

                  [latex]\\begin{align*}\r\n&PV_1=\\frac{$2,000}{1+0.07\\times \\frac{7}{12}}=$1,921.54\\\\\r\n&PV_2=\\frac{$4,000}{1+0.07\\times \\frac{13}{12}}=$3,718.05\r\n\\end{align*}[\/latex]<\/p>\r\n\r\nTotal Equivalent Debt Now = $5,639.59\r\n

                  Knowledge Check 3.8<\/h2>\r\n\"Timeline\r\n\r\n \r\n\r\nValue of \"old\" payments at the focal date :\r\n

                  [latex]$3,500\\left(1+0.08\\times \\frac{8}{12}\\right)+$5,500\\left(1+0.08\\times \\frac{1}{12}\\right)[\/latex]<\/p>\r\n

                  [latex]=$3,686.67+$5,536.67=$9,223.33[\/latex]<\/p>\r\n \r\n\r\nValue of \"replacement\" payments at the focal date:\r\n

                  [latex]$3,000\\left(1+0.08\\times \\frac{6}{12}\\right)+x=$3,120+x[\/latex]<\/p>\r\n

                  Therefore:<\/p>\r\n

                  [latex]$3,120+x=$9,223.33[\/latex]<\/p>\r\n

                  [latex]x =$9,223.33 -$3,120.00=$6,103.33[\/latex]<\/p>\r\n \r\n

                  Knowledge Check 3.9<\/h2>\r\n\"Timeline\r\n\r\nLet the amount of two equal payments be x. Value of \"old\" payments at the focal date:\r\n

                  [latex]$10,000\\left(1+0.09\\times \\frac{4}{12}\\right)+\\frac{$9,000}{\\left(1+0.09\\times \\frac{2}{12}\\right)}[\/latex]<\/p>\r\nValue of \"replacement\" payments at the focal date:\r\n

                  [latex]x+\\frac{x}{\\left(1+0.09\\times \\frac{6}{12}\\right)}[\/latex]<\/p>\r\nAnd we can set these to be equal:\r\n

                  [latex]x+\\frac{x}{\\left(1+0.09\\times \\frac{6}{12}\\right)} = $10,000\\left(1+0.09\\times \\frac{4}{12}\\right)+\\frac{$9,000}{\\left(1+0.09\\times \\frac{2}{12}\\right)}[\/latex]<\/p>\r\nAnd solve, remembering to store all intermediate values in the calculator:\r\n

                  [latex]\\begin{align*}\r\nx+\\frac{x}{1.045} &=$10,000(1.03)+\\frac{$9,000}{1.015}\\\\\r\nx\\left(1+\\frac{1}{1.045}\\right)&=$10,300+$8866.995074\\\\\r\nx&=\\frac{$19,166.995074}{1.956937799}=$9,794.38\r\n\\end{align*}[\/latex]\r\n\r\nTherefore: The size of the two equal payments is $9,794.38.","rendered":"

                  Knowledge Check 3.1<\/h2>\n
                    \n
                  1. P\u00a0 = $3,000;\u00a0 r<\/em>\u00a0 =\u00a0 6%\u00a0 = 0.06;\u00a0 t<\/em>\u00a0 =180\/365 years [latex]I=Prt=$3,000\\times0.06\\times\\frac{180}{365}=$88.77[\/latex]<\/li>\n
                  2. I=$55; r<\/em> =5.5%=0.055; t =125\/365 years [latex]P=\\frac{I}{rt}=\\frac{$55}{0.055\\left(\\frac{125}{365}\\right)} =$2920.00[\/latex]<\/li>\n
                  3. P = $900,\u00a0\u00a0 I = $65,\u00a0\u00a0 r<\/em> = 7.5% = 0.075<\/li>\n<\/ol>\n

                    [latex]t=\\frac{I}{Pr}=\\frac{$65}{$900\\times 0.075}=0.962963\\text{ years}[\/latex]<\/p>\n

                    To convert this answer to days, you must multiply by 365. To eliminate a rounding error, be sure to use the exact value from your calculator, i.e., just multiply the above value by 365 without re-entering the displayed value .<\/p>\n

                    [latex]\u2234 t =\\frac{$65}{$900\\times0.075}\\text{years}\\times \\frac{365\\text{ days}}{1\\text{ years}} = 351.48\\text{ days}[\/latex]<\/p>\n

                    This should now be rounded up to 352 days.<\/p>\n

                      \n
                    1. \u00a0P\u00a0 =\u00a0 $975, I\u00a0 = $36.73, t<\/em>\u00a0 =220\/365\u00a0 years<\/li>\n<\/ol>\n

                      [latex]r=\\frac{I}{Pt}=\\frac{$36.73}{$975\\left(\\frac{220}{365}\\right)} = 0.062500932 = 6.25%[\/latex]<\/p>\n

                       <\/p>\n

                      Knowledge Check 3.2<\/h2>\n
                        \n
                      1. P = $4,000, r\u00a0 = 8%\u00a0 = 0.08,\u00a0 t = 210\/365\u00a0 years<\/li>\n<\/ol>\n

                        \"Timeline<\/p>\n

                        [latex]FV=$4,000 \\left[1 +0.08\\left(\\frac{210}{365}\\right)\\right][\/latex]<\/p>\n

                          \n
                        1. P = $1,250,\u00a0 r<\/em> = 6.75% = 0.0675; [latex]t = (251 - 69)+365 =\\frac{182}{365}[\/latex] years (from Table 3- 1)<\/li>\n<\/ol>\n

                          \"Timeline<\/p>\n

                          [latex]FV=$1,250\\left[1 +0.0675\\left(\\frac{182}{365}\\right)\\right]=$1,292.07[\/latex]<\/p>\n

                            \n
                          1. P = $2,500, r = 3.75% = 0.0375, t = 2 years<\/li>\n<\/ol>\n

                            \"Timeline<\/p>\n

                            [latex]FV =$2,500 (1 +0.0375\\times 2)= $2,687.50[\/latex]<\/p>\n

                             <\/p>\n

                            Knowledge Check 3.3<\/h2>\n
                              \n
                            1. P = $2,000, FV = $2,210, t<\/em> = 1.5 years<\/li>\n<\/ol>\n

                              \"Timeline<\/p>\n

                              Either of the two following approaches is acceptable:<\/p>\n

                              Approach A:<\/p>\n

                              [latex]I = FV - P =$2,210 -$2,000 =$210[\/latex]<\/p>\n

                              So<\/p>\n

                              [latex]r=\\frac{I}{Pt}=\\frac{$210}{$2,000\\times1.5}=0.07=7%[\/latex]<\/p>\n

                              Approach B:<\/p>\n

                              [latex]r = \\frac{\\frac{FV}{P}-1}{t} = \\frac{\\frac{$2,210}{$2,000}-1}{1.5}= 0.07[\/latex]<\/p>\n

                                \n
                              1. FV = $1,871.25, r<\/em> = 9% = 0.09, t<\/em> = 33 months =33\/12\u00a0 years.<\/li>\n<\/ol>\n

                                \"Timeline<\/p>\n

                                [latex]P=\\frac{FV}{1+rt}=\\frac{$1,871.25}{1+0.09\\times \\left(\\frac{33}{12}\\right)}=$1,500.00[\/latex]<\/p>\n

                                 <\/p>\n

                                Knowledge Check 3.4<\/h2>\n

                                a. Find FV at 7%:<\/p>\n

                                \"Timeline<\/p>\n

                                [latex]FV = P(1+rt) = $10,000 \\left(1+0.07\u00d7\\frac{6}{12}\\right) = $10,350[\/latex]<\/p>\n

                                Conclusion: The value of $10,000 now, in six months\u2019 time is $10,350. Since this is $25 greater than $10,325, you would prefer $10,000 now from a purely financial point of view.<\/p>\n

                                b. Find FV at 6%:<\/p>\n

                                \"Timeline<\/p>\n

                                 <\/p>\n

                                [latex]FV = P(1+rt) = $10,000 \\left(1+0.06\u00d7\\frac{6}{12}\\right) = $10,300[\/latex]<\/p>\n

                                Conclusion: The value of $10,000 now, in six months time is $10,300. Since this is less than $10,325, you would prefer $10,325 in six months time.<\/p>\n

                                Knowledge Check 3.5<\/h2>\n

                                \"Timeline<\/p>\n

                                [latex]\\begin{align*} &FV_1=$20,000\\left(1+0.08\\times\\frac{15}{12}\\right)=$22,000.00\\\\ &FV_2=$5,000\\left(1+0.08\\times\\frac{10}{12}\\right)=$5,333.33\\\\ &FV_3=$10,000\\left(1+0.08\\times \\frac{6}{12}\\right)=$10,400.00\\\\ \\end{align*}[\/latex]<\/p>\n

                                Total Debt Outstanding = $37,733.33(6 months from now)<\/p>\n

                                 <\/p>\n

                                Knowledge Check 3.6<\/h2>\n

                                1.<\/p>\n

                                \"Timeline<\/p>\n

                                 <\/p>\n

                                [latex]PV=\\frac{$5,000}{1+0.09\\times \\frac{19}{12}}=$4,376.37[\/latex]<\/p>\n

                                  \n
                                1. <\/li>\n<\/ol>\n

                                  \"Timeline<\/p>\n

                                  [latex]\\begin{align*} &PV_1=\\frac{$2,000}{1+0.07\\times \\frac{7}{12}}=$1,921.54\\\\ &PV_2=\\frac{$4,000}{1+0.07\\times \\frac{13}{12}}=$3,718.05 \\end{align*}[\/latex]<\/p>\n

                                  Total Equivalent Debt Now = $5,639.59<\/p>\n

                                  Knowledge Check 3.8<\/h2>\n

                                  \"Timeline<\/p>\n

                                   <\/p>\n

                                  Value of “old” payments at the focal date :<\/p>\n

                                  [latex]$3,500\\left(1+0.08\\times \\frac{8}{12}\\right)+$5,500\\left(1+0.08\\times \\frac{1}{12}\\right)[\/latex]<\/p>\n

                                  [latex]=$3,686.67+$5,536.67=$9,223.33[\/latex]<\/p>\n

                                   <\/p>\n

                                  Value of “replacement” payments at the focal date:<\/p>\n

                                  [latex]$3,000\\left(1+0.08\\times \\frac{6}{12}\\right)+x=$3,120+x[\/latex]<\/p>\n

                                  Therefore:<\/p>\n

                                  [latex]$3,120+x=$9,223.33[\/latex]<\/p>\n

                                  [latex]x =$9,223.33 -$3,120.00=$6,103.33[\/latex]<\/p>\n

                                   <\/p>\n

                                  Knowledge Check 3.9<\/h2>\n

                                  \"Timeline<\/p>\n

                                  Let the amount of two equal payments be x. Value of “old” payments at the focal date:<\/p>\n

                                  [latex]$10,000\\left(1+0.09\\times \\frac{4}{12}\\right)+\\frac{$9,000}{\\left(1+0.09\\times \\frac{2}{12}\\right)}[\/latex]<\/p>\n

                                  Value of “replacement” payments at the focal date:<\/p>\n

                                  [latex]x+\\frac{x}{\\left(1+0.09\\times \\frac{6}{12}\\right)}[\/latex]<\/p>\n

                                  And we can set these to be equal:<\/p>\n

                                  [latex]x+\\frac{x}{\\left(1+0.09\\times \\frac{6}{12}\\right)} = $10,000\\left(1+0.09\\times \\frac{4}{12}\\right)+\\frac{$9,000}{\\left(1+0.09\\times \\frac{2}{12}\\right)}[\/latex]<\/p>\n

                                  And solve, remembering to store all intermediate values in the calculator:<\/p>\n

                                  [latex]\\begin{align*} x+\\frac{x}{1.045} &=$10,000(1.03)+\\frac{$9,000}{1.015}\\\\ x\\left(1+\\frac{1}{1.045}\\right)&=$10,300+$8866.995074\\\\ x&=\\frac{$19,166.995074}{1.956937799}=$9,794.38 \\end{align*}[\/latex]<\/p>\n

                                  Therefore: The size of the two equal payments is $9,794.38.<\/p>\n","protected":false},"author":883,"menu_order":13,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-461","chapter","type-chapter","status-publish","hentry"],"part":42,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/pressbooks\/v2\/chapters\/461","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/wp\/v2\/users\/883"}],"version-history":[{"count":25,"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/pressbooks\/v2\/chapters\/461\/revisions"}],"predecessor-version":[{"id":3867,"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/pressbooks\/v2\/chapters\/461\/revisions\/3867"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/pressbooks\/v2\/parts\/42"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/pressbooks\/v2\/chapters\/461\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/wp\/v2\/media?parent=461"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/pressbooks\/v2\/chapter-type?post=461"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/wp\/v2\/contributor?post=461"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/wp\/v2\/license?post=461"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}