{"id":1651,"date":"2017-02-02T22:32:26","date_gmt":"2017-02-02T22:32:26","guid":{"rendered":"http:\/\/mathwise.net\/?page_id=1651"},"modified":"2017-02-03T05:12:19","modified_gmt":"2017-02-03T05:12:19","slug":"finance-word-problems","status":"publish","type":"page","link":"http:\/\/mathwise.net\/?page_id=1651","title":{"rendered":"Finance Word Problems"},"content":{"rendered":"

Finance Word Problems<\/strong><\/p>\n

    \n
  1. Jack Potts invested $10,000. Part of it he put in the bank at 5 percent interest. The remainder he put in bonds which pay a 9 percent yearly return. Find the absolute value of the difference of the two investments in each vehicle if his yearly income from the two investments was $660?<\/li>\n<\/ol>\n

    Solution<\/p>\n

    Let<\/p>\n

    x = amount in dollars invested at 5 percent.<\/p>\n

    10,000 – x\u00a0 = amount in dollars invested at 9 percent.<\/p>\n

    Convert above information\u00a0 into income.<\/p>\n

    ie\u00a0 I = P x R x T<\/p>\n

    Interest = Principal x Rate x Time<\/p>\n

    If you put $1,000 in the bank at 5 percent, your interest would be 1000(0.05); remember 5 percent = 0.05. Setting up our equation,<\/p>\n

    0.05x = interest earned from bank investment<\/p>\n

    0.09(10,000 – x) = interest earned from bond investment<\/p>\n

    The interest earned from the bank plus the interest raened from the bond equals the total interest earned.<\/p>\n

    0.05x + 0.09(10,000 – x) = 660<\/p>\n

    0.05x + 900 – 0.09x = 600<\/p>\n

    Multiply both sides by 100<\/p>\n

    5x + 90,000 – 9x = 66,000<\/p>\n

    -4x = -24,000<\/p>\n

    x = 6,000<\/p>\n

    10,000 – x = 4,000<\/p>\n

    So,<\/p>\n

    6,000 was invested in the bank<\/p>\n

    4,000 was invested in the bond.<\/p>\n

     <\/p>\n

     <\/p>\n

      \n
    1. Mr. Gold invested $50,000, part at 6 percent and part at 8 percent. The annual interest on the 6 percent investment was $480 more than that from the 8 percent investment. How much was invested at each rate?<\/li>\n<\/ol>\n

      Solution<\/p>\n

      Let x = amount in dollars invested at 6 percent.<\/p>\n

      50,000 – x = amount in dollars invested at 8 percent.<\/p>\n

      Then<\/strong><\/p>\n

      0.06x = income on first investment<\/p>\n

      0.08(50,000 – x) = income on second investment<\/p>\n

      The interest on the 6% investment was $480 more than the interest on the 8% investment.<\/p>\n

      Equation<\/strong><\/p>\n

      The first income equals the second income plus 480.<\/p>\n

      0.06x = 0.08(50,000 – x) + 480<\/p>\n

      0.06x = 4,000 – 0.08x + 480<\/p>\n

      0.06x = 4480 – 0.08x<\/p>\n

      6x = 44,8000 – 8x<\/p>\n

      14x = 448,000<\/p>\n

      Answers<\/strong><\/p>\n

      x = 32,000<\/p>\n

      50,000 – x = 18,000<\/p>\n

      So,<\/strong><\/p>\n

      $32,000 was invested at 6% and $18,000 was invested at 8%<\/p>\n

       <\/p>\n

        \n
      1. A store advertised dresses on sale at 20 percent off. The sale price was $76. What was the original price of the dress?<\/li>\n<\/ol>\n

        Solution<\/p>\n

        Let x = original price in dollars<\/p>\n

        If the sale price is 20% off the original price, the sale price is 80% of the original price.<\/p>\n

        Equation<\/strong><\/p>\n

        0.08x = 76<\/p>\n

        80x = 7600<\/p>\n

        x = 95<\/p>\n

        So,<\/strong><\/p>\n

        The original price was $95.<\/p>\n

         <\/p>\n

          \n
        1. Tickets to the school play sold at $4 each for adults and $1.50 each for children. If there were four times as many adult tickets sold as children\u2019s tickets, and the total receipts were $3500, how many children\u2019s tickets were sold?<\/li>\n<\/ol>\n

          Solution<\/p>\n

          Let x = number of children’s ticket at $ 1.50 each.<\/p>\n

          4x = number of adult tickets at $ 4 each.<\/p>\n

          Then<\/strong><\/p>\n

          1.50x = total amount of money received from children’s tickets in dollars<\/p>\n

          4(4x) = total amount of money received from adult tickets in dollars<\/p>\n

          Equation<\/strong><\/p>\n

          1.50x + 4(4x) = 3500<\/p>\n

          1.50x + 16x = 3500<\/p>\n

          15x + 160x = 35,000<\/p>\n

          175x = 35,000<\/p>\n

          x = 200<\/p>\n

          So,<\/strong><\/p>\n

          200 tickets were sold at $1.50 and 800 tickets were sold at $4.00<\/p>\n

           <\/p>\n

            \n
          1. Art Tillery owns a jewelry store. He marks up all merchandise 50 percent of cost. If he sells a diamond ring for $1500, what did he pay the wholesaler for it?<\/li>\n<\/ol>\n

            Solution<\/p>\n

            Let x = amount in dollars he paid wholesaler.<\/p>\n

            0.50x = his markup (50%)<\/p>\n

            Cost plus markup equals selling price.<\/p>\n

            Equation: \u00a0\u00a0\u00a0\u00a0x + 0.50x = 1500<\/p>\n

            Multiply by 10 to clear decimals.<\/p>\n

            10x + 5x = 15,000<\/p>\n

            15x + 15,000<\/p>\n

            x = 1000<\/p>\n

            So,<\/strong><\/p>\n

            The wholesaler paid $1,000 for the diamond ring.<\/p>\n

             <\/p>\n

              \n
            1. What amount of money invested at 8 \u00bc percent yields a $2475 return per year?<\/li>\n<\/ol>\n

              Solution<\/p>\n

              Let x = amount of money invested in dollars<\/p>\n

              0.0825x = interest of 8 1\/4 percent per year.<\/p>\n

              Equation<\/strong><\/p>\n

              0.0825 = 2475<\/p>\n

              x = 30,000<\/p>\n

              So,<\/strong><\/p>\n

              $30,000 was invested.<\/p>\n

               <\/p>\n

                \n
              1. Orney and Sly Company often price miscellaneous at a price they know will sell fast and determine what they can pay for it by this selling price. The Company bought men\u2019s t-shirts to sell at $10 each. If they allow 40 percent of the selling price<\/em> for expenses and profit, what will they be willing to pay for the shirts?<\/li>\n<\/ol>\n

                Solution<\/p>\n

                Cost plus the markup equals the selling price.<\/p>\n

                Let x = cost in dollars they will pay for each shirt<\/p>\n

                0.40(10) = markup (40% of selling price)<\/p>\n

                Equation<\/strong><\/p>\n

                x + 0.40(10) = 10<\/p>\n

                x + 4 = 10<\/p>\n

                x = 6<\/p>\n

                So,<\/strong><\/p>\n

                The Market would be willing to pay $6 each for the shirts.<\/p>\n

                 <\/p>\n

                  \n
                1. Rhoda Davidson invested $50,000. Part of it she put in a gold mine stock from which she hoped to receive a 20 percent return per year. The rest he invested in a bank stock which was paying 6 percent per year. If she received $400 more the first year from the bank stock than from the mining stock, how much did she invest in each stock?<\/li>\n<\/ol>\n

                  Solution<\/p>\n

                  Let\u00a0x = amount of dollars invested at 20 percent.<\/p>\n

                  50,000 – x = amount of dollars invested at 6 percent (total minus x)<\/p>\n

                  0.20x = interest on mining stock (smaller income)<\/p>\n

                  0.06(50,000 – x) = interest on bank stock (larger income)<\/p>\n

                  Equation<\/strong><\/p>\n

                  Income from bank stock is interest on the mining stock plus $400.<\/p>\n

                  0.06(50,000 – x) = 0.20x + 400<\/p>\n

                  3000 – 0.06x\u00a0= 0.20x\u00a0+ 400<\/p>\n

                  300,000 – 6x\u00a0= 20x\u00a0 + 40,000<\/p>\n

                  -26x =\u00a0-260,000<\/p>\n

                  Answer<\/strong><\/p>\n

                  x = 10,000<\/p>\n

                  50,000 – x = 40,000<\/p>\n

                  Check<\/strong><\/p>\n

                  0.06(40,000) = 0.20(10,000) + 400<\/p>\n

                  2400 = 2000 + 400 = 2400<\/p>\n

                   <\/p>\n

                   <\/p>\n

                    \n
                  1. Jim Shortz wished to invest a sum of money so that the interest each year would pay his son\u2019s college expenses. If the money was invested at 8 percent and the college expenses were $10,000 per year, how much should Jim invest?<\/li>\n<\/ol>\n

                    Solution<\/p>\n

                    Let x = amount in dollars Jim should invest.<\/p>\n

                    0.08 = interest per year (assume simple interest)<\/p>\n<\/a>

                    Equation<\/strong><\/h2>\n

                    0.08 = 10,000<\/p>\n

                    8x = 1,000,000<\/p>\n

                    Answer<\/strong><\/p>\n

                    x = 125,000<\/p>\n<\/a>

                    Check<\/strong><\/h2>\n

                    0.08(125,000) = 10,000<\/p>\n

                    10,000\u00a0= 10,000<\/p>\n

                     <\/p>\n

                     <\/p>\n

                      \n
                    1. The Worn & Torn Store had a sale on T-shirts at which all shirts were sold at 15 percent off the original price. Casey bought a shirt for $7.65.What was the original selling price of the shirt? (Assume no taxes.)<\/li>\n<\/ol>\n

                      Solution<\/p>\n

                      Let x = original selling price in dollars<\/p>\n

                      0.15x = discount<\/p>\n

                      The original price minus the discount equals the scale price.<\/p>\n

                      Equation<\/strong><\/p>\n

                      x – 0.15x = 7.65<\/p>\n

                      0.85x = 7.65<\/p>\n

                      85x = 765<\/p>\n

                      Answer<\/strong><\/p>\n

                      x = 9<\/p>\n

                      Check<\/strong><\/p>\n

                      9 – 0.15(9) = 7.65<\/p>\n

                      9 – 1.35 = 7.65<\/p>\n

                      7.65 = 7.65<\/p>\n

                       <\/p>\n

                       <\/p>\n

                        \n
                      1. Stan Back inherited two different stocks whose yearly income was $2 100. The total appraised value of the stocks was $40,000; one was paying 4 percent and one 6 percent per year. What was the value of each stock?<\/li>\n<\/ol>\n

                        Solution<\/p>\n

                        Let x = value in dollars of stock paying 4%<\/p>\n

                        40000 – = value in dollars of stock paying 6%<\/p>\n

                        0.04x = interest on stock paying 4%<\/p>\n

                        0.06(40000 – x) = interest on stock paying 6%<\/p>\n

                        Equation<\/strong><\/p>\n

                        0.04x + 0.06(40000 – x) = 2100<\/p>\n

                        0.04x = 2400 – 0.06x = 2100<\/p>\n

                        Multiply through by 100<\/p>\n

                        4x + 240000 – 6x = 210000<\/p>\n

                        -2x = -30000<\/p>\n

                        Answers<\/strong><\/p>\n

                        x = 15000<\/p>\n

                        40000 – x = 25000<\/p>\n

                        Check<\/strong><\/p>\n

                        0.04(15,000) + 0.06(25,000) = 2100<\/p>\n

                        600 + 1500 = 2100<\/p>\n

                        2100 = 2100<\/p>\n

                         <\/p>\n

                         <\/p>\n

                          \n
                        1. A men\u2019s store bought 500 suits, some at $125 each and the rest at $200 each. If the total cost of the suits was $77,500, how many suits were purchased at each price?<\/li>\n<\/ol>\n

                          Solution<\/p>\n

                          Let x = number of suits purchased at $125 each.<\/p>\n

                           <\/p>\n

                          500 – x = number of suits purchased at $200 each.<\/p>\n

                           <\/p>\n

                          The total value equals the price times the number at $125 plus the price times the number at $200.<\/p>\n

                           <\/p>\n

                          Equation<\/strong><\/p>\n

                           <\/p>\n

                          125x + 200(500 – x) = 77,500<\/p>\n

                           <\/p>\n

                          125x + 100,000 – 200x = 77,500<\/p>\n

                           <\/p>\n

                          -75x = 22,500<\/p>\n

                           <\/p>\n

                          Answer<\/strong><\/p>\n

                           <\/p>\n

                          x = 300 (suits at $125 each)<\/p>\n

                           <\/p>\n

                          500 – x = 200 (suits at $200 each)<\/p>\n

                           <\/p>\n

                          Check<\/strong><\/p>\n

                           <\/p>\n

                          125(300) + 200(200) = 77,500<\/p>\n

                           <\/p>\n

                          37,500 + 40,000 = 77,500<\/p>\n

                           <\/p>\n

                          77,500 = 77,500<\/p>\n

                           <\/p>\n

                            \n
                          1. When Mary Thon sold her house recently, she received $210,000 for it. This was 40 percent more than she paid for it 10 years ago. What was the original purchase price?<\/li>\n<\/ol>\n

                            Solution<\/p>\n

                            Let x = original purchase price in dollars<\/p>\n

                            0.40x= increase in value in dollars<\/p>\n

                            Equation<\/strong><\/p>\n

                            x + 0.40x = 210,000<\/p>\n

                            1.40x = 210,000<\/p>\n

                            14x = 2,100,000<\/p>\n

                            Answer<\/strong><\/p>\n

                            x = 150,000 (original purchase price)<\/p>\n

                            Check<\/strong><\/p>\n

                            150,000 = 0.4(150,000) = 210,000<\/p>\n

                            150,000 + 60,000 = 210,000<\/p>\n

                            210,000 = 210,000<\/p>\n

                             <\/p>\n

                             <\/p>\n

                              \n
                            1. Minnie Sota inherited $20,000 which she invested in stocks and bonds. The stocks returned 6 percent and the bonds 8 percent. If the return on the bonds was $80 less than the return on the stocks, how much did she invest in each?<\/li>\n<\/ol>\n

                              Solution<\/p>\n

                              Let x = amount in dollars invested in stock at 6%<\/p>\n

                              20000 – x = amount in dollars invested in bonds at 8%<\/p>\n

                              0.06x = interest on stocks<\/p>\n

                              0.08(20000 – x) = interest on bonds<\/p>\n

                              Equation<\/strong><\/p>\n

                              Interest on stock equals interest on bonds plus $80<\/p>\n

                              0.06x = 0.08(20000 – x) + 80<\/p>\n

                              0.06x = 1600 – 0.08 + 80<\/p>\n

                              Multiply through by 100<\/p>\n

                              6x = 160000 – 8x + 8000<\/p>\n

                              14x = 168000<\/p>\n

                              Answers<\/strong><\/p>\n

                              x = 12000 (stocks)<\/p>\n

                              20000 – x = 8000 (bonds)<\/p>\n

                              Check<\/strong><\/p>\n

                              0.06(12,000) = 0.08(8000) + 80<\/p>\n

                              720 = 640 + 80 = 720<\/p>\n

                               <\/p>\n

                               <\/p>\n

                                \n
                              1. The total of two investments is $25,000. One amount is invested at 7 percent and one at 9 percent. The annual interest from the 7 percent investment is $470 more than from the 9 percent invest\u00adment. How much is invested at each rate?<\/li>\n<\/ol>\n

                                Solution<\/p>\n

                                Let x = amount in dollars invested at 7 percent.<\/p>\n

                                25,000 -x = amount in dollars invested at 9 percent.<\/p>\n

                                Equation<\/strong><\/p>\n

                                The amount at 7 percent equals the amount at 9 percent plus $470.<\/p>\n

                                0.07x = 0.09 (25,000 – x) + 470<\/p>\n

                                0.07x = 2250 – 0.09x + 470<\/p>\n

                                7x = 225,000 – 9x + 47,000<\/p>\n

                                16x = 272,000<\/p>\n

                                x = 17,000 \u00a0\u00a0\u00a0(amount invested at 7 percent)<\/p>\n

                                25,000 – x = 8000 (amount invested at 9 percent)<\/p>\n

                                So,<\/strong><\/p>\n

                                $17,000 was invested at 7 percent and $8,000 was invested at 9 percent.<\/p>\n

                                Check<\/strong><\/p>\n

                                0.07(17,000) = 0.09(8000) + 470<\/p>\n

                                1190 = 720 = 470<\/p>\n

                                1190 = 1190<\/p>\n

                                 <\/p>\n

                                  \n
                                1. A taxpayer\u2019s state and federal income taxes plus an inheritance tax totaled $14,270. His California state income tax was $5780 less than his federal tax. His inheritance tax was $2750. How much did he pay in state and federal taxes?<\/li>\n<\/ol>\n

                                  Solution<\/p>\n

                                  Let x = amount in dollars of state tax.<\/p>\n

                                  x + 5780 = amount in dollars of federal tax.<\/p>\n

                                  Equation<\/strong><\/p>\n

                                  x + x + 5780 + 2750 = 14,270<\/p>\n

                                  2x + 8530 = 14,270<\/p>\n

                                  2x = 5740<\/p>\n

                                  x = 2870 (state tax)<\/p>\n

                                  x + 5780 = 8650 (federal tax)<\/p>\n

                                  So,<\/strong><\/p>\n

                                  He paid $2,870 in state tax and $8,650 in federal tax.<\/p>\n

                                  Check<\/strong><\/p>\n

                                  2870 + 2870 + 5780 + 2750 = 14,270<\/p>\n

                                  14,270 = 14,270<\/p>\n

                                   <\/p>\n

                                    \n
                                  1. Ann Athlete had saved $6000 which she wished to invest. She put part of it in a term Certificate of Deposit (CD) at 8 percent and part in a regular savings account at 5 1\/2 percent. How much was invested in each account if her total yearly income amounted to $425?<\/li>\n<\/ol>\n

                                    Solution<\/p>\n

                                    Let x = amount in dollars invested at 8 percent.<\/p>\n

                                    6000 – x = amount in dollars invested at 51\/2 percent.<\/p>\n

                                    0.08x = interest on 8 percent investment.<\/p>\n

                                    0.055(6000 – x) = interest on 5 \u00bd percent investment.<\/p>\n

                                    Equation<\/strong><\/p>\n

                                    Total interest (income) was $425.<\/p>\n

                                    0.08x + 0.055(6000 – x) = 420<\/p>\n

                                    0.08x + 330 – 0.055x = 425<\/p>\n

                                    Multiply by 1000 to clear decimals.<\/p>\n

                                    80x + 330,000 – 55x = 425,000<\/p>\n

                                    25x = 95,000<\/p>\n

                                    x = 3800<\/p>\n

                                    6000 – x = 2200<\/p>\n

                                    So,<\/strong><\/p>\n

                                    $3,800 was invested at 8% and $2,200 was invested at 5 1\/2%<\/p>\n

                                    Check<\/strong><\/p>\n

                                    0.08(3800) + 0.055(2200) = 425<\/p>\n

                                    304 + 121 = 425<\/p>\n

                                    425 = 425<\/p>\n

                                     <\/p>\n

                                      \n
                                    1. Mel Ting had $10,000 invested at 5 percent. How many dollars more would he have to invest at 8 percent so that his total interest per year would equal 7 percent of the two investments?<\/li>\n<\/ol>\n

                                      Solution<\/p>\n

                                      Let x = amount in dollars at 8 percent.<\/p>\n

                                      10,000 = total amount invested at 5 percent.<\/p>\n

                                      x + 10,000 = total amount invested at 7 percent.<\/p>\n

                                      0.08x = interest on amount at 8 percent.<\/p>\n

                                      0.05(10,000) = interest on amount at 5 percent.<\/p>\n

                                      0.07 (x + 10,000) = interest on entire investment at 7 percent.<\/p>\n

                                      Equation<\/strong><\/p>\n

                                      Total interest per year equals 7 percent of the entire investment.<\/p>\n

                                      0.08x + 0.05(10,000) = 0.07(x + 10,000)<\/p>\n

                                      0.08x + 500 = 0.07x + 700<\/p>\n

                                      Multiply both sides by 100.<\/p>\n

                                      8x + 50,000 = 7x + 70,000<\/p>\n

                                      x = 20,000<\/p>\n

                                      So,<\/strong><\/p>\n

                                      $20,000 would have to be invested at 8%<\/p>\n

                                      Check<\/strong><\/p>\n

                                      0.08(20,000) + 0.05(10,000) = 0.07(30,000)<\/p>\n

                                      1600 + 500 = 2100<\/p>\n

                                      2100 = 2100<\/p>\n

                                       <\/p>\n

                                        \n
                                      1. Ms. Twinkle invested part of her savings at 6% and the rest at 9%. The amount at 9% was twice the amount at 6%. If her total return after one year was $72, find the amount invested at each rate.<\/li>\n<\/ol>\n

                                        Solution<\/p>\n

                                        Let x = amount in dollars at 6 percent.<\/p>\n

                                        Let 2x = amount in dollars at 9 percent.<\/p>\n

                                        Equation<\/strong><\/p>\n

                                        0.06x + 0.09(2x) = 72<\/p>\n

                                        Multiply both sides by 100.<\/p>\n

                                        6x + 9(2x) = 7,200<\/p>\n

                                        6x + 18x = 7,200<\/p>\n

                                        24x = 7,200<\/p>\n

                                        x = 300<\/p>\n

                                        and<\/p>\n

                                        2x = 600<\/p>\n

                                        So,<\/strong><\/p>\n

                                        $300 would have to be invested at 6%<\/p>\n

                                        $600 would have to be invested at 9%<\/p>\n

                                        Check<\/strong><\/p>\n

                                        0.06(300) + 0.09(600)<\/p>\n

                                        $18 + $54 = $72<\/p>\n

                                         <\/p>\n

                                          \n
                                        1. Rockjaw invested part of his savings at 7% and the rest at 13%. The amount at 7% was $200 more than the amount at 13%. If his total return after one year was $84, find the amount invested at each rate.<\/li>\n<\/ol>\n

                                          Solution<\/p>\n

                                          Let x = amount in dollars at 13 percent.<\/p>\n

                                          Let x +200 = amount in dollars at 7 percent.<\/p>\n

                                          Equation<\/strong><\/p>\n

                                          0.13x + 0.07(x + 200) = 84<\/p>\n

                                          Multiply both sides by 100.<\/p>\n

                                          13x + 7(x + 200) = 8,400<\/p>\n

                                          13x + 7x + 1400 = 8400<\/p>\n

                                          20x + 1400 = 8,400<\/p>\n

                                          20x = 7,000<\/p>\n

                                          x = 350<\/p>\n

                                          So,<\/strong><\/p>\n

                                          $350 would have to be invested at 13%<\/p>\n

                                          $550 would have to be invested at 7%<\/p>\n

                                          Check<\/strong><\/p>\n

                                          0.13(350) + 0.07(550)<\/p>\n

                                          $45.5 + $38.5 = $84<\/p>\n

                                           <\/p>\n

                                            \n
                                          1. Carol invested part of her savings at 10% and the rest at 8%. The amount at 8% was $1500 more than the amount at 10%. If the total annual income is $480, how much was invested at each rate?<\/li>\n<\/ol>\n

                                            Solution<\/p>\n

                                            Let x = amount in dollars at 10 percent.<\/p>\n

                                            Let x +1500 = amount in dollars at 8 percent.<\/p>\n

                                            Equation<\/strong><\/p>\n

                                            0.10x + 0.08(x + 1500) = 480<\/p>\n

                                            Multiply both sides by 100.<\/p>\n

                                            10x + 8(x + 1500) = 48,000<\/p>\n

                                            10x + 8x + 12,000 = 48,000<\/p>\n

                                            18x + 12,000 = 48,000<\/p>\n

                                            18x = 36,000<\/p>\n

                                            x = 2,000<\/p>\n

                                            So,<\/strong><\/p>\n

                                            $2,000 would have to be invested at 10%<\/p>\n

                                            $3,500 would have to be invested at 8%<\/p>\n

                                            Check<\/strong><\/p>\n

                                            0.10(2,000) + 0.08(3,500)<\/p>\n

                                            $200 + $280 = $480<\/p>\n

                                             <\/p>\n

                                              \n
                                            1. Patty Wack had $900. She invested part of it at 12% and the rest at 9%. If her total annual return was $96, how much did she invest at each rate?<\/li>\n<\/ol>\n

                                              Solution<\/p>\n

                                              Let x = amount in dollars at 12 percent.<\/p>\n

                                              Let 900 – x = amount in dollars at 9 percent.<\/p>\n

                                              Equation<\/strong><\/p>\n

                                              0.12x + 0.09(900 – x) = 96<\/p>\n

                                              Multiply both sides by 100.<\/p>\n

                                              12x + 9(900 – x) = 9,600<\/p>\n

                                              12x + 8,100 – 9x = 9,600<\/p>\n

                                              3x + 8,100 = 9,600<\/p>\n

                                              3x = 1,500<\/p>\n

                                              x = 500<\/p>\n

                                              So,<\/strong><\/p>\n

                                              $500 would have to be invested at 12%<\/p>\n

                                              $400 would have to be invested at 9%<\/p>\n

                                              Check<\/strong><\/p>\n

                                              0.12(500) + 0.09(400)<\/p>\n

                                              $60 + $36 = $96<\/p>\n

                                               <\/p>\n

                                               <\/p>\n

                                                \n
                                              1. Dr. Beaker invested $3000, part at 8% and the rest at 7 \u00bd%The total return for one year was $231 How much was invested at each rate?<\/li>\n<\/ol>\n

                                                Solution<\/p>\n

                                                Let x = amount in dollars at 8 percent.<\/p>\n

                                                Let 3000 – x = amount in dollars at 7.5 percent.<\/p>\n

                                                Equation<\/strong><\/p>\n

                                                0.08x + 0.075(3000 – x) = 231<\/p>\n

                                                Multiply both sides by 1000.<\/p>\n

                                                80x + 75(3000 – x) = 231,000<\/p>\n

                                                80x + 225,000 – 75x = 231,000<\/p>\n

                                                5x + 225,000 = 231,000<\/p>\n

                                                5x = 6,000<\/p>\n

                                                x = 1,200<\/p>\n

                                                So,<\/strong><\/p>\n

                                                $1,200 would have to be invested at 8%<\/p>\n

                                                $1,800 would have to be invested at 7.5%<\/p>\n

                                                Check<\/strong><\/p>\n

                                                0.08(1200) + 0.075(1800)<\/p>\n

                                                $96 + $135 = $231<\/p>\n

                                                 <\/p>\n

                                                24. A scholarship fund raised $7000 in contributions. Part was invested in bonds paying 6% interest, and the rest was invested in bank certificates paying 8 \u00bd%. It the total annual income is $520, find the amount invested at each rate.<\/p>\n

                                                Solution<\/p>\n

                                                Let x = amount in dollars at 6 percent.<\/p>\n

                                                Let 7000 – x = amount in dollars at 8.5 percent.<\/p>\n

                                                Equation<\/strong><\/p>\n

                                                0.06x + 0.085(7000 – x) = 520<\/p>\n

                                                Multiply both sides by 1000.<\/p>\n

                                                60x + 85(7000 – x) = 520,000<\/p>\n

                                                60x + 595,000 – 85x = 520,000<\/p>\n

                                                -25x + 595,000 = 520,000<\/p>\n

                                                -25x = -75,000<\/p>\n

                                                x = 3,000<\/p>\n

                                                So,<\/strong><\/p>\n

                                                $3,000 would have to be invested at 6%<\/p>\n

                                                $4,000 would have to be invested at 8.5%<\/p>\n

                                                Check<\/strong><\/p>\n

                                                0.06(3000) + 0.085(4000)<\/p>\n

                                                $180 + $340 = $520<\/p>\n

                                                 <\/p>\n

                                                  \n
                                                1. Sam Quirk invested $7000, part at 7% and the rest at 11%. If his total return for one year was $690, how much was invested at each rate?<\/li>\n<\/ol>\n

                                                  Solution<\/p>\n

                                                  Let x = amount in dollars at 7 percent.<\/p>\n

                                                  Let 7000 – x = amount in dollars at 11 percent.<\/p>\n

                                                  Equation<\/strong><\/p>\n

                                                  0.07x + 0.11(7000 – x) = 690<\/p>\n

                                                  Multiply both sides by 100.<\/p>\n

                                                  7x + 11(7000 – x) = 69,000<\/p>\n

                                                  7x + 77,000 – 11x = 69,000<\/p>\n

                                                  -4x +\u00a0 = -8,000<\/p>\n

                                                  x = 2,000<\/p>\n

                                                  So,<\/strong><\/p>\n

                                                  $2,000 would have to be invested at 7%<\/p>\n

                                                  $5,000 would have to be invested at 11%<\/p>\n

                                                  Check<\/strong><\/p>\n

                                                  0.07(2000) + 0.11(5000)<\/p>\n

                                                  $140 + $550 = $690<\/p>\n

                                                   <\/p>\n

                                                    \n
                                                  1. An investment fund has $3000 more invested at 8% than it does at 10%. If the annual return from the 8% investment is the same as the annual return from the 10% investment, how much is invested at each rate?<\/li>\n<\/ol>\n

                                                    Solution<\/p>\n

                                                    Let x = amount in dollars at 8 percent.<\/p>\n

                                                    Let x – 3000 = amount in dollars at 10 percent.<\/p>\n

                                                    Equation<\/strong><\/p>\n

                                                    0.08x = 0.10(x – 3000)<\/p>\n

                                                    Multiply both sides by 100.<\/p>\n

                                                    8x = 10(x – 3000)<\/p>\n

                                                    8x = 10x – 30,000<\/p>\n

                                                    -2x +\u00a0 = -30,000<\/p>\n

                                                    x = 15,000<\/p>\n

                                                    So,<\/strong><\/p>\n

                                                    $15,000 would have to be invested at 8%<\/p>\n

                                                    $12,000 would have to be invested at 10%<\/p>\n

                                                    Check<\/strong><\/p>\n

                                                    0.08(15000) = 0.10(12000)<\/p>\n

                                                    1200 = 1200<\/p>\n

                                                     <\/p>\n

                                                      \n
                                                    1. Ms. Smyle has $200 less invested at 9% than she does at 6 \u00bd%. If the annual return from the two investments is the same, how much is invested at each rate?<\/li>\n<\/ol>\n

                                                      Solution<\/p>\n

                                                      Let x = amount in dollars at 9 percent.<\/p>\n

                                                      Let x + 200 = amount in dollars at 6.5 percent.<\/p>\n

                                                      Equation<\/strong><\/p>\n

                                                      0.09x = 0.065(x + 200)<\/p>\n

                                                      Multiply both sides by 1000.<\/p>\n

                                                      90x = 65(x + 200)<\/p>\n

                                                      90x = 65x + 13000<\/p>\n

                                                      25x +\u00a0 = 13,000<\/p>\n

                                                      x = 520<\/p>\n

                                                      So,<\/strong><\/p>\n

                                                      $520 would have to be invested at 9%<\/p>\n

                                                      $720 would have to be invested at 6.5%<\/p>\n

                                                      Check<\/strong><\/p>\n

                                                      0.09(520) = 0.065(720)<\/p>\n

                                                      46.8 = 48.6<\/p>\n

                                                       <\/p>\n

                                                        \n
                                                      1. Sally Snuggle has $1600 more invested at 5% than she does at 8%. The annual return from the 5% investment is $17 more than the annual return from the 8% investment. How much is invested at each rate?<\/li>\n<\/ol>\n

                                                        Solution<\/p>\n

                                                        Let x = amount in dollars at 5 percent.<\/p>\n

                                                        Let x – 1600 = amount in dollars at 8 percent.<\/p>\n

                                                        Equation<\/strong><\/p>\n

                                                        0.05x -17 = 0.08(x – 1600)<\/p>\n

                                                        Multiply both sides by 100.<\/p>\n

                                                        5x – 1700= 8(x – 1600)<\/p>\n

                                                        5x – 1700 = 8x – 12800<\/p>\n

                                                        3x = 11100<\/p>\n

                                                        x = 3700<\/p>\n

                                                        So,<\/strong><\/p>\n

                                                        $3700 would have to be invested at 5%<\/p>\n

                                                        $2100 would have to be invested at 8%<\/p>\n

                                                        Check<\/strong><\/p>\n

                                                        0.05(3700)\u00a0 – 17 = 0.08(2100)<\/p>\n

                                                        185 – 17 = 168<\/p>\n

                                                        168 = 168<\/p>\n

                                                         <\/p>\n

                                                          \n
                                                        1. Merlin invested half of his money at 12%, one fourth at 8%, and the rest at 6%. If the total annual income is $570, how much was invested altogether?<\/li>\n<\/ol>\n

                                                          Solution<\/p>\n

                                                          Let x = amount in dollars at 6 percent.<\/p>\n

                                                          Let x = amount in dollars at 8 percent.<\/p>\n

                                                          Let 2x = amount in dollars at 12 percent.<\/p>\n

                                                          Equation<\/strong><\/p>\n

                                                          .06x + .08x + .12(2x) = 570<\/p>\n

                                                          Multiply through by 100<\/p>\n

                                                          6x + 8x + 12(2x) = 57000<\/p>\n

                                                          6x + 8x + 24x = 57000<\/p>\n

                                                          38x = 57000<\/p>\n

                                                          x = 1500<\/p>\n

                                                          So,<\/strong><\/p>\n

                                                          $1500 would have to be invested at 6%<\/p>\n

                                                          $1500 would have to be invested at 8%<\/p>\n

                                                          $3000 would have to be invested at 12%<\/p>\n

                                                          Check<\/strong><\/p>\n

                                                          0.06(1500)\u00a0 +\u00a0 0.08(1500)\u00a0 + .12(3000) = 570<\/p>\n

                                                          90 + 120 + 360 = 570<\/p>\n

                                                          570 = 570<\/p>\n

                                                           <\/p>\n

                                                           <\/p>\n","protected":false},"excerpt":{"rendered":"

                                                          Finance Word Problems Jack Potts invested $10,000. Part of it he put in the bank at 5 percent interest. The remainder he put in bonds which pay a 9 percent yearly return. Find the absolute value of the difference of the two investments in each vehicle if his yearly income from the two investments was … Continue reading Finance Word Problems<\/span> →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":37,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-1651","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"http:\/\/mathwise.net\/index.php?rest_route=\/wp\/v2\/pages\/1651","targetHints":{"allow":["GET"]}}],"collection":[{"href":"http:\/\/mathwise.net\/index.php?rest_route=\/wp\/v2\/pages"}],"about":[{"href":"http:\/\/mathwise.net\/index.php?rest_route=\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"http:\/\/mathwise.net\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/mathwise.net\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1651"}],"version-history":[{"count":1,"href":"http:\/\/mathwise.net\/index.php?rest_route=\/wp\/v2\/pages\/1651\/revisions"}],"predecessor-version":[{"id":1652,"href":"http:\/\/mathwise.net\/index.php?rest_route=\/wp\/v2\/pages\/1651\/revisions\/1652"}],"up":[{"embeddable":true,"href":"http:\/\/mathwise.net\/index.php?rest_route=\/wp\/v2\/pages\/37"}],"wp:attachment":[{"href":"http:\/\/mathwise.net\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1651"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}