{"id":1643,"date":"2017-02-02T20:20:40","date_gmt":"2017-02-02T20:20:40","guid":{"rendered":"http:\/\/mathwise.net\/?p=1643"},"modified":"2017-02-02T20:20:40","modified_gmt":"2017-02-02T20:20:40","slug":"coin-word-problems","status":"publish","type":"post","link":"http:\/\/mathwise.net\/?p=1643","title":{"rendered":"Coin Word Problems"},"content":{"rendered":"

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

    \n
  1. Phil T. Rich has some coins in his pocket consisting of dimes, nickels, and pennies. He has two more nickels than dimes, and three times as many pennies as nickels. How many of each kind of coin does he have if the total value is 52 cents?<\/li>\n<\/ol>\n

    Solution<\/p>\n

    First try to determine which type of coin he has fewest of. This is often a good way to find what to let x stand for. Here he has fewer dimes than nickels or pennies.<\/p>\n

    The question asks how many of each kind of coin (not how much they are worth). That is, what number of each kind of coin does he have? So, let<\/p>\n

    x = number of dimes<\/p>\n

    Go back to one fact at a time. He has two more nickels than dimes.<\/p>\n

    x + 2 = number of nickels<\/p>\n

    Another fact is that he has three times as many pennies as nickels.<\/p>\n

    3(x + 2) = number of pennies<\/p>\n

    Next the information left for the equation is that the total value is 52 cents. You can’t say that the total number of coins equals 52 cents. Number of must be changed to value. If you had two dimes, you would have 20 cents. You multiplied how many coins by how much each is worth. So let’s change our how many to how much. If you have the number of dimes, multiply by 10 to change to cents. Multiply the number of nickels by 5 to change to cents. The number of pennies is the same as the number of cents.<\/p>\n\n\n\n\n\n\n
    Number of Coins<\/td>\nValue in Cents<\/td>\n<\/tr>\n
    x = number of dimes<\/td>\n\u00a010x = number of cents in dimes<\/td>\n<\/tr>\n
    x + 2 = number of nickels<\/td>\n5(x + 2) = number of cents in nickels<\/td>\n<\/tr>\n
    3(x + 2) = number of pennies<\/td>\n3(x+ 2) = number of cents in pennies<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    Now we can add the amounts of money. If you make it all pennies, there are no decimals.<\/p>\n

    10x + 5(x + 2) + 3(x + 2) = 52<\/p>\n

    10x +5x + 10 + 3x + 6 = 52<\/p>\n

    18x + 16 = 52<\/p>\n

    18x = 36<\/p>\n

    x = 2<\/p>\n

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

    x = 2<\/p>\n

    x + 2 = 4<\/p>\n

    3(x + 2) = 12<\/p>\n

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

    2 dimes<\/p>\n

    4 nickels<\/p>\n

    12 pennies<\/p>\n

     <\/p>\n

      \n
    1. Ernest Worker had a collection of silver coins worth $205. There were five times as many quarters as half-dollars (50-cent pieces) and 200 fewer dimes than quarters. How many of each kind of coin did the Ernest have?<\/li>\n<\/ol>\n

      Solution<\/p>\n

      First try to determine which type of coin he has fewest of. This is often a good way to find what to let x stand for. Here he has fewer dimes than nickels or pennies.<\/p>\n

      The question asks how many of each kind of coin (not how much they are worth). That is, what number of each kind of coin does he have? So, let<\/p>\n

      x = number of dimes<\/p>\n\n\n\n\n\n\n
      Number of Coins<\/strong><\/td>\nValue in Cents<\/strong><\/td>\n<\/tr>\n
      x = number of half-dollars<\/td>\n50x = number of cents in half-dollars<\/td>\n<\/tr>\n
      5x = number of quarters<\/td>\n25(5x) = number of cents in quarters<\/td>\n<\/tr>\n
      5x – 200 = number of dimes<\/td>\n10(5x – 200) = number of cents in dimes<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

       <\/p>\n

      Remember, everything has been changed to cents. So $205 has to be changed to cents by multiplying by 100. $205 has to be changed to cents by multiplying by 100. $205 = 20,500 cents.<\/p>\n

      50x + 25(5x) + 10(5x – 200) = 20,500<\/p>\n

      50x + 125x + 50x – 2000 = 20,500<\/p>\n

      225x = 22,500<\/p>\n

      x = 100<\/p>\n

      5x = 500<\/p>\n

      5x – 200 = 300<\/p>\n

      So, Ernest has 100 half-dollars, 500 quarters and 300 dimes.<\/p>\n

       <\/p>\n

        \n
      1. Kay Oss bought $32.56 worth of stamps. She bought 20 more 19-cent stamps than 50-cent stamps. She bought twice as many 32-cent stamps as 19-cent stamps. How many of each kind did she buy?<\/li>\n<\/ol>\n

        Solution<\/p>\n\n\n\n\n\n\n
        Number of Stamps<\/td>\nValue in Cents<\/td>\n<\/tr>\n
        x = number of 50-cent stamps<\/td>\n50x<\/td>\n<\/tr>\n
        x + 20 = number of 19-cents<\/td>\n19(x + 20)<\/td>\n<\/tr>\n
        2(x + 20) = number of 32-cent stamps<\/td>\n32[2(x+ 20)]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

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

        Remember, $32.56 equals 3256 cents.<\/p>\n

        50x + 19(x + 20) + 32[2(x + 20)] = 3256<\/p>\n

        50x + 19x + 380 + 64x + 1160 = 3256<\/p>\n

        133x = 1596<\/p>\n

        x = 12<\/p>\n

        x + 20 = 32<\/p>\n

        2(x + 20) = 64<\/p>\n

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

        Kay purchased 12 50-cent stamps, 32 19-cent stamps and 64 32-cent stamps<\/p>\n

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

        12(50) + 19(32) + 64(32) = 3256<\/p>\n

        600 + 608 + 2048 = 3256<\/p>\n

        3256 = 3256<\/p>\n

         <\/p>\n

          \n
        1. A collection of coins has a value of 64 cents. There are two more nickels than dimes and three times as many pennies as dimes. How many of each kind of coin are there?<\/li>\n<\/ol>\n

          Solution<\/p>\n\n\n\n\n\n\n
          Number of Coins<\/td>\nValue in Cents<\/td>\n<\/tr>\n
          x = number of dimes<\/td>\n10x = value of dimes<\/td>\n<\/tr>\n
          x + 2 = number of nickels<\/td>\n5(x + 2) = value of nickels<\/td>\n<\/tr>\n
          3x = number of pennies<\/td>\n3x = number of pennies<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

          The total value of the dimes, nickels, and quarters equals 64 cents.<\/p>\n

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

          10x + 5(x + 2) + 3x = 64<\/p>\n

          10x + 5x + 10 + 3x = 64<\/p>\n

          18x = 54<\/p>\n

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

          x = 3<\/p>\n

          x + 2 = 5<\/p>\n

          3x = 9<\/p>\n

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

          There are 3 dimes, 5 nickels and 9 pennies.<\/p>\n

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

          3(10 cents) = 30<\/p>\n

          5(5 cents) = 25<\/p>\n

          9(1 cent) = 9<\/p>\n

          ______<\/p>\n

          Total = 64 cents<\/p>\n

           <\/p>\n

            \n
          1. Lotta Spences has ten bills in her wallet. She has a total of $40. If she has one more $5 bill than $10 bills, and two more $ bills than $5 bills, how many of each does she have? (There are two ways of working this problem. See if you can do it both ways.)<\/li>\n<\/ol>\n

            Solution<\/p>\n\n\n\n\n\n\n
            Number of Bills<\/td>\nValues in dollars<\/td>\n<\/tr>\n
            x = number of $10 bills<\/td>\n10x = value of $10 bills<\/td>\n<\/tr>\n
            x + 1 = numbers of $5 bills<\/td>\n5(x +1) = value of $5 bills<\/td>\n<\/tr>\n
            (x + 1) + 2 = number of $1 bills<\/td>\n1(x + 3) = value of $1 bills<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

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

            10x + 5(x + 1) + (x + 3) = 40<\/p>\n

            10x + 5x + 5 + x + 3 = 40<\/p>\n

            16x = 32<\/p>\n

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

            x = 2<\/p>\n

            x + 1 = 3<\/p>\n

            x + 3 = 5<\/p>\n

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

            Lotta has 2 ten dollar bills, 3 five dollar bills and 5 one dollar bills.<\/p>\n

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

            2($10) = $20<\/p>\n

            3($5) = 15<\/p>\n

            5($1) = 5<\/p>\n

            ____<\/p>\n

            Total = $40<\/p>\n

             <\/p>\n

              \n
            1. Dan D. Lyons bought $21.44 worth of stamps at the post office. He bought 10 more 4-cent stamps than 19-cent stamps. The number of 32-cent stamps was three times the number of 19-cent stamps. He also bought two $1 stamps. How many of each kind of stamp did he purchase?<\/li>\n<\/ol>\n

              Solution<\/p>\n\n\n\n
              Number of Stamps<\/td>\nValue in Cents<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n\n\n\n\n
              x = number of 19-cent stamps<\/td>\n19x<\/td>\n<\/tr>\n
              x + 10 = number of 4-cent stamps<\/td>\n4(x + 10)<\/td>\n<\/tr>\n
              3x = number of 32-cent stamps<\/td>\n32(3x)<\/td>\n<\/tr>\n
              2 = number of $1 stamps<\/td>\n2(100)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

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

              19x + 4(x + 10) + 32(3x) + 2(100) = 2144<\/p>\n

              19x + 4x + 40 +96x + 200 = 2144<\/p>\n

              119x + 240 = 2144<\/p>\n

              119x = 1904<\/p>\n

              199x = 1904<\/p>\n

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

              x = 16<\/p>\n

              x + 10 = 26<\/p>\n

              3x = 48<\/p>\n

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

              Dan purchased 16 19-cent stamps, 26 4-cent stamps and 48 32-cent stamps.<\/p>\n

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

              19(16) + 4(26) + 32(48) + 200 =2144<\/p>\n

              304 + 104 + 1536 + 200 = 2144<\/p>\n

              2144 = 2144<\/p>\n

               <\/p>\n

                \n
              1. Nick O\u2019 Time purchases a selection of wrenches for his shop. His bill is $78. He buys the same number of $1.50 and $2.50 wrench\u00ades, and half that many of $4 wrenches. The number of $3 wrenches is one more than the number of $4 wrenches. How many of each did he purchase? (Hint: if you have not worked with fractions, use decimals for all fractional parts.)<\/li>\n<\/ol>\n

                Solution<\/p>\n

                Number of Wrenches<\/strong><\/p>\n

                x = number of $4.00 wrenches<\/p>\n

                2x = number of $1.50 wrenches<\/p>\n

                2x = number of $2.50 wrenches<\/p>\n

                x + 1 = number of $3.00 wrenches<\/p>\n

                Value in Dollars<\/strong><\/p>\n

                (x)(4.00) = value of $4.00 wrenches<\/p>\n

                (2x) (1.50) = value of $1.50 wrenches<\/p>\n

                (2x)(2.50) = value of $2.50 wrenches<\/p>\n

                (x + 1)(3.00) = value of $3.00 wrenches<\/p>\n

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

                x(4.00) + 2x(1.50) + 2x(2.50) + (x + 1)(3.00) = 78<\/p>\n

                Note in this problem that the decimal is removed when you clear parentheses.<\/p>\n

                4x + 3x + 5x + 3x + 3 = 78<\/p>\n

                15x = 75<\/p>\n

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

                x = 5<\/p>\n

                2x = 10<\/p>\n

                x + 1 = 6<\/p>\n

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

                Nick purchased 5 $4.00 wrenches,10 $1.50 wrenches, 10 $2.50 wrenches and 6 $3.00 wrenches.<\/p>\n

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

                5($4.00) = $20<\/p>\n

                10($1.50) = 15<\/p>\n

                10($2.50) = 25<\/p>\n

                6($3.00) = 18<\/p>\n

                ____<\/p>\n

                Total = $78<\/p>\n

                 <\/p>\n

                  \n
                1. Phoebe Small at the XYZ Department Store receives $15 in change for her cash drawer at the start of each day. She receives twice as many dimes as fifty-cent pieces, and the same number of quarters as dimes. She has twice as many nickels as dimes and a dollar\u2019s worth of pennies. How many of each kind of coin does she receive?<\/li>\n<\/ol>\n

                  Solution<\/p>\n\n\n\n\n
                  Number of Coins<\/td>\nValue in Cents<\/td>\n<\/tr>\n
                  x = number of 50-cent pieces<\/td>\n50x = value of 50-cent pieces<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n\n\n\n\n
                  2x = number of 10-cent pieces<\/td>\n10(2x) = value of 10-cent pieces<\/td>\n<\/tr>\n
                  2x = number of 25-cent pieces<\/td>\n25(2x) = value of 25-cent pieces<\/td>\n<\/tr>\n
                  4x = number of 5-cent pieces<\/td>\n5(4x) = value of 5-cent pieces<\/td>\n<\/tr>\n
                  100 = number of 1-cent pieces<\/td>\n1(100) = value of 1-cent pieces<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

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

                  Fifteen dollars equals 1500 cents.<\/p>\n

                  50x + 2x(10) + 2x(25) + 4x(5) + 100 =1500<\/p>\n

                  50x + 20x + 50x 20x + 100 = 1500<\/p>\n

                  140x = 1400<\/p>\n

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

                  x = 10<\/p>\n

                  2x = 20<\/p>\n

                  2x = 20<\/p>\n

                  4x = 40<\/p>\n

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

                  Phoebe has 10 50 cents pieces, 20 10 cents pieces, 20 25 cents pieces, 40 5 cents pieces and 100 1 cent pieces.<\/p>\n

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

                  10(50 cents) = $5<\/p>\n

                  20(10 cents) = 2<\/p>\n

                  20(25 cents) = 5<\/p>\n

                  40(5 cents) = 2<\/p>\n

                  100(1 cent) = 1<\/p>\n

                  ____<\/p>\n

                  Total = $15<\/p>\n

                   <\/p>\n

                    \n
                  1. A collection of 36 coins consists of nickels, dimes, and quarters. There are three fewer quarters than nickels and six more dimes than quarters. How many of each kind of coin are there?<\/li>\n<\/ol>\n

                    Let\u00a0 x = number of quarters (there are fewer quarters)<\/p>\n

                    x + 3 = number of nickels<\/p>\n

                    x + 6 = number of dimes<\/p>\n

                    Here we do not change to cents because the number of coins is given.<\/p>\n

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

                    x + (x + 3) + (x + 6) = 36<\/p>\n

                    3x + 9 = 36<\/p>\n

                    3x = 27<\/p>\n

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

                    x = 9 quarters<\/p>\n

                    x + 3 = 12 nickels<\/p>\n

                    x + 6 = 15 dimes<\/p>\n

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

                    9 +12 + 15 = 36<\/p>\n

                    36 = 36<\/p>\n

                     <\/p>\n

                     <\/p>\n

                      \n
                    1. The cash drawer of the Greasy Spoon Cafe contains $227 in bills. There are six more $5 bills than $10 bills. The number of $1 bills is two more than 24 times the number of $10 bills. How many bills of each kind are there?<\/li>\n<\/ol>\n

                      Let: x= number of dollars in $10 bills.<\/p>\n

                      5(x + 6) = number of dollars in $5 bills.<\/p>\n

                      24x + 2 = number of dollars in $1 bills.<\/p>\n

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

                      10x + 5(x + 6) + 24x + 2 = 227<\/p>\n

                      10x + 5x + 30 + 24x + 2 = 227<\/p>\n

                      39x + 32 = 227<\/p>\n

                      39x = 195<\/p>\n

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

                      x = 5<\/p>\n

                      x + 6 = 11<\/p>\n

                      24x + 2 = 122<\/p>\n

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

                      10(5) + 5(11) + 24(5) + 2 = 227<\/p>\n

                      50 + 55 + 120 + 2 = 227<\/p>\n

                      227 = 227<\/p>\n

                       <\/p>\n

                        \n
                      1. Lois Lane went to the drugstore. She bought a bottle of aspirin and a bottle of Tylenol. The aspirin cost $1.25 more than the Tylenol. She also bought cologne which cost twice as much as the total of the other two combined. How much did each cost if her total (without tax) was $24.75?<\/li>\n<\/ol>\n

                        Let x = cost in cents of Tylenol<\/p>\n

                        x + 125 = cost in cents of aspirin<\/p>\n

                        2(x + x + 125) = cost in cents of cologne<\/p>\n

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

                        x + x + 125 + 2(2x + 125) = 2475<\/p>\n

                        2x + 125 +4x + 250 + 2475<\/p>\n

                        6x + 375 + 2475<\/p>\n

                        6x = 2100<\/p>\n

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

                        x = 350<\/p>\n

                        x + 125 = 475<\/p>\n

                        2(2x + 125) = 1650<\/p>\n

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

                        3.50 + 4.75 + 16.50 = 24.75<\/p>\n

                        24.75 = 24.75<\/p>\n

                         <\/p>\n

                         <\/p>\n

                          \n
                        1. Superman bought some gum and some candy. The number of packages of gum was one more than the number of mints. The number of mints was three times the number of candy bars. If gum was 24 cents a package, mints were 10 cents each, and candy bars were 35 cents each, how many of each did he get for $5.72?<\/li>\n<\/ol>\n

                          Let x = number of 35 cent candy bars<\/p>\n

                          3x = number of 10 cent mints<\/p>\n

                          3x + 1 = the number of 24 cent packages of gum<\/p>\n

                          Value in Cents<\/strong><\/p>\n

                          35x = cents for candy bar<\/p>\n

                          10(3x) = cents for mints<\/p>\n

                          24(3x+1)= cents for gum<\/p>\n

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

                          35x + 10(3x) + 24(3x+1) = 572<\/p>\n

                          35x + 30x + 72x + 24= 572<\/p>\n

                          137x + 24 = 572<\/p>\n

                          137x = 548<\/p>\n

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

                          x = 4<\/p>\n

                          3x = 12<\/p>\n

                          3x + 1 = 13<\/p>\n

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

                          4(35) + 12(10) + 13(24) = 572<\/p>\n

                          140 + 120 + 312 + = 572<\/p>\n

                          572 = 572<\/p>\n

                           <\/p>\n

                           <\/p>\n

                            \n
                          1. Clay Potts had $50 to buy his groceries. He needed milk at $1.95 a carton, bread at $2.39 a loaf, breakfast cereal at $3.00 a box, and meat at $5.39 a pound. He bought twice as many cartons of milk as loaves of bread and one more package of cereal than loaves of bread. He also bought the same number of pounds of meat as packages of cereal. How many of each item did he pur\u00adchase if he received $12.25 in change?<\/li>\n<\/ol>\n

                            Let x <\/strong>= number of loaves of bread at $2.39 each<\/p>\n

                            2x <\/strong>= number of cartons of milk at $1.95 each<\/p>\n

                            x + 1\u00a0<\/strong>=number of packages of cereal at $3.00 each<\/p>\n

                            x + 1 <\/strong>= number of pounds of meat at $5.39 each<\/p>\n

                            Cost in cents<\/strong><\/p>\n

                            239x<\/strong> = cost of bread<\/p>\n

                            195(2x) <\/strong>= cost of milk<\/p>\n

                            300(x + 1) = <\/strong>cost of cereal<\/p>\n

                            539(x + 1) <\/strong>= cost of meat<\/p>\n

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

                            239x + 195(2x) + 300(x + 1) + 539(x + 1) = 5000 – 1225<\/strong><\/p>\n

                            239x + 390x + 300x + 300 + 539x + 539 = 3775<\/strong><\/p>\n

                            1468x + 839 = 3775<\/strong><\/p>\n

                            1468x = 2936<\/strong><\/p>\n

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

                            x = 2<\/strong><\/p>\n

                            2x = 4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/strong><\/p>\n

                            x + 1 = 3<\/strong><\/p>\n

                            x + 1 = 3<\/strong><\/p>\n

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

                            169(2) + 130(4) + 200(3) + 359(3) = 3500 – 1225<\/strong><\/p>\n

                            338 + 520 + 600 + 1077 = 2535<\/strong><\/p>\n

                            2535 = 2535<\/strong><\/p>\n

                             <\/p>\n

                             <\/p>\n

                              \n
                            1. Mr. Merrill has 3 times as many nickels as dimes. The coins have a total value of $1.50. How many of each coin does he have?<\/li>\n<\/ol>\n

                              Let n <\/strong>= number of nickels.<\/p>\n

                              Let d <\/strong>= number of dimes.<\/p>\n

                              n = 3d<\/p>\n

                              .05n + .10d = 1.50<\/p>\n

                              Multiply through by 100<\/p>\n

                              5n + 10d = 150<\/p>\n

                              Use (n = 3d) and sub in 3d for n<\/p>\n

                              5n + 10d = 150<\/p>\n

                              5(3d) + 10d = 150<\/p>\n

                              15d + 10d = 150<\/p>\n

                              25d = 150<\/p>\n

                              d = 6<\/p>\n

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

                              There are 6 dimes and 18 nickels<\/p>\n

                              6 dimes equals 60 cents and 18 nickels equals 90 cents<\/p>\n

                              60 cents and 90 cents = $1.50<\/p>\n

                               <\/p>\n

                                \n
                              1. Ms. Lynch has 21 coins in nickels and dimes. Their total value is $1.65. How many of each coin does she have?<\/li>\n<\/ol>\n

                                Let n <\/strong>= number of nickels.<\/p>\n

                                Let d <\/strong>= number of dimes.<\/p>\n

                                n + d = 21<\/p>\n

                                .05n + .10d = $1.65<\/p>\n

                                Multiply through by 100<\/p>\n

                                5n + 10d = 165<\/p>\n

                                Multiply first equation by -5<\/p>\n

                                -5n + -5d = -105<\/p>\n

                                Add both equations<\/p>\n

                                5n + 10d = 165<\/p>\n

                                +<\/p>\n

                                -5n -5d = -105<\/p>\n

                                =<\/p>\n

                                5d = 60<\/p>\n

                                d = 12<\/p>\n

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

                                Ms. Lynch has 12 dimes and 9 nickels.<\/p>\n

                                12 dimes equals $1.20 and 9 nickels equals 45 cents<\/p>\n

                                $1.20 and 45 cents = $1.65.<\/p>\n

                                 <\/p>\n

                                 <\/p>\n

                                  \n
                                1. A vending machine that takes only dimes and quarters contains 30 coins, with a total value of $4.20. How many of each coin are there?<\/li>\n<\/ol>\n

                                  Let d <\/strong>= number of dimes.<\/p>\n

                                  Let q <\/strong>= number of quarters.<\/p>\n

                                  d + q = 30<\/p>\n

                                  .10d + .25q = $4.20<\/p>\n

                                  Multiply through by 100<\/p>\n

                                  10d + 25q = 420<\/p>\n

                                  Multiply first equation by -10<\/p>\n

                                  -10d + -10q = -300<\/p>\n

                                  Add both equations<\/p>\n

                                  10d + 25q = 420<\/p>\n

                                  +<\/p>\n

                                  -10d + -10q = -300<\/p>\n

                                  =<\/p>\n

                                  15q = 120<\/p>\n

                                  q = 8<\/p>\n

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

                                  The vending machine has 8 quarters and 22 dimes.<\/p>\n

                                  8 quarters equals $2.00 and 22 dimes equals $2.20<\/p>\n

                                  $2.00 and\u00a0 $2.20 equals $4.20<\/p>\n

                                   <\/p>\n

                                    \n
                                  1. The total value of the $1 bills and $5 bills in a cash box is $124. There are 8 more $5 bills than $1 bills. How many of each are there?<\/li>\n<\/ol>\n

                                    Let x <\/strong>= number of one dollar bills<\/p>\n

                                    Let f <\/strong>= number of five dollar bills.<\/p>\n

                                    1x + 5f = 124<\/p>\n

                                    x + 8 = f<\/p>\n

                                    Substitute (x+ 8) for f in the first equation.<\/p>\n

                                    1x + 5(x + 8) = 124<\/p>\n

                                    x + 5x + 40 = 124<\/p>\n

                                    6x + 40 = 124<\/p>\n

                                    6x = 84<\/p>\n

                                    x = 14<\/p>\n

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

                                    There are 14 one dollar bills and 22 five dollar bills.<\/p>\n

                                    $14 + $110 = $124<\/p>\n

                                     <\/p>\n

                                      \n
                                    1. A collection of nickels and quarters amounts to $2.60. There are 16 coins in all. How many of each coin are there?<\/li>\n<\/ol>\n

                                      Let n <\/strong>= number of nickels<\/p>\n

                                      Let q <\/strong>= number of quarters.<\/p>\n

                                      n + q = 16<\/p>\n

                                      .05n + .25q = $2.60<\/p>\n

                                      Multiply through by 100<\/p>\n

                                      5n + 25q = 260<\/p>\n

                                      Multiply first equation by -5<\/p>\n

                                      -5n + – 5q = -80<\/p>\n

                                      Add both equations<\/p>\n

                                      5n + 25q = 260<\/p>\n

                                      +<\/p>\n

                                      -5n + – 5q = -80<\/p>\n

                                      =<\/p>\n

                                      20q = 180<\/p>\n

                                      q = 9 and n = 7<\/p>\n

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

                                      The collection contains 9 quarters ($2.25) and 7 nickels (.35)<\/p>\n

                                      $2.25<\/p>\n

                                      +<\/p>\n

                                      .35<\/p>\n

                                      =<\/p>\n

                                      $2.60<\/p>\n

                                       <\/p>\n

                                        \n
                                      1. Joe Lick bought some 20-cent and 25-cent stamps. He bought 32 stamps in all, and paid $7.40 for them. How many stamps of each kind did he buy?<\/li>\n<\/ol>\n

                                        Let x <\/strong>= number of 20 cent stamps<\/p>\n

                                        Let y <\/strong>= number of 25 cent stamps<\/p>\n

                                        x + y = 32<\/p>\n

                                        .20x + .25y = $7.40<\/p>\n

                                        Multiply through by 100<\/p>\n

                                        20x + 25y = 740<\/p>\n

                                        Multiply first equation by -20<\/p>\n

                                        -20x + -20y = -640<\/p>\n

                                        Add both equations<\/p>\n

                                        20x + 25y = 740<\/p>\n

                                        +<\/p>\n

                                        -20x + -20y = -640<\/p>\n

                                        =<\/p>\n

                                        5y = 100<\/p>\n

                                        y = 20 and x = 12<\/p>\n

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

                                        Joe lick has 20 25 cent stamps (5.00) and 12 20 cent stamps ($2.40)<\/p>\n

                                        $5.00<\/p>\n

                                        +<\/p>\n

                                        $2.40<\/p>\n

                                        =<\/p>\n

                                        $7.40<\/p>\n

                                         <\/p>\n

                                         <\/p>\n

                                          \n
                                        1. For a school play, 340 tickets valued at $810 were sold. Some cost $2 and some cost $3. How many tickets of each kind were sold?<\/li>\n<\/ol>\n

                                          Let x <\/strong>= number of two dollar tickets.<\/p>\n

                                          Let y <\/strong>= number of three dollar tickets.<\/p>\n

                                          x + y = 340<\/p>\n

                                          $2x + $3y = $810<\/p>\n

                                          Multiply first equation by -2<\/p>\n

                                          -2x + -2y = -680<\/p>\n

                                          Add both equations<\/p>\n

                                          2x + 3y = 810<\/p>\n

                                          +<\/p>\n

                                          -2x + -2y = -680<\/p>\n

                                          =<\/p>\n

                                          y = 130 and x = 210<\/p>\n

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

                                          The school play sold 210 two dollar tickets ($420) and 130 three dollar tickets ($390)<\/p>\n

                                          $420<\/p>\n

                                          +<\/p>\n

                                          $390<\/p>\n

                                          =<\/p>\n

                                          $810<\/p>\n

                                           <\/p>\n

                                            \n
                                          1. Romeo bought a mixture of 20-cent, 35-cent, and 50-cent valentines. The number of 20-cent valentines was 1 more than twice the number of 35-cent valentines, and the number of 50-cent valentines was 2 less than the number of 35-cent ones. If he spent $4.20 all together, how many valentines of each kind did he buy?<\/li>\n<\/ol>\n

                                            Let x <\/strong>= number of 20 cent stamps.<\/p>\n

                                            Let y <\/strong>= number of 35 cent stamps<\/p>\n

                                            Let z = number of 50 cent stamps<\/p>\n

                                            x = 2y + 1<\/p>\n

                                            z = y – 2<\/p>\n

                                            .20x + .35y + .50z = 4.20<\/p>\n

                                            Multiply third equation through by 100<\/p>\n

                                            20x + 35y + 50z = 420<\/p>\n

                                            Substitute x = 2y + 1 into third equation<\/p>\n

                                            Substitute z = y – 2 into third equation<\/p>\n

                                            20x + 35y + 50z = 420<\/p>\n

                                            20(2y + 1) + 35y + 50(y – 2) = 420<\/p>\n

                                            40y + 20 + 35y + 50y – 100 = 420<\/p>\n

                                            125y – 80 = 420<\/p>\n

                                            125y = 500<\/p>\n

                                            y = 4 and x = 9\u00a0 and z = 2<\/p>\n

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

                                            Romeo purchased nine 20 cent stamps (1.80), four 35 cent stamps (1.40), and two 50 cent stamps (1.00).<\/p>\n

                                            1.80<\/p>\n

                                            +<\/p>\n

                                            1.40<\/p>\n

                                            +<\/p>\n

                                            1.00<\/p>\n

                                            =<\/p>\n

                                            4.20 or $4.20<\/p>\n

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

                                            Coin Word Problems Phil T. Rich has some coins in his pocket consisting of dimes, nickels, and pennies. He has two more nickels than dimes, and three times as many pennies as nickels. How many of each kind of coin does he have if the total value is 52 cents? Solution First try to determine … Continue reading Coin Word Problems<\/span> →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-1643","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"http:\/\/mathwise.net\/index.php?rest_route=\/wp\/v2\/posts\/1643","targetHints":{"allow":["GET"]}}],"collection":[{"href":"http:\/\/mathwise.net\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/mathwise.net\/index.php?rest_route=\/wp\/v2\/types\/post"}],"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=1643"}],"version-history":[{"count":1,"href":"http:\/\/mathwise.net\/index.php?rest_route=\/wp\/v2\/posts\/1643\/revisions"}],"predecessor-version":[{"id":1644,"href":"http:\/\/mathwise.net\/index.php?rest_route=\/wp\/v2\/posts\/1643\/revisions\/1644"}],"wp:attachment":[{"href":"http:\/\/mathwise.net\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1643"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/mathwise.net\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1643"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/mathwise.net\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1643"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}