{"id":1649,"date":"2017-02-02T22:31:39","date_gmt":"2017-02-02T22:31:39","guid":{"rendered":"http:\/\/mathwise.net\/?page_id=1649"},"modified":"2017-02-03T04:52:37","modified_gmt":"2017-02-03T04:52:37","slug":"digit-word-problems","status":"publish","type":"page","link":"http:\/\/mathwise.net\/?page_id=1649","title":{"rendered":"Digit Word Problems"},"content":{"rendered":"

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

    \n
  1. The tens digit of a certain number is 3 less than the units digit. The sum of the digits is 11. What is the number?<\/li>\n<\/ol>\n

    Solution<\/p>\n

    Let x = units digit<\/p>\n

    x – 3 = tens digit<\/p>\n

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

    x + (x – 3) = 11<\/p>\n

    2x – 3 = 11<\/p>\n

    2x = 14<\/p>\n

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

    x = 7 (units digit)<\/p>\n

    x – 3 = 4 (tens digit)<\/p>\n

    Using the values above for units and tens, we find the number 47.<\/p>\n

     <\/p>\n

      \n
    1. The tens digit of a number is twice the units digit. If the digits are reversed, the new number is 27 less than the original. Find the original number.<\/li>\n<\/ol>\n

      Solution<\/p>\n

      Let x = units digit
      \n2x = tens digits<\/p>\n

      Then the original number is 10(2x) + x, the reserved number is 10(x) + 2x, and the new number is the original number less than 27.<\/p>\n

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

      10(x) + 2x = 10(2x) + x \u2013 27<\/p>\n

      12x = 21x \u2013 27<\/p>\n

      -9x = -27<\/p>\n

      x = 3 \u00a0(units digit)<\/p>\n

      2x = 6 (tens digit)<\/p>\n

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

      The number is (6 X 10) + 3 or 63.<\/p>\n

       <\/p>\n

        \n
      1. The sum of the digits in a two-digit number is 12. If the digits are reversed, the number is 18 greater than the original number. What is the number?<\/li>\n<\/ol>\n

        Solution<\/p>\n

        Let x = units digit
        \n12 – x = tens digit<\/p>\n

        Then the original number is 10(12 – x)\u00a0+\u00a0x and the reserved \u00a0number is 10(x) + (12 – x).<\/p>\n

        Equation:<\/strong> \u00a0The reserved number is the original number plus 18.<\/p>\n

        10(x) + (12 \u2013 x) = 10(12 – x) + (x) + 18<\/p>\n

        10x + 12 – x = 120 – 10x + x + 18<\/p>\n

        9x + 12 = 138 – 9x<\/p>\n

        18x = 126<\/p>\n

        x = 7 (units digit)<\/p>\n

        12 – x = 5 (tens digit)<\/p>\n

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

        The number is ( 5 x 10) + 7 or 57<\/p>\n

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

        10(7) + (12 \u2013 7) = 10(12- 7) + (7) + 18<\/p>\n

        70 + 5 = 50 + 7 + 18<\/p>\n

        75 = 75<\/p>\n

         <\/p>\n

          \n
        1. The tens digit of a certain number is 5 more than the units digit. The sum of the digits is 9. Find the number.<\/li>\n<\/ol>\n

          Solution<\/p>\n

          Let x = units digit<\/p>\n

          x + 5 = tens digit<\/p>\n

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

          x + (x + 5) \u00a0= 9<\/p>\n

          2x + 5 = 9<\/p>\n

          2x = 4<\/p>\n

          x = 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0(unit digit)<\/p>\n

          x + 5 = 7 \u00a0\u00a0(tens digit)<\/p>\n

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

          The number is (7 X 10) + 2 or 72.<\/p>\n

           <\/p>\n

            \n
          1. The tens digit of a two-digit number is twice the units digit. If the digits are reversed, the new number is 36 less than the original number. Find the number.<\/li>\n<\/ol>\n

            Solution<\/p>\n

            Let x = units digit<\/p>\n

            2x = tens digit<\/p>\n

            Then the number is 10(2x) + x and the reversed number is 10(x)+ 2x.<\/p>\n

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

            10(x) + 2x= 10(2x) + x \u2013 36<\/p>\n

            12x \u00a0= 21x \u2013 36<\/p>\n

            -9x = -36<\/p>\n

            x = 4 (units digit)<\/p>\n

            2x = 8\u00a0(tens digit)<\/p>\n

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

            The number is (8 X 10) + 4 or 84.<\/p>\n

             <\/p>\n

              \n
            1. The sum of the digits of a two-digit number is 13. The units digit is 1 more than twice the tens digit. Find the number.<\/li>\n<\/ol>\n

              Solution<\/p>\n

              Let x = units digit<\/p>\n

              13 \u2013 x = tens digit<\/p>\n

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

              The units digit is twice the tens digit plus 1.<\/p>\n

              x = 2(13 \u2013 x) \u00a0+ 1<\/p>\n

              x = 26 – 2x + 1<\/p>\n

              x = 27 – 2x<\/p>\n

              3x = 27<\/p>\n

              x = 9 \u00a0(units digit)<\/p>\n

              13 \u2013 x = 4 (tens digit)<\/p>\n

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

              The number is (4 x 10) \u00a0+ 9 or 49.<\/p>\n

               <\/p>\n

                \n
              1. The sum of the digits of a three-digit number is 6. The hundreds digit is twice the units digit, and the tens digit equals the sum of the other two. Find the number.<\/li>\n<\/ol>\n

                Solution<\/p>\n

                Let x = units digit<\/p>\n

                2x = hundreds digit<\/p>\n

                x + 2x = tens digit<\/p>\n

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

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

                6x = 6<\/p>\n

                x = 1 (units digit)<\/p>\n

                2x = 2 (hundreds digit)<\/p>\n

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

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

                The number is (2 x 100) + (3 x 10) + 1 or 231.<\/p>\n

                 <\/p>\n

                  \n
                1. The units digit is twice the tens digit. If the number is doubled, it will be 12 more than the reversed number. Find the number.<\/li>\n<\/ol>\n

                  Solution<\/p>\n

                  Let x = tens digit.<\/p>\n

                  2x = units digit.<\/p>\n

                  Then the number is 10(x) + 2x and the reversed number is 10(2) = x.<\/p>\n

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

                  Two times the number equals 12 \u00a0more than the reversed number.<\/p>\n

                  2[10(x) + (2x)] = 10(2x) + x + 12<\/p>\n

                  2(12x) = 21x + 12<\/p>\n

                  24x = 21x + 12<\/p>\n

                  3x = 12<\/p>\n

                  x = 4 (tens digit)<\/p>\n

                  2x = 8 (units digit)<\/p>\n

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

                  The number is (4 x 10) + 8 or 48.<\/p>\n

                   <\/p>\n

                    \n
                  1. Eight times the sum of the digits of a certain two-digit number exceeds the number by 19.The tens digit is 3 more than the units digit. Find the number.<\/li>\n<\/ol>\n

                    Solution<\/p>\n

                    Let x = units digit (smaller)<\/p>\n

                    x + 3 = tens digit<\/p>\n

                    then the number is 10(x + 3) + x.<\/p>\n

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

                    Eight times the sum of the digits exceeds the number by 19.<\/p>\n

                    8[ x + (x + 3) ] – [ 10(x + 3) + x ] = 19<\/p>\n

                    8[ 2x + 3 ] – [ 10x+30 + x ] = 19<\/p>\n

                    16x + 24 – [ 9x + 30 ] = 19<\/p>\n

                    16x + 24 – 9x – 30 = 19<\/p>\n

                    5x – 6 = 19<\/p>\n

                    5x = 25<\/p>\n

                    x = 5 (units digit)<\/p>\n

                    x + 3 = 8 (tens digit)<\/p>\n

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

                    The number is (8 x 10) + 5 or 85.<\/p>\n

                     <\/p>\n

                      \n
                    1. The ratio of the units digit to the tens digit of a two-digit number one-half. The tens digit is 2 more than the units digit. Find the number.<\/li>\n<\/ol>\n

                      Solution<\/p>\n

                      Let x = units digit<\/p>\n

                      x + 2 = tens digit<\/p>\n

                      The ratio is a fractional relationship.<\/p>\n

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

                      x\/(x + 2) = 1\/2<\/p>\n

                      Multiply by the LCD, 2(x + 2)<\/p>\n

                      x = 2 (units digit)<\/p>\n

                      x + 2 = 4 (tens digit)<\/p>\n

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

                      The number is (4 x 10) + 2 or 42.<\/p>\n

                       <\/p>\n

                        \n
                      1. There is a two-digit number whose units digit is 6 less than the tens digit. Four times the tens digit plus five times the units digit equals 51. Find the digits.<\/li>\n<\/ol>\n

                        Solution<\/p>\n

                        Let x = units digit<\/p>\n

                        x + 6 = tens digit<\/p>\n

                        Four times the tens digit plus five times the units digit equal 51.<\/p>\n

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

                        4(x + 6) + 5x = 51<\/p>\n

                        4x + 24 + 5x = 51<\/p>\n

                        9x + 24 = 51<\/p>\n

                        9x = 27<\/p>\n

                        x = 3 (units digit)<\/p>\n

                        x + 6 = 9 (tens digit)<\/p>\n

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

                        The number is (9 x 10) + 3 or 93.<\/p>\n

                         <\/p>\n

                          \n
                        1. The tens digit is 2 less than the units digit. If the digits are reversed, the sum of the reversed number and the original number is 154. Find the original number.<\/li>\n<\/ol>\n

                          Solution<\/p>\n

                          Let\u00a0 x = units digit<\/p>\n

                          x – 2 = tens digit<\/p>\n

                          Then the number is 10(x – 2) + x and the reversed number is 10(x) + (x – 2)<\/p>\n

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

                          The reversed number plus the original number equal 154.<\/p>\n

                          10(x) + (x – 2) + 10(x – 2) + x = 154<\/p>\n

                          10x + x – 2 + 10x -20 + x = 154<\/p>\n

                          22x – 22 = 154<\/p>\n

                          22x = 176<\/p>\n

                          x = 8 (units digit)<\/p>\n

                          x – 2 = 6 (tens digit)<\/p>\n

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

                          The number is (6 X 10) + 8 or 68.<\/p>\n

                           <\/p>\n

                            \n
                          1. A three-digit number has a tens digit 2 greater than the units digit and a hundreds digit 1 greater than the tens digit. The sum of the tens and hundreds digits is three times the units digit. What is the number?<\/li>\n<\/ol>\n

                            Solution<\/p>\n

                            Let x = units digit<\/p>\n

                            x + 2 = tens digit<\/p>\n

                            (x + 2) + 1 = hundreds digit<\/p>\n

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

                            The sum of the tens and hundreds digits is three times the units digit.<\/p>\n

                            (x + 2) + (x + 2) + 1 = 3x<\/p>\n

                            2x + 5 = 3x<\/p>\n

                            -x = -5<\/p>\n

                            x = 5 (units digit)<\/p>\n

                            x + 2 = 7 (tens digit)<\/p>\n

                            (x + 2) + 1 = 8 (hundreds digit)<\/p>\n

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

                            The number is (8 x 100) + (7 x 10) + 5 or 875.<\/p>\n

                             <\/p>\n

                              \n
                            1. The sum of the digits of a two-digit number is 9. The value of the number is 12 times the tens digit. Find the number.<\/li>\n<\/ol>\n

                              Solution<\/p>\n

                              Let u = units digit<\/p>\n

                              Let t = tens digit<\/p>\n

                              Let u + 10t = the number itself<\/p>\n

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

                              u + t = 9\u00a0 sum of the digits is 9<\/p>\n

                              u + 10t = 12t\u00a0\u00a0 value of the number is 12 times the tens digit<\/p>\n

                              u = 2t\u00a0\u00a0 sub this into u + t = 9<\/p>\n

                              2t = t = 9<\/p>\n

                              3t = 9<\/p>\n

                              t = 3<\/p>\n

                              So the tens digit is 3 and the units must be 6<\/p>\n

                              So the number is 36<\/p>\n

                              Check<\/p>\n

                              3 + 6 = 9<\/p>\n

                              36 = 3(12)<\/p>\n

                              36 = 36<\/p>\n

                               <\/p>\n

                                \n
                              1. The sum of the digits of a two-digit number is 12. If 15 is added to the number, the result is 6 times the units digit. Find the number.<\/li>\n<\/ol>\n

                                Solution<\/p>\n

                                Let u = units digit<\/p>\n

                                Let t = tens digit<\/p>\n

                                Let u + 10t = the number itself<\/p>\n

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

                                u + t = 12\u00a0\u00a0\u00a0\u00a0\u00a0 sum of the digits is 12<\/p>\n

                                u + 10t + 15 = 6u\u00a0\u00a0\u00a0\u00a0\u00a0 if 15 is added to the number, the result is 6 times the units digit<\/p>\n

                                10t + 15 = 5u\u00a0\u00a0 Simplify<\/p>\n

                                2t + 3 = u\u00a0\u00a0 divide through by 5<\/p>\n

                                u = 2t + 3\u00a0\u00a0 sub this into u + t = 12<\/p>\n

                                2t + 3 + t = 12<\/p>\n

                                3t + 3 = 12<\/p>\n

                                3t = 9<\/p>\n

                                t = 3<\/p>\n

                                So the tens digit is 3 and the units must be 9<\/p>\n

                                So the number is 39<\/p>\n

                                Check<\/p>\n

                                3 + 9 = 12<\/p>\n

                                39 + 15 = 6(9)<\/p>\n

                                54 = 54<\/p>\n

                                 <\/p>\n

                                  \n
                                1. The sum of the digits of a two-digit number is 8. If the digits of the number are reversed, the new number is 18 less than the original number. Find the number.<\/li>\n<\/ol>\n

                                  Solution<\/p>\n

                                  Let u = units digit<\/p>\n

                                  Let t = tens digit<\/p>\n

                                  Let u + 10t = the original number<\/p>\n

                                  10u + t = the number with the digits reversed<\/p>\n

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

                                  u + t = 8\u00a0\u00a0\u00a0\u00a0\u00a0 sum of the digits is 8.<\/p>\n

                                  10u + t + 18 = u + 10t\u00a0\u00a0\u00a0\u00a0 if the digits of the number are reversed, the new number is 18 less than the original number<\/p>\n

                                  9u + 18 = 9t\u00a0\u00a0\u00a0 simplify<\/p>\n

                                  u + 2 = t\u00a0\u00a0\u00a0\u00a0\u00a0 divide through by 9<\/p>\n

                                  t = u + 2\u00a0\u00a0\u00a0\u00a0 sub this into u + t = 8<\/p>\n

                                  u + (u + 2) = 8<\/p>\n

                                  2u + 2 = 8<\/p>\n

                                  2u = 6<\/p>\n

                                  u = 3<\/p>\n

                                  So the units digit is 3 and the tens must be 5<\/p>\n

                                  So the number is 53<\/p>\n

                                  Check<\/p>\n

                                  3 + 5 = 8<\/p>\n

                                  35 + 18 = 53<\/p>\n

                                  53 = 53<\/p>\n

                                   <\/p>\n

                                    \n
                                  1. The tens digit of a two-digit number is twice the units digit. If the digits are reversed, the new number is 36 less than the original number. Find the number.<\/li>\n<\/ol>\n

                                    Solution<\/p>\n

                                    Let u = units digit<\/p>\n

                                    Let t = tens digit<\/p>\n

                                    Let u + 10t = the original number<\/p>\n

                                    10u + t = the number with the digits reversed<\/p>\n

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

                                    t = 2u<\/p>\n

                                    10u + t + 36 = u + 10t<\/p>\n

                                    9u + 36 = 9t\u00a0\u00a0\u00a0 simplify<\/p>\n

                                    u + 4 = t\u00a0\u00a0\u00a0\u00a0\u00a0 divide through by 9<\/p>\n

                                    t = u + 4\u00a0\u00a0\u00a0\u00a0 sub this into t = 2u<\/p>\n

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

                                    u = 4<\/p>\n

                                    So the units digit is 4 and the tens must be 8<\/p>\n

                                    So the number is 48 and the number with the digits reversed is 84<\/p>\n

                                    Check<\/p>\n

                                    8 = 2(4)<\/p>\n

                                    48 + 36 = 84<\/p>\n

                                     <\/p>\n

                                      \n
                                    1. The units digit of a two-digit number is 4 times the tens digit. If the digits are reversed, the new number is 54 more than the original number. Find the number.<\/li>\n<\/ol>\n

                                      Solution<\/p>\n

                                      Let u = units digit<\/p>\n

                                      Let t = tens digit<\/p>\n

                                      Let u + 10t = the original number<\/p>\n

                                      10u + t = the number with the digits reversed<\/p>\n

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

                                      u = 4t<\/p>\n

                                      10u + t – 54 = u + 10t<\/p>\n

                                      9u – 54 = 9t\u00a0\u00a0\u00a0 simplify<\/p>\n

                                      u – 6 = t\u00a0\u00a0\u00a0\u00a0\u00a0 divide through by 9<\/p>\n

                                      u = t + 6\u00a0\u00a0\u00a0\u00a0 solve for u<\/p>\n

                                      u = t + 6\u00a0\u00a0\u00a0\u00a0 sub this into u = 4t<\/p>\n

                                      t + 6 = 4t<\/p>\n

                                      6 = 3t<\/p>\n

                                      t = 2<\/p>\n

                                      So the tens digit is 2 and the units digit must be 8<\/p>\n

                                      So the number is 28 and the number with the digits reversed is 82<\/p>\n

                                      Check<\/p>\n

                                      8 = 4(2)<\/p>\n

                                      82 = 28 + 54<\/p>\n

                                       <\/p>\n

                                       <\/p>\n

                                        \n
                                      1. The sum of the digits of a two-digit number is 11. If 27 is added to the number, the digits will be reversed. Find the number.<\/li>\n<\/ol>\n

                                        Solution<\/p>\n

                                        Let u = units digit<\/p>\n

                                        Let t = tens digit<\/p>\n

                                        Let u + 10t = the original number<\/p>\n

                                        10u + t = the number with the digits reversed<\/p>\n

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

                                        u + t = 11<\/p>\n

                                        u + 10t + 27 = 10u + t<\/p>\n

                                        9t + 27 = 9u\u00a0\u00a0\u00a0 simplify<\/p>\n

                                        t + 3 = u\u00a0\u00a0\u00a0\u00a0\u00a0 divide through by 9<\/p>\n

                                        u = t + 3<\/p>\n

                                        u = t + 3\u00a0\u00a0\u00a0\u00a0 sub this into u + t = 11<\/p>\n

                                        (t + 3) + t = 11<\/p>\n

                                        2t + 3 = 11<\/p>\n

                                        2t = 8<\/p>\n

                                        t = 4<\/p>\n

                                        So the tens digit is 4 and the units digit must be 7<\/p>\n

                                        So the number is 47 and the number with the digits reversed is 74<\/p>\n

                                        Check<\/p>\n

                                        7 + 4 = 11<\/p>\n

                                        47 + 27 = 74<\/p>\n

                                        74 = 74<\/p>\n

                                         <\/p>\n

                                         <\/p>\n

                                          \n
                                        1. The units digit of a two-digit number is 1 less than 3 times the tens digit. It the digits are reversed, the new number is 45 more than the original number. Find the number.<\/li>\n<\/ol>\n

                                          Solution<\/p>\n

                                          Let u = units digit<\/p>\n

                                          Let t = tens digit<\/p>\n

                                          Let u + 10t = the original number<\/p>\n

                                          10u + t = the number with the digits reversed<\/p>\n

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

                                          u + 1 = 3t<\/p>\n

                                          10u + t – 45 = u + 10t<\/p>\n

                                          9u – 45 = 9t\u00a0\u00a0\u00a0 simplify<\/p>\n

                                          u – 5 = t\u00a0\u00a0\u00a0\u00a0\u00a0 divide through by 9<\/p>\n

                                          u = t + 5\u00a0\u00a0\u00a0\u00a0 solve for u<\/p>\n

                                          u = t + 5\u00a0\u00a0\u00a0\u00a0 sub this into u + 1 = 3t<\/p>\n

                                          (t + 5) + 1 = 3t<\/p>\n

                                          t + 6 = 3t<\/p>\n

                                          2t = 6<\/p>\n

                                          t = 3<\/p>\n

                                          So the tens digit is 3 and the units digit must be 8<\/p>\n

                                          So the number is 38 and the number with the digits reversed is 83<\/p>\n

                                          Check<\/p>\n

                                          8 + 1 = 3(3)<\/p>\n

                                          38 + 45 = 83<\/p>\n

                                          83 = 83<\/p>\n

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

                                          Digit Word Problems The tens digit of a certain number is 3 less than the units digit. The sum of the digits is 11. What is the number? Solution Let x = units digit x – 3 = tens digit Equation x + (x – 3) = 11 2x – 3 = 11 2x = … Continue reading Digit 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-1649","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"http:\/\/mathwise.net\/index.php?rest_route=\/wp\/v2\/pages\/1649","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=1649"}],"version-history":[{"count":1,"href":"http:\/\/mathwise.net\/index.php?rest_route=\/wp\/v2\/pages\/1649\/revisions"}],"predecessor-version":[{"id":1650,"href":"http:\/\/mathwise.net\/index.php?rest_route=\/wp\/v2\/pages\/1649\/revisions\/1650"}],"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=1649"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}