Page 47 - Computer Science Class 11 With Functions
P. 47

Step 2: Binary to octal
                     added one 0
               0    1    1  1 0    0 1    1 1

                  3        4        7

            Octal equivalent of (11100111)  is (347) 8
                                       2
            Alternatively,  you  may  first  convert  the  hexadecimal  number  to  its  decimal  equivalent  and  then  find  the  octal
            representation of the decimal number.



                    Find octal representation of (14) , (23) and (81) .
                                              16    16      16


            Do the following conversions:
              a.  (436)  to binary
                      10
              b.  (11011)  to decimal
                        2
              c.  (195)  to octal
                      10
              d.  (706) to decimal
                      8
              e.  (942)  to hexadecimal
                      10
              f.  (A19)  to decimal
                      16
              g.  (1010110)   to octal
                           2
              h.  (732)  to binary
                      8
              i.  (101011100)  to hexadecimal
                             2
              j.  (F31)  to binary
                      16
              k.  (474)  to hexadecimal
                      8
              l.  (C8E)  to octal
                      16
            Ans.  a. (436)  to binary
                         10
                   2   436         Remainder

                   2   218            0
                   2   109            0
                   2     54           1
                   2     27           0

                   2     13           1
                   2      6           1
                   2      3           0
                   2      1           1
                   2      0           1

                 (436)  = (110110100)
                      10             2
            b.  (11011)  to decimal
                     2
              Face Value       1     1    0     1    1
              Place value      2 4   2 3  2 2   2 1  2 0
                                          1
                                   2
                    4
              = 1 × 2  + 1 × 2  + 0 × 2  + 1 × 2  + 1 × 2 0
                            3

                                                                            Number Systems and Encoding Schemes  45
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