Page 77 - Computer Science V2.0 Class 11
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iii.

                       Digits          4                 5                  5
                       Position        2                 1                  0
                       Weight          8 2               8 1                8 0
                      Therefore,
                                       2
                      Decimal number = 4×8  +5×8 + 5×8 0
                                            1
                                                            = 4×64 + 5×8 + 5× 1
                                                            = 256 + 40 + 5
                                                            = (301) 10
                      iv.

                       Digits       1               0              7              5
                       Position     1               0              -1             -2

                       Weight       8 1             8 0            8 -1           8 -2
                      Therefore,
                                           0
                                                -1
                                       1
                      Decimal number = 1×8 +0×8 +7×8 +5×8 -2
                                                     = 1×8+0×1+7×0.125+5×0.015625
                                                     = 8+0+0.875+0.078125
                                                     = (8.953125) 10
                    6.  Express the following decimal numbers into hexadecimal numbers.

                      i.  548
                      ii.  4052
                      iii.  58
                      iv.  100.25
                  Ans.  i.

                        16     548
                        16       34         4
                        16         2        2

                                     0      2
                      = (224) 16
                      ii.
                        16     4052
                        16     243          4

                        16       15        D
                                     0      F

                      = (FD4) 16
                      iii.

                        16      58
                        16        3         A
                                 0          3

                      = (3A) 16






                                                                                 Number Systems and Encoding Schemes  63
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