### Digital Electronics Objectives Part 05

81 . 1011101.1 – 101010.11 ≠ 110010.1.

82 . 22.7 × 147.5 = 1111010.110 is incorrect.

83 . Decimal number = 5436 to 9’s compliment= 4652 is incorrect.

84 . Binary = 45.15 to 10’s compliment = 54.58 is incorrect.

85 . Binary = 11011.01 to 1’s compliment 00111.10 is incorrect.

86 . For the binary number 101101110 the equivalent octal number is 556.

87 . For the octal number 66.3 the equivalent binary number is 110110.011.

88 . For the octal number 3.554 the equivalent binary number is 011.1011011.

89 . The decimal number 932 when converted to octal number will be equal to 1644.

90 . The decimal number 4429.625 when converted into octal number will be equal to 10515.5.

91 . The octal number 1170, 76051 when converted to decimal number will be 632.97.

92 . The binary number 0.10110111101 is equal to hexadecimal number B7A.

93 . The hexadecimal number 5F is equal to binary number 101111.

94 . The binary number 0.01111110 is equivalent to hexadecimal number 0.7F.

95 . The decimal number 3964 is equal to the octal number 7574.

96 . The hexadecimal number AB6 is equal to decimal number 2742.

97 . In digital computer the number 127 is stored as 1111111.

98. 10110 – 10011 = 00011

99 . 11111 + 00001 = 100000.

100 . The decimal equivalent of the hexadecimal number E5 is 229.