### 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 a**s 1111111.**

**98.** 10110 – 10011 = 00011

**99 . ** 11111 + 00001 = 100000.

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