Compute 110101.01102-10110.10102 in octal number system
Conversion 110101.01102 in Octal:
Divide the source code of the integer part of the number into groups of 3 digits:
1101012 = 110 1012
Then replace each group with binary code:
110 1012 = 658
Conversion the fractional part of the number. To do this, divide the source code into groups of 3 digits:
01102 = 011 0002 = 508
Result: 110101.01102 = 65. 508
Conversion 10110.10102 in Octal:
Divide the source code of the integer part of the number into groups of 3 digits:
101102 = 010 1102
Then replace each group with binary code:
010 1102 = 268
Conversion the fractional part of the number. To do this, divide the source code into groups of 3 digits:
10102 = 101 0002 = 508
Result: 10110.10102 = 26. 508
65. 508
- 26. 508
36.608
Answer: 36.608
Compute the value of E7BAD16-E5ACF16 in 15’s complement and write your final answer in binary.
Compute in Hex:
E7BAD16
-E5ACF16
20DE16
Compute using 15's complement.
1) 15's complement of a number is obtained by subtracting all bits from FFFFF16:
15's complement of E5ACF16 is:
FFFFFF16 – E5ACF16 = 1A53016
FFFFFF16
– E5ACF16
1A53016
2) Now Add this 15's complement
E7BAD16
+ 1A53016
1020DD16
---------------------
1020DD16
+ 1 (cary)
= 20DE16
Answer: 20DE16
Compute the value of 5510-4510 using 7-bit 2’s complement sign magnitude number.
Compute in Dec:
5510 - 4510 = 1010
Compute in Bin:
1101112
- 1011012
= 10102
Compute using 7-bit 2’s complement sign magnitude number:
2's complement of -45 is: 1 0100112 (Invert 0 and 1 of a given binary number. Then add 1.)
Add this 2's complement of 4510 to 5510
0 1101112 (5510)
+
1 0100112 (2’s complement 4510)
---------------------------------------------
00010102
end-around-carry-bit addition does not occur in 2's complement arithmetic operations. It is ignored.
Answer: 00010102
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