By substituting the value of x in the following, first compute y and then represent y in Radix and Diminished-Radix complement formats (for respective Radix),
iii (x)10 + (1011)2 = ( y )8
iv (x)10 × (15)8 = ( y )2Â
value of x is 49
Part iii
(1011)2 = 1110
"\\underline{\\begin{array}{cc}\n \\space \\space \\space \\space 490_{10} \\\\\n + 011_{10}\n\\end{array}}"Â "\\underline{\\begin{array}{cc}\n X \\\\\n + Y\n\\end{array}}"
Z
"\\underline{\\begin{array}{cc}\n \\space \\space \\space \\space 490_{10} \\\\\n + 011_{10}\n\\end{array}}"
50110
Hence discarding the leading 1 the result is 501.
Part iv
"\\underline{\\begin{array}{cc}\n \\space \\space \\space \\space 490_{10} \\\\\n + 120_{10}\n\\end{array}}"Â "\\underline{\\begin{array}{cc}\n X \\\\\n + Y\n\\end{array}}"
Z
"\\underline{\\begin{array}{cc}\n \\space \\space \\space \\space 490_{10} \\\\\n + 120_{10}\n\\end{array}}"
61010
Hence the result is 610
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