By substituting the value of x in the following, compute y and z
i (x)10 + (35) 8 = ( y ) 2 = ( z ) 16
ii (x)10 × (1010110) 2 = ( y ) 4 = ( z ) 10
value of x is 49
Part i
49010+28010‾\underline{\begin{array}{cc} \space \space \space \space 490_{10} \\ + 280_{10} \end{array}} 49010+28010 X+Y‾\underline{\begin{array}{cc} X \\ + Y \end{array}}X+Y
Z
49010+72010‾\underline{\begin{array}{cc} \space \space \space \space 490_{10} \\ + 720_{10} \end{array}} 49010+72010
121010
Hence discarding the leading 1 the result is 210.
Part ii
(x)10 × (1010110) 2 = ( y ) 4 = ( z ) 10
(1010110)2=8610
49010+08610‾\underline{\begin{array}{cc} \space \space \space \space 490_{10} \\ + 086_{10} \end{array}} 49010+08610 X+Y‾\underline{\begin{array}{cc} X \\ + Y \end{array}}X+Y
49010+08610‾\underline{\begin{array}{cc} \space \space \space \space 490_{10} \\ + 086_{10} \end{array}} 49010+08610
57610
Hence the result is 576
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