Question on: JAMB Mathematics - 2015
100112 + *****2 + 111002 + 1012 = 10011112
A
11112
B
110112
C
101112
D
110012
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Correct Option: B
Convert the binary to base 10 and they convert back to base two
100112 + xxxxx2 + 111002 + 1012 = 10011112
(1 × 24 + 0 × 23 + 1 × 22 + 1 × 21 + 1 × 20) + xxxxx2 +(1 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 0 × 20) + (1 × 22 + 0 × 21 + 1 × 20)
= (16 + 0 + 0 + 2 + 1) + xxxxx2 + (16 + 8 + 4 + 0 + 0 ) + (4 + 0 + 1)
=(64 + 0 + 0 + 8 + 4 + 2 + 1)
19 + xxxxx2 + 33 = 79
xxxxx2 + 52 = 79
xxxxx2 = 79 − 52
xxxxx2 = 2710
\( \begin{array}{c|c}
2 & 27 \\ \hline
2 & 13 \text{ rem 1}\\ 2 & 6 \text{ rem 1}\\ 2 & 3 \text{ rem 0}\\ 2 & 1 \text{ rem 1}\\ & 0 \text{ rem 1}\\ \end{array}\uparrow \)
2710 = 110112
Therefore xxxxx2 = 2710 = 110112
100112 + xxxxx2 + 111002 + 1012 = 10011112
(1 × 24 + 0 × 23 + 1 × 22 + 1 × 21 + 1 × 20) + xxxxx2 +(1 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 0 × 20) + (1 × 22 + 0 × 21 + 1 × 20)
= (16 + 0 + 0 + 2 + 1) + xxxxx2 + (16 + 8 + 4 + 0 + 0 ) + (4 + 0 + 1)
=(64 + 0 + 0 + 8 + 4 + 2 + 1)
19 + xxxxx2 + 33 = 79
xxxxx2 + 52 = 79
xxxxx2 = 79 − 52
xxxxx2 = 2710
\( \begin{array}{c|c}
2 & 27 \\ \hline
2 & 13 \text{ rem 1}\\ 2 & 6 \text{ rem 1}\\ 2 & 3 \text{ rem 0}\\ 2 & 1 \text{ rem 1}\\ & 0 \text{ rem 1}\\ \end{array}\uparrow \)
2710 = 110112
Therefore xxxxx2 = 2710 = 110112
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