JAMB - Mathematics (1998 - No. 1)
If \(1011_2\) + \(X_7\) = \(25_{10}\), solve for X.
207
\(14_7\)
\(20_7\)
\(24_7\)
Explanation
\(1011_2 + X_7 = 25_{10} = 1011_2 = 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0\)
= 8 + 0 + 2 + 1
= \(11_{10}\)
\(X_7 = 25_{10} - 11_{10}\)
= \(14_{10}\)
\(\begin{array}{c|c}
7 & 14 \\ 7 & 2 R 0 \\ & 0 R 2
\end{array}\)
X = \(20_7\)
= 8 + 0 + 2 + 1
= \(11_{10}\)
\(X_7 = 25_{10} - 11_{10}\)
= \(14_{10}\)
\(\begin{array}{c|c}
7 & 14 \\ 7 & 2 R 0 \\ & 0 R 2
\end{array}\)
X = \(20_7\)
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