JAMB 1985 · UME · Q3

In the equation $\dfrac{11}{x} = \dfrac{1000}{x + 101}$ , all the numbers are in base 2. Solve for $x$ .

Worked solution (try it first)

Change to base ten: $11_2 = 3$ , $1000_2 = 8$ and $101_2 = 5$ , so $\frac3x = \frac{8}{x + 5}$ .
Cross-multiply: $3x + 15 = 8x$ , so $5x = 15$ and $x = 3$ .
Change back to base two: $3 = 2 + 1 = 11_2$ , option B.