$C_4H_X + (4+\frac{X}{4})O_2 \rightarrow 4CO_2 + \frac{X}{2} H_2O$
combustion of 1 volume $C_4H_X$ with $(4 + \frac{X}{4})$ volumes $O_2$ give 4 volumes $CO_2 + \frac{X}{2}$ volumes $H_2O$
$\frac{10}{1} = \frac{55}{(4+\frac{X}{4})}$
$\therefore 10(4 +\frac{X}{4}) = 55$
$\therefore 40 + \frac{10X}{4} = 55$
$ \therefore \frac{10X}{4} = 55-40$
$ \therefore \frac{10X}{4} = 15$
$\therefore 10X = 60$
Hence X = 6