SOLUTION : Suppose the volume of $CO =x mL , CH _{4}=y mL . He$ gas is inert in reaction.
Reactions :
Given, for contraction, $\frac{x}{2}+2 y=13 \ldots$ (I) for $CO _{2}, x+y=14$
From (I) and (II) $, x=10, y=4$; hence vol. of $He =6 mL$
$\%$ of $CO =\frac{10}{20} \times 100=50 \%, \%$ of $CH _{4}=20 \%, \%$ of $He =30 \%$.