A gas mixture contains acetylene and carbon-dioxide. $20$ lit of this mixture  requires $20$ Lit of oxygen under the same conditions for complete combustion. The percentage by volume of acetylene in the mixture is 

Between mixture of acetylene and carbon-dioxide, acetylene undergo combustion with   Oxygen to produce water. Hence balanced equation is

$C_2{H}_2+5/2O_2\to 2C{O}_2+H_{2}O$         

There are five oxygens in the products side (right hand side), and hence to balance the 

equation oxygen on the left hand side is multiplied with $\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

One mole of acetylene is equal to $22.4$ liters of volume.

$\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ moles of oxygen is equals to $5/2\times 22.4=56$ liters of oxygen.  

It requires only $20$ liters of oxygen for complete combustion.

Therefore $20$ liters of oxygen react with $22.4/56\times 20=8$ liters of acetylene.

Upon converting $8$ into percentage we get $8/20\times 100=40\%$ of acetylene.

Hence the correct option is (B).