JEE Main 2016 · 3 Apr · Shift 1, Chemistry question 31

375 mL of air holds 75 mL of $\ce{O2}$, so for $\ce{C_xH_y}$: $15\left(x+\frac y4\right)=75$, i.e. $x+\frac y4=5$. Of the options only $\ce{C3H8}$ fits ($3+2=5$).

Note: with that formula the gases after combustion would be 300 mL $\ce{N2}$ + 45 mL $\ce{CO2}$ = 345 mL, not the 330 mL printed. The 330 mL would need $x=2$, which no option allows, so the oxygen figure decides.