Suppose, volume of propane in mixture = xL
Volume of butane in mixture = (3 - x) L

1L C3Hg gives CO2 = 3 L.
x L C3H8 gives CO2 = 3x L

1L C4H10 give CO2 = 4L
(3 - x) LC4H10 gives CO2 = 4 × (3 - x) L
Volume of free CO2 = 10 L
3x + 4(3 - x) = 10
3x + 12 4x = 40
- x = 10 -12
x = 2
Volume of propane = 2 L
Volume of butane = (3 - 2) = 1L