Question

An oxygen tank contained 212 12/3 liters of oxygen before 27 1/3 liters were used. If the tank can hold 240 3/8 liters, how much space in the tank is unused

Answer 1

So, initially   240$\frac{3}{8}$ -212 $\frac{12}{3}$ were empty:

this is :

$240 \frac{3}{8} - 212 \frac{12}{3} = 240 \frac{3}{8} - 212 +4 = 240 \frac{3}{8} - 216= 24 \frac{3}{8}$

12/4 is 4 so that's why i substituted it with 4.

now, later 27 1/3 were used so we add this tho the original empty space

 $24 \frac{3}{8}+ 27 \frac{1}{3}$ =  $24 \frac{9}{24}+ 27 \frac{8}{24}$ = $51 \frac{17}{24}$


which is the result!