A student goes to school at a speed of $$5\frac{1}{2}$$ km/h and returns at a speed of 4 km/h. If he takes $$4\frac{3}{4}$$ hours for the entire journey, then total distance covered by the student (in km) is:

$$\eqalign{ & {\text{Total distance}} = \frac{{{S_1} \times {S_2}}}{{{S_1} + {S_2}}} \times {\text{Total time}} \cr & = \frac{{5\frac{1}{2} \times 4}}{{5\frac{1}{2} + 4}} \times 4\frac{3}{4} \cr & = \frac{{\frac{{11}}{2} \times 4}}{{\frac{{11}}{2} + 4}} \times \frac{{19}}{4} \cr & = \frac{{22}}{{\frac{{19}}{2}}} \times \frac{{19}}{4} \cr & = \frac{{44}}{{19}} \times \frac{{19}}{4} \cr & = 11{\text{ km}} \cr & {\text{One side distance}} = 11{\text{ km}} \cr & {\text{Both side distance}} = 2 \times 11 = 22{\text{ km}} \cr} $$