## Elementary Technical Mathematics

35$\frac{11}{20}$ gal
Let pilot used 13$\frac{3}{4}$ gal, 11$\frac{2}{5}$ gal, 10$\frac{2}{5}$ gal of aviation fuel to reach the first, second and third location respectively Therefore total aviation fuel used = 13$\frac{3}{4}$ gal+ 11$\frac{2}{5}$ gal + 10$\frac{2}{5}$ gal = $\frac{55}{4}$ + $\frac{57}{5}$ + $\frac{52}{5}$ = $\frac{55}{4} \times$ $\frac{5}{5}$+ $\frac{57}{5}\times$ $\frac{4}{4}$+ $\frac{52}{5} \times$ $\frac{4}{4}$ =$\frac{275}{20}$ + $\frac{228}{20}$ + $\frac{208}{20}$ = $\frac{711}{20}$ = 35$\frac{11}{20}$ gal